That the constant force act on the object is conservative.

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Answer 1

Question

Chapter 6, Problem 44P

(a)

To determine

To Show: That the constant force act on the object is conservative.

(a)

Expert Solution

Answer to Problem 44P

the constant force applied on the object is conservative in nature.

Explanation of Solution

The work done by a conservative force on a particle moving between any two points is independent of the path taken by the particle. Its only depends upon the end points of the path taken by the particle to move.

Given Information:

The general definition for work done by a force is,

$W=∫ifF→⋅dr→$ .

Formula to calculate the work done by the force on the object is,

$W=∫ifF→⋅dr→$

$W$ is the work done by the force on the particle.
$F→$ is the applied force vector on the object.
$dr→$ is the displacement vector of the object.

Since the force is constant that does not vary with respect to time or the position or the velocity of the object. So, the value of force can be taken out from the integration since it is constant quantity.

$W=F→∫ifdr→=F→[r→]if=F→[r→f−r→i]$

Now, here the force is constant so, the work done by this force on the object in only depends upon the end points of the displace object that shows the work done is independent of the path taken by the object to displace between the end points. But the work done is independent of the path only when the force is conservative.

Conclusion:

Therefore, the constant force applied on the object is conservative in nature.

(b)

To determine

The work done by the force $F→=(3i^+4j^) N$ on the particle along each one of the three paths shown in the figure and show that the work done along the three path is identical.

(b)

Expert Solution

Answer to Problem 44P

The work done by the force $F→=(3i^+4j^) N$ on the particle along all the three path are $35 J$ and the work done by the force is identical in all the three paths.

Explanation of Solution

Section-1:

To determine: The work done by the force $F→=(3i^+4j^) N$ on the particle along the path $OA$ .

Answer: The work done by the force on the particle as it goes from O to C along the path $OA$ is $35 J$ .

Given Information:

The force acting on the particle is $F→=(3i^+4j^) N$ and the three different path of the particle moves is shown in figure (I).

Principles of Physics: A Calculus-Based Text, Hybrid (with Enhanced WebAssign Printed Access Card), Chapter 6, Problem 44P

Figure (I)

Formula to calculate the work done by the force on the particle is,

$W=∫F→⋅dr→$

Substitute $(3i^+4j^) N$ for $F→$ and $dxi^$ for $dr^$ .

$W=∫((3i^+4j^) N)⋅(dxi^) m=∫(3dx) N⋅m×1 J1 N⋅m=[∫3dx] J$

Since along the path $OA$ the particle moves from $x=0$ to $x=5.00 m$ and there is no displacement in y-direction that is $y=0$ .

Taking the limits of the integration,

$WOA=[∫05.003dx] J=(3[x]05.00) J=(3[5.00−0]) J=15 J$

Section-2:

To determine: The work done by the force $F→=(3i^+4j^) N$ on the particle along the path $AC$ .

Answer: The work done by the force on the particle as it goes from O to C along the path $AC$ is $35 J$ .

Given Information:

The force acting on the particle is $F→=(3i^+4j^) N$ and the three different path of the particle moves is shown in figure (I).

In the path $AC$ , the particle moves from $y=0$ to $y=5.00 m$ and there is a displacement of particle in x-direction that is $x=5.00 m$ .

Formula to calculate the work done by the force on the particle is,

$W=∫F→⋅dr→$

Substitute $(3i^+4j^) N$ for $F→$ and $dyj^$ for $dr^$ .

$W=∫((3i^+4j^) N)⋅(dyj^) m=∫(4dy) N⋅m×1 J1 N⋅m=[∫4dy] J$

Taking the limits of the integration,

$WAC=[∫05.004dy] J=(4[y]05.00) J=4[5.00−0]=20 J$

Section-3:

To determine: The work done by the force $F→=(3i^+4j^) N$ on the particle along the purple path.

Answer: The work done by the force on the particle as it goes from O to C along the purple path is $35 J$ .

Given Information:

The force acting on the particle is $F→=(3i^+4j^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle along the purple path $OAC$ is,

$Wpurple=WOA+WAC$

$Wpurple$ is the work done along the purple path.
$WOA$ is the work done along the path $OA$ .
$WAC$ is the work done along the path $AC$ .

Substitute $15 J$ for $WOA$ and $20 J$ for $WAC$ .

$Wpurple=15 J+20 J=35 J$

Conclusion:

Therefore, the work done by the force on the particle as it goes from O to C along the purple path is $35 J$ .

Section-4:

To determine: The work done by the force $F→=(3i^+4j^) N$ on the particle along the path $OB$ .

Answer: The work done by the force on the particle as it goes from O to C along the path $OB$ is $35 J$ .

Given Information:

The force acting on the particle is $F→=(3i^+4j^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle is,

$W=∫F→⋅dr→$

Substitute $(3i^+4j^) N$ for $F→$ and $dyj^$ for $dr^$ .

$W=∫((3i^+4j^) N)⋅(dyj^) m=∫(4dy) N⋅m×1 J1 N⋅m=[∫4dy] J$

Since along the path $OB$ the particle moves from $y=0$ to $y=5.00 m$ and there is no displacement in x-direction that is $x=0$ .

Taking the limits of the integration,

$WOB=[∫05.004dy] J=4[y]05.00=(4[5.00−0]) J=20 J$

Section-5:

To determine: The work done by the force $F→=(3i^+4j^) N$ on the particle along the path $BC$ .

Answer: The work done by the force on the particle as it goes from O to C along the path $BC$ is $35 J$ .

Given Information:

The force acting on the particle is $F→=(3i^+4j^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle is,

$W=∫F→⋅dr→$

Substitute $(3i^+4j^) N$ for $F→$ and $dxi^$ for $dr^$ .

$W=∫((3i^+4j^) N)⋅(dxi^) m=∫(3dx) N⋅m×1 J1 N⋅m=[∫3dx] J$

In the path $BC$ , the particle moves from $x=0$ to $x=5.00 m$ and there is a displacement of particle in y-direction that is $y=5.00 m$ .

Taking the limits of the integration,

$WBC=[∫05.003dx+∫004dy] J=(3[x]05.00+0) J=(3[5.00−0]) J=15 J$

Section-6:

To determine: The work done by the force $F→=(3i^+4j^) N$ on the particle along the red path.

Answer: The work done by the force on the particle as it goes from O to C along the red path is $35 J$ .

Given Information:

The force acting on the particle is $F→=(3i^+4j^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle along the red path $OBC$ is,

$Wred=WOB+WBC$

$Wred$ is the work done along the red path.
$WOB$ is the work done along the path $OB$ .
$WBC$ is the work done along the path $BC$ .

Substitute $20 J$ for $WOB$ and $15 J$ for $WBC$ .

$Wred=20+15 J=35 J$

Conclusion:

Therefore, the work done by the force on the particle as it goes from O to C along the red path is $35 J$ .

Section-7:

To determine: The work done by the force $F→=(3i^+4j^) N$ on the particle along the blue path.

Answer: The work done by the force on the particle as it goes from O to C along the blue path is $35 J$ .

Given Information:

The force acting on the particle is $F→=(3i^+4j^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle is,

$W=[∫3dx+∫4dy] J$

The path $OC$ is a straight line passes through a origin. In this blue path the particle moves in both the direction by $5.00 m$ .

Taking the limits on integration,

$Wblue=[∫05.003dx+∫05.004dx] J=(3[x]05.00+4[y]05.00) J=(3[5.00−0]+4[5.00−0]) J=35 J$

Since the work done by the force $F→=(3i^+4j^) N$ in all three purple, red and blue path is $35 J$ hence the force is conservative and the work done is identical in all the three paths.

Conclusion:

Therefore, the work done by the force on the particle as it goes from O to C along the blue path is $35 J$ and the work done by the force is same in all the three paths.

(c)

To determine

Whether the work done by the force $F→=(4xi^+3yj^) N$ on the particle along each one of the three paths shown in the figure is identical.

(c)

Expert Solution

Answer to Problem 44P

The work done by the force $F→=(4xi^+3yj^) N$ on the particle along all the three path are $35 J$ and the work done by the force is identical in all the three paths.

Explanation of Solution

Section-1:

To determine: The work done by the force $F→=(4xi^+3yj^) N$ on the particle along the path $OA$ .

Answer: The work done by the force on the particle as it goes from O to C along the path $OA$ is $35 J$ .

Given Information:

The force acting on the particle is $F→=(4xi^+3yj^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle is,

$W=∫F→⋅dr→$

Substitute $(4xi^+3yj^) N$ for $F→$ and $dxi^$ for $dr^$ .

$W=∫((4xi^+3yj^) N)⋅(dxi^) m=∫(4xdx) N⋅m×1 J1 N⋅m=[∫4xdx] J$

Since along the path $OA$ the particle moves from $x=0$ to $x=5.00 m$ and there is no displacement in y-direction that is $y=0$ .

Taking the limits of the integration,

$WOA=[∫05.004xdx] J=(4[x]05.00) J=(4[5.00−0]) J=20 J$

Section-2:

To determine: The work done by the force $F→=(4xi^+3yj^) N$ on the particle along the path $AC$ .

Answer: The work done by the force on the particle as it goes from O to C along the path $AC$ is $35 J$ .

Given Information:

The force acting on the particle is $F→=(4xi^+3yj^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle is,

$W=∫F→⋅dr→$

Substitute $(4xi^+3yj^) N$ for $F→$ and $dyj^$ for $dr^$ .

$W=∫((4xi^+3yj^) N)⋅(dyj^) m=∫(3ydy) N⋅m×1 J1 N⋅m=[∫3ydy] J$

In the path $AC$ , the particle moves from $y=0$ to $y=5.00 m$ and there is a displacement of particle in x-direction that is $x=5.00 m$ .

Taking the limits of the integration,

$WAC=[∫05.003ydy] J=(3[y]05.00) J=3[5.00−0]=15 J$

Section-3:

To determine: The work done by the force $F→=(4xi^+3yj^) N$ on the particle along the purple path.

Answer: The work done by the force on the particle as it goes from O to C along the purple path is $35 J$ .

Given Information:

The force acting on the particle is $F→=(4xi^+3yj^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle along the purple path $OAC$ is,

$Wpurple=WOA+WAC$ (III)

$Wpurple$ is the work done along the purple path.
$WOA$ is the work done along the path $OA$ .
$WAC$ is the work done along the path $AC$ .

Substitute $20 J$ for $WOA$ and $15 J$ for $WAC$ in equation (II).

$Wpurple=20 J+15 J=35 J$

Conclusion:

Therefore, the work done by the force on the particle as it goes from O to C along the purple path is $35 J$ .

Section-4:

To determine: The work done by the force $F→=(4xi^+3yj^) N$ on the particle along the path $OB$ .

Answer: The work done by the force on the particle as it goes from O to C along the path $OB$ is $35 J$ .

Given Information:

The force acting on the particle is $F→=(4xi^+3yj^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle is,

$W=∫F→⋅dr→$

Substitute $(4xi^+3yj^) N$ for $F→$ and $dyj^$ for $dr^$ .

$W=∫((4xi^+3yj^) N)⋅(dyj^) m=∫(3ydy) N⋅m×1 J1 N⋅m=[∫3ydy] J$

Since along the path $OB$ the particle moves from $y=0$ to $y=5.00 m$ and there is no displacement in x-direction that is $x=0$ .

Taking the limits of the integration,

$WOB=[∫05.003ydy] J=3[y]05.00=(3[5.00−0]) J=15 J$

Section-5:

To determine: The work done by the force $F→=(4xi^+3yj^) N$ on the particle along the path $BC$ .

Answer: The work done by the force on the particle as it goes from O to C along the path $BC$ is $20 J$ .

Given Information:

The force acting on the particle is $F→=(4xi^+3yj^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle is,

$W=∫F→⋅dr→$

Substitute $(4xi^+3yj^) N$ for $F→$ and $dxi^$ for $dr^$ .

$W=∫((4xi^+3yj^) N)⋅(dxi^) m=∫(4xdx) N⋅m×1 J1 N⋅m=[∫4xdx] J$

In the path $BC$ , the particle moves from $x=0$ to $x=5.00 m$ and there is a displacement of particle in y-direction that is $y=5.00 m$ .

Taking the limits of the integration,

$WBC=[∫05.004xdx] J=(4[x]05.00) J=(4[5.00−0]) J=20 J$

Section-6:

To determine: The work done by the force $F→=(4xi^+3yj^) N$ on the particle along the red path.

Answer: The work done by the force on the particle as it goes from O to C along the red path is $35 J$ .

Given Information:

The force acting on the particle is $F→=(4xi^+3yj^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle along the red path $OBC$ is,

$Wred=WOB+WBC$

$Wred$ is the work done along the red path.
$WOB$ is the work done along the path $OB$ .
$WBC$ is the work done along the path $BC$ .

Substitute $15 J$ for $WOB$ and $20 J$ for $WBC$ .

$Wred=15 J+20 J=35 J$

Conclusion:

Therefore, the work done by the force on the particle as it goes from O to C along the red path is $35 J$ .

Section-7:

To determine: The work done by the force $F→=(4xi^+3yj^) N$ on the particle along the blue path.

Answer: The work done by the force on the particle as it goes from O to C along the blue path is $35 J$ .

Given Information:

The force acting on the particle is $F→=(4xi^+3yj^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle is,

$W=[∫4xdx+∫3ydy] J$

The path $OC$ is a straight line passes through a origin. In this blue path the particle moves in both the direction by $5.00 m$ .

Taking the limits on integration,

$Wred=[∫05.004xdx+∫05.003ydx] J=(4[x]05.00+3[y]05.00) J=(4[5.00−0]+3[5.00−0]) J=35 J$

Since the work done by the force $F→=(3i^+4j^) N$ in all three purple, red and blue path is $35 J$ hence the force is conservative and the work done is identical in all the three paths.

Conclusion:

Therefore, the work done by the force on the particle as it goes from O to C along the blue path is $35 J$ and the work done by the force is same in all the three paths.

(d)

To determine

Whether the work done by the force $F→=(yi^−xj^) N$ on the particle along each one of the three paths shown in the figure is identical.

(d)

Expert Solution

Answer to Problem 44P

The work done by the force $F→=(yi^−xj^) N$ on the particle along all the three path are not identical.

Explanation of Solution

Section-1:

To determine: The work done by the force $F→=(yi^−xj^) N$ on the particle along the path $OA$ .

Answer: The work done by the force on the particle as it goes from O to C along the path $OA$ is $−25 J$ .

Given Information:

The force acting on the particle is $F→=(yi^−xj^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle is,

$W=∫F→⋅dr→$

Substitute $(yi^−xj^) N$ for $F→$ and $dxi^$ for $dr^$ .

$W=∫((yi^−xj^) N)⋅(dxi^) m=∫(ydx) N⋅m×1 J1 N⋅m=[∫ydx] J$

Since along the path $OA$ the particle moves from $x=0$ to $x=5.00 m$ and there is no displacement in y-direction that is $y=0$ .

Substitute $0$ for $y$ and $0$ for $dy$ .

$WOA=[∫0dx] J=0 J$

Section-2:

To determine: The work done by the force $F→=(yi^−xj^) N$ on the particle along the path $AC$ .

Answer: The work done by the force on the particle as it goes from O to C along the path $AC$ is $−25 J$ .

Given Information:

The force acting on the particle is $F→=(yi^−xj^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle is,

$W=∫F→⋅dr→$

Substitute $(yi^−xj^) N$ for $F→$ and $dyj^$ for $dr^$ .

$W=∫((yi^−xj^) N)⋅(dyj) m=−∫(xdy) N⋅m×1 J1 N⋅m=−[∫xdy] J$

In the path $AC$ , the particle moves from $y=0$ to $y=5.00 m$ and there is a displacement of particle in x-direction that is $x=5.00 m$ .

Substitute $5.00$ for $x$ and $0$ for $dx$ .

$WAC=[−∫(5.00)dy] J$

Taking the limits of the integration,

$WAC=[∫05.00(5.00)dy] J=(−5[y]05.00) J=−5[5.00−0]=−25 J$

Section-3:

To determine: The work done by the force $F→=(yi^−xj^) N$ on the particle along the path $AC$ .

Answer: The work done by the force on the particle as it goes from O to C along the path $AC$ is $−25 J$ .

Given Information:

The force acting on the particle is $F→=(yi^−xj^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle along the purple path $OAC$ is,

$Wpurple=WOA+WAC$

$Wpurple$ is the work done along the purple path.
$WOA$ is the work done along the path $OA$ .
$WAC$ is the work done along the path $AC$ .

Substitute $0$ for $WOA$ and $−25 J$ for $WAC$ .

$Wpurple=0 J−25J=−25 J$

Conclusion:

Therefore, the work done by the force on the particle as it goes from O to C along the purple path is $−25 J$ .

Section-4:

To determine: The work done by the force $F→=(yi^−xj^) N$ on the particle along the path $OB$ .

Answer: The work done by the force on the particle as it goes from O to C along the path $OB$ is $25 J$ .

Given Information:

The force acting on the particle is $F→=(yi^−xj^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle is,

$W=∫F→⋅dr→$

Substitute $(yi^−xj^) N$ for $F→$ and $dyj^$ for $dr^$ .

$W=∫((yi^−xj^) N)⋅(dyj) m=−∫(xdy) N⋅m×1 J1 N⋅m=−[∫xdy] J$

Since along the path $OB$ the particle moves from $y=0$ to $y=5.00 m$ and there is no displacement in x-direction that is $x=0$ .

Substitute $0$ for $x$ in above integration.

$WOB=[∫05.000dy] J=0$

Section-5:

To determine: The work done by the force $F→=(yi^−xj^) N$ on the particle along the path $BC$ .

Answer: The work done by the force on the particle as it goes from O to C along the path $BC$ is $25 J$ .

Given Information:

The force acting on the particle is $F→=(yi^−xj^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle is,

$W=∫F→⋅dr→$

Substitute $(yi^−xj^) N$ for $F→$ and $dxi^$ for $dr^$ .

$W=∫((yi^−xj^) N)⋅(dxi^) m=∫(ydx) N⋅m×1 J1 N⋅m=[∫ydx] J$

In the path $BC$ , the particle moves from $x=0$ to $x=5.00 m$ and there is a displacement of particle in y-direction that is $y=5.00 m$ .

Substitute $5.00$ for $y$ and $0$ for $dy$ .

Taking the limits of the integration,

$WBC=[∫05.005dx] J=(5[x]05.00) J=(5[5.00−0]) J=25 J$

Section-6:

To determine: The work done by the force $F→=(yi^−xj^) N$ on the particle along the red path.

Answer: The work done by the force on the particle as it goes from O to C along the red path is $25 J$ .

Given Information:

The force acting on the particle is $F→=(yi^−xj^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle along the red path $OBC$ is,

$Wred=WOB+WBC$ (VI)

$Wred$ is the work done along the red path.
$WOB$ is the work done along the path $OB$ .
$WBC$ is the work done along the path $BC$ .

Substitute $0$ for $WOB$ and $25 J$ for $WBC$ in equation (VI).

$Wred=0 J+25 J=25 J$

Conclusion:

Therefore, the work done by the force on the particle as it goes from O to C along the red path is $25 J$ .

Section-7:

To determine: The work done by the force $F→=(yi^−xj^) N$ on the particle along the blue path.

Answer: The work done by the force on the particle as it goes from O to C along the blue path is $0$ .

Given Information:

The force acting on the particle is $F→=(yi^−xj^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle is,

$W=∫F→⋅dr→$

Substitute $(yi^−xj^) N$ for $F→$ and $dxi^+dyj^$ for $dr^$ .

$W=∫((yi^−xj^) N)⋅(dxi^+dyj^) m=∫(ydx−xdy) N⋅m×1 J1 N⋅m=[∫ydx−xdy] J$

The path $OC$ is a straight line passes through a origin. In this blue path the particle moves in both the direction by $5.00 m$ . Hence equation of line become $y=x$ .

Substitute $x$ for $y$ and $dx$ for $dy$ .

Taking the limits on integration,

$Wblue=[∫xdx−∫xdx] J=0$

Conclusion:

Therefore, the work done by the force on the particle as it goes along the three paths is not same.

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Answer 2

Question

Chapter 6, Problem 44P

(a)

To determine

To Show: That the constant force act on the object is conservative.

(a)

Expert Solution

Answer to Problem 44P

the constant force applied on the object is conservative in nature.

Explanation of Solution

The work done by a conservative force on a particle moving between any two points is independent of the path taken by the particle. Its only depends upon the end points of the path taken by the particle to move.

Given Information:

The general definition for work done by a force is,

$W=∫ifF→⋅dr→$ .

Formula to calculate the work done by the force on the object is,

$W=∫ifF→⋅dr→$

$W$ is the work done by the force on the particle.
$F→$ is the applied force vector on the object.
$dr→$ is the displacement vector of the object.

Since the force is constant that does not vary with respect to time or the position or the velocity of the object. So, the value of force can be taken out from the integration since it is constant quantity.

$W=F→∫ifdr→=F→[r→]if=F→[r→f−r→i]$

Now, here the force is constant so, the work done by this force on the object in only depends upon the end points of the displace object that shows the work done is independent of the path taken by the object to displace between the end points. But the work done is independent of the path only when the force is conservative.

Conclusion:

Therefore, the constant force applied on the object is conservative in nature.

(b)

To determine

The work done by the force $F→=(3i^+4j^) N$ on the particle along each one of the three paths shown in the figure and show that the work done along the three path is identical.

(b)

Expert Solution

Answer to Problem 44P

The work done by the force $F→=(3i^+4j^) N$ on the particle along all the three path are $35 J$ and the work done by the force is identical in all the three paths.

Explanation of Solution

Section-1:

To determine: The work done by the force $F→=(3i^+4j^) N$ on the particle along the path $OA$ .

Answer: The work done by the force on the particle as it goes from O to C along the path $OA$ is $35 J$ .

Given Information:

The force acting on the particle is $F→=(3i^+4j^) N$ and the three different path of the particle moves is shown in figure (I).

Principles of Physics: A Calculus-Based Text, Hybrid (with Enhanced WebAssign Printed Access Card), Chapter 6, Problem 44P

Figure (I)

Formula to calculate the work done by the force on the particle is,

$W=∫F→⋅dr→$

Substitute $(3i^+4j^) N$ for $F→$ and $dxi^$ for $dr^$ .

$W=∫((3i^+4j^) N)⋅(dxi^) m=∫(3dx) N⋅m×1 J1 N⋅m=[∫3dx] J$

Since along the path $OA$ the particle moves from $x=0$ to $x=5.00 m$ and there is no displacement in y-direction that is $y=0$ .

Taking the limits of the integration,

$WOA=[∫05.003dx] J=(3[x]05.00) J=(3[5.00−0]) J=15 J$

Section-2:

To determine: The work done by the force $F→=(3i^+4j^) N$ on the particle along the path $AC$ .

Answer: The work done by the force on the particle as it goes from O to C along the path $AC$ is $35 J$ .

Given Information:

The force acting on the particle is $F→=(3i^+4j^) N$ and the three different path of the particle moves is shown in figure (I).

In the path $AC$ , the particle moves from $y=0$ to $y=5.00 m$ and there is a displacement of particle in x-direction that is $x=5.00 m$ .

Formula to calculate the work done by the force on the particle is,

$W=∫F→⋅dr→$

Substitute $(3i^+4j^) N$ for $F→$ and $dyj^$ for $dr^$ .

$W=∫((3i^+4j^) N)⋅(dyj^) m=∫(4dy) N⋅m×1 J1 N⋅m=[∫4dy] J$

Taking the limits of the integration,

$WAC=[∫05.004dy] J=(4[y]05.00) J=4[5.00−0]=20 J$

Section-3:

To determine: The work done by the force $F→=(3i^+4j^) N$ on the particle along the purple path.

Answer: The work done by the force on the particle as it goes from O to C along the purple path is $35 J$ .

Given Information:

The force acting on the particle is $F→=(3i^+4j^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle along the purple path $OAC$ is,

$Wpurple=WOA+WAC$

$Wpurple$ is the work done along the purple path.
$WOA$ is the work done along the path $OA$ .
$WAC$ is the work done along the path $AC$ .

Substitute $15 J$ for $WOA$ and $20 J$ for $WAC$ .

$Wpurple=15 J+20 J=35 J$

Conclusion:

Therefore, the work done by the force on the particle as it goes from O to C along the purple path is $35 J$ .

Section-4:

To determine: The work done by the force $F→=(3i^+4j^) N$ on the particle along the path $OB$ .

Answer: The work done by the force on the particle as it goes from O to C along the path $OB$ is $35 J$ .

Given Information:

The force acting on the particle is $F→=(3i^+4j^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle is,

$W=∫F→⋅dr→$

Substitute $(3i^+4j^) N$ for $F→$ and $dyj^$ for $dr^$ .

$W=∫((3i^+4j^) N)⋅(dyj^) m=∫(4dy) N⋅m×1 J1 N⋅m=[∫4dy] J$

Since along the path $OB$ the particle moves from $y=0$ to $y=5.00 m$ and there is no displacement in x-direction that is $x=0$ .

Taking the limits of the integration,

$WOB=[∫05.004dy] J=4[y]05.00=(4[5.00−0]) J=20 J$

Section-5:

To determine: The work done by the force $F→=(3i^+4j^) N$ on the particle along the path $BC$ .

Answer: The work done by the force on the particle as it goes from O to C along the path $BC$ is $35 J$ .

Given Information:

The force acting on the particle is $F→=(3i^+4j^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle is,

$W=∫F→⋅dr→$

Substitute $(3i^+4j^) N$ for $F→$ and $dxi^$ for $dr^$ .

$W=∫((3i^+4j^) N)⋅(dxi^) m=∫(3dx) N⋅m×1 J1 N⋅m=[∫3dx] J$

In the path $BC$ , the particle moves from $x=0$ to $x=5.00 m$ and there is a displacement of particle in y-direction that is $y=5.00 m$ .

Taking the limits of the integration,

$WBC=[∫05.003dx+∫004dy] J=(3[x]05.00+0) J=(3[5.00−0]) J=15 J$

Section-6:

To determine: The work done by the force $F→=(3i^+4j^) N$ on the particle along the red path.

Answer: The work done by the force on the particle as it goes from O to C along the red path is $35 J$ .

Given Information:

The force acting on the particle is $F→=(3i^+4j^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle along the red path $OBC$ is,

$Wred=WOB+WBC$

$Wred$ is the work done along the red path.
$WOB$ is the work done along the path $OB$ .
$WBC$ is the work done along the path $BC$ .

Substitute $20 J$ for $WOB$ and $15 J$ for $WBC$ .

$Wred=20+15 J=35 J$

Conclusion:

Therefore, the work done by the force on the particle as it goes from O to C along the red path is $35 J$ .

Section-7:

To determine: The work done by the force $F→=(3i^+4j^) N$ on the particle along the blue path.

Answer: The work done by the force on the particle as it goes from O to C along the blue path is $35 J$ .

Given Information:

The force acting on the particle is $F→=(3i^+4j^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle is,

$W=[∫3dx+∫4dy] J$

The path $OC$ is a straight line passes through a origin. In this blue path the particle moves in both the direction by $5.00 m$ .

Taking the limits on integration,

$Wblue=[∫05.003dx+∫05.004dx] J=(3[x]05.00+4[y]05.00) J=(3[5.00−0]+4[5.00−0]) J=35 J$

Since the work done by the force $F→=(3i^+4j^) N$ in all three purple, red and blue path is $35 J$ hence the force is conservative and the work done is identical in all the three paths.

Conclusion:

Therefore, the work done by the force on the particle as it goes from O to C along the blue path is $35 J$ and the work done by the force is same in all the three paths.

(c)

To determine

Whether the work done by the force $F→=(4xi^+3yj^) N$ on the particle along each one of the three paths shown in the figure is identical.

(c)

Expert Solution

Answer to Problem 44P

The work done by the force $F→=(4xi^+3yj^) N$ on the particle along all the three path are $35 J$ and the work done by the force is identical in all the three paths.

Explanation of Solution

Section-1:

To determine: The work done by the force $F→=(4xi^+3yj^) N$ on the particle along the path $OA$ .

Answer: The work done by the force on the particle as it goes from O to C along the path $OA$ is $35 J$ .

Given Information:

The force acting on the particle is $F→=(4xi^+3yj^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle is,

$W=∫F→⋅dr→$

Substitute $(4xi^+3yj^) N$ for $F→$ and $dxi^$ for $dr^$ .

$W=∫((4xi^+3yj^) N)⋅(dxi^) m=∫(4xdx) N⋅m×1 J1 N⋅m=[∫4xdx] J$

Since along the path $OA$ the particle moves from $x=0$ to $x=5.00 m$ and there is no displacement in y-direction that is $y=0$ .

Taking the limits of the integration,

$WOA=[∫05.004xdx] J=(4[x]05.00) J=(4[5.00−0]) J=20 J$

Section-2:

To determine: The work done by the force $F→=(4xi^+3yj^) N$ on the particle along the path $AC$ .

Answer: The work done by the force on the particle as it goes from O to C along the path $AC$ is $35 J$ .

Given Information:

The force acting on the particle is $F→=(4xi^+3yj^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle is,

$W=∫F→⋅dr→$

Substitute $(4xi^+3yj^) N$ for $F→$ and $dyj^$ for $dr^$ .

$W=∫((4xi^+3yj^) N)⋅(dyj^) m=∫(3ydy) N⋅m×1 J1 N⋅m=[∫3ydy] J$

In the path $AC$ , the particle moves from $y=0$ to $y=5.00 m$ and there is a displacement of particle in x-direction that is $x=5.00 m$ .

Taking the limits of the integration,

$WAC=[∫05.003ydy] J=(3[y]05.00) J=3[5.00−0]=15 J$

Section-3:

To determine: The work done by the force $F→=(4xi^+3yj^) N$ on the particle along the purple path.

Answer: The work done by the force on the particle as it goes from O to C along the purple path is $35 J$ .

Given Information:

The force acting on the particle is $F→=(4xi^+3yj^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle along the purple path $OAC$ is,

$Wpurple=WOA+WAC$ (III)

$Wpurple$ is the work done along the purple path.
$WOA$ is the work done along the path $OA$ .
$WAC$ is the work done along the path $AC$ .

Substitute $20 J$ for $WOA$ and $15 J$ for $WAC$ in equation (II).

$Wpurple=20 J+15 J=35 J$

Conclusion:

Therefore, the work done by the force on the particle as it goes from O to C along the purple path is $35 J$ .

Section-4:

To determine: The work done by the force $F→=(4xi^+3yj^) N$ on the particle along the path $OB$ .

Answer: The work done by the force on the particle as it goes from O to C along the path $OB$ is $35 J$ .

Given Information:

The force acting on the particle is $F→=(4xi^+3yj^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle is,

$W=∫F→⋅dr→$

Substitute $(4xi^+3yj^) N$ for $F→$ and $dyj^$ for $dr^$ .

$W=∫((4xi^+3yj^) N)⋅(dyj^) m=∫(3ydy) N⋅m×1 J1 N⋅m=[∫3ydy] J$

Since along the path $OB$ the particle moves from $y=0$ to $y=5.00 m$ and there is no displacement in x-direction that is $x=0$ .

Taking the limits of the integration,

$WOB=[∫05.003ydy] J=3[y]05.00=(3[5.00−0]) J=15 J$

Section-5:

To determine: The work done by the force $F→=(4xi^+3yj^) N$ on the particle along the path $BC$ .

Answer: The work done by the force on the particle as it goes from O to C along the path $BC$ is $20 J$ .

Given Information:

The force acting on the particle is $F→=(4xi^+3yj^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle is,

$W=∫F→⋅dr→$

Substitute $(4xi^+3yj^) N$ for $F→$ and $dxi^$ for $dr^$ .

$W=∫((4xi^+3yj^) N)⋅(dxi^) m=∫(4xdx) N⋅m×1 J1 N⋅m=[∫4xdx] J$

In the path $BC$ , the particle moves from $x=0$ to $x=5.00 m$ and there is a displacement of particle in y-direction that is $y=5.00 m$ .

Taking the limits of the integration,

$WBC=[∫05.004xdx] J=(4[x]05.00) J=(4[5.00−0]) J=20 J$

Section-6:

To determine: The work done by the force $F→=(4xi^+3yj^) N$ on the particle along the red path.

Answer: The work done by the force on the particle as it goes from O to C along the red path is $35 J$ .

Given Information:

The force acting on the particle is $F→=(4xi^+3yj^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle along the red path $OBC$ is,

$Wred=WOB+WBC$

$Wred$ is the work done along the red path.
$WOB$ is the work done along the path $OB$ .
$WBC$ is the work done along the path $BC$ .

Substitute $15 J$ for $WOB$ and $20 J$ for $WBC$ .

$Wred=15 J+20 J=35 J$

Conclusion:

Therefore, the work done by the force on the particle as it goes from O to C along the red path is $35 J$ .

Section-7:

To determine: The work done by the force $F→=(4xi^+3yj^) N$ on the particle along the blue path.

Answer: The work done by the force on the particle as it goes from O to C along the blue path is $35 J$ .

Given Information:

The force acting on the particle is $F→=(4xi^+3yj^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle is,

$W=[∫4xdx+∫3ydy] J$

The path $OC$ is a straight line passes through a origin. In this blue path the particle moves in both the direction by $5.00 m$ .

Taking the limits on integration,

$Wred=[∫05.004xdx+∫05.003ydx] J=(4[x]05.00+3[y]05.00) J=(4[5.00−0]+3[5.00−0]) J=35 J$

Since the work done by the force $F→=(3i^+4j^) N$ in all three purple, red and blue path is $35 J$ hence the force is conservative and the work done is identical in all the three paths.

Conclusion:

Therefore, the work done by the force on the particle as it goes from O to C along the blue path is $35 J$ and the work done by the force is same in all the three paths.

(d)

To determine

Whether the work done by the force $F→=(yi^−xj^) N$ on the particle along each one of the three paths shown in the figure is identical.

(d)

Expert Solution

Answer to Problem 44P

The work done by the force $F→=(yi^−xj^) N$ on the particle along all the three path are not identical.

Explanation of Solution

Section-1:

To determine: The work done by the force $F→=(yi^−xj^) N$ on the particle along the path $OA$ .

Answer: The work done by the force on the particle as it goes from O to C along the path $OA$ is $−25 J$ .

Given Information:

The force acting on the particle is $F→=(yi^−xj^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle is,

$W=∫F→⋅dr→$

Substitute $(yi^−xj^) N$ for $F→$ and $dxi^$ for $dr^$ .

$W=∫((yi^−xj^) N)⋅(dxi^) m=∫(ydx) N⋅m×1 J1 N⋅m=[∫ydx] J$

Since along the path $OA$ the particle moves from $x=0$ to $x=5.00 m$ and there is no displacement in y-direction that is $y=0$ .

Substitute $0$ for $y$ and $0$ for $dy$ .

$WOA=[∫0dx] J=0 J$

Section-2:

To determine: The work done by the force $F→=(yi^−xj^) N$ on the particle along the path $AC$ .

Answer: The work done by the force on the particle as it goes from O to C along the path $AC$ is $−25 J$ .

Given Information:

The force acting on the particle is $F→=(yi^−xj^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle is,

$W=∫F→⋅dr→$

Substitute $(yi^−xj^) N$ for $F→$ and $dyj^$ for $dr^$ .

$W=∫((yi^−xj^) N)⋅(dyj) m=−∫(xdy) N⋅m×1 J1 N⋅m=−[∫xdy] J$

In the path $AC$ , the particle moves from $y=0$ to $y=5.00 m$ and there is a displacement of particle in x-direction that is $x=5.00 m$ .

Substitute $5.00$ for $x$ and $0$ for $dx$ .

$WAC=[−∫(5.00)dy] J$

Taking the limits of the integration,

$WAC=[∫05.00(5.00)dy] J=(−5[y]05.00) J=−5[5.00−0]=−25 J$

Section-3:

To determine: The work done by the force $F→=(yi^−xj^) N$ on the particle along the path $AC$ .

Answer: The work done by the force on the particle as it goes from O to C along the path $AC$ is $−25 J$ .

Given Information:

The force acting on the particle is $F→=(yi^−xj^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle along the purple path $OAC$ is,

$Wpurple=WOA+WAC$

$Wpurple$ is the work done along the purple path.
$WOA$ is the work done along the path $OA$ .
$WAC$ is the work done along the path $AC$ .

Substitute $0$ for $WOA$ and $−25 J$ for $WAC$ .

$Wpurple=0 J−25J=−25 J$

Conclusion:

Therefore, the work done by the force on the particle as it goes from O to C along the purple path is $−25 J$ .

Section-4:

To determine: The work done by the force $F→=(yi^−xj^) N$ on the particle along the path $OB$ .

Answer: The work done by the force on the particle as it goes from O to C along the path $OB$ is $25 J$ .

Given Information:

The force acting on the particle is $F→=(yi^−xj^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle is,

$W=∫F→⋅dr→$

Substitute $(yi^−xj^) N$ for $F→$ and $dyj^$ for $dr^$ .

$W=∫((yi^−xj^) N)⋅(dyj) m=−∫(xdy) N⋅m×1 J1 N⋅m=−[∫xdy] J$

Since along the path $OB$ the particle moves from $y=0$ to $y=5.00 m$ and there is no displacement in x-direction that is $x=0$ .

Substitute $0$ for $x$ in above integration.

$WOB=[∫05.000dy] J=0$

Section-5:

To determine: The work done by the force $F→=(yi^−xj^) N$ on the particle along the path $BC$ .

Answer: The work done by the force on the particle as it goes from O to C along the path $BC$ is $25 J$ .

Given Information:

The force acting on the particle is $F→=(yi^−xj^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle is,

$W=∫F→⋅dr→$

Substitute $(yi^−xj^) N$ for $F→$ and $dxi^$ for $dr^$ .

$W=∫((yi^−xj^) N)⋅(dxi^) m=∫(ydx) N⋅m×1 J1 N⋅m=[∫ydx] J$

In the path $BC$ , the particle moves from $x=0$ to $x=5.00 m$ and there is a displacement of particle in y-direction that is $y=5.00 m$ .

Substitute $5.00$ for $y$ and $0$ for $dy$ .

Taking the limits of the integration,

$WBC=[∫05.005dx] J=(5[x]05.00) J=(5[5.00−0]) J=25 J$

Section-6:

To determine: The work done by the force $F→=(yi^−xj^) N$ on the particle along the red path.

Answer: The work done by the force on the particle as it goes from O to C along the red path is $25 J$ .

Given Information:

The force acting on the particle is $F→=(yi^−xj^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle along the red path $OBC$ is,

$Wred=WOB+WBC$ (VI)

$Wred$ is the work done along the red path.
$WOB$ is the work done along the path $OB$ .
$WBC$ is the work done along the path $BC$ .

Substitute $0$ for $WOB$ and $25 J$ for $WBC$ in equation (VI).

$Wred=0 J+25 J=25 J$

Conclusion:

Therefore, the work done by the force on the particle as it goes from O to C along the red path is $25 J$ .

Section-7:

To determine: The work done by the force $F→=(yi^−xj^) N$ on the particle along the blue path.

Answer: The work done by the force on the particle as it goes from O to C along the blue path is $0$ .

Given Information:

The force acting on the particle is $F→=(yi^−xj^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle is,

$W=∫F→⋅dr→$

Substitute $(yi^−xj^) N$ for $F→$ and $dxi^+dyj^$ for $dr^$ .

$W=∫((yi^−xj^) N)⋅(dxi^+dyj^) m=∫(ydx−xdy) N⋅m×1 J1 N⋅m=[∫ydx−xdy] J$

The path $OC$ is a straight line passes through a origin. In this blue path the particle moves in both the direction by $5.00 m$ . Hence equation of line become $y=x$ .

Substitute $x$ for $y$ and $dx$ for $dy$ .

Taking the limits on integration,

$Wblue=[∫xdx−∫xdx] J=0$

Conclusion:

Therefore, the work done by the force on the particle as it goes along the three paths is not same.

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Answer 3

Question

Chapter 6, Problem 44P

(a)

To determine

To Show: That the constant force act on the object is conservative.

(a)

Expert Solution

Answer to Problem 44P

the constant force applied on the object is conservative in nature.

Explanation of Solution

The work done by a conservative force on a particle moving between any two points is independent of the path taken by the particle. Its only depends upon the end points of the path taken by the particle to move.

Given Information:

The general definition for work done by a force is,

$W=∫ifF→⋅dr→$ .

Formula to calculate the work done by the force on the object is,

$W=∫ifF→⋅dr→$

$W$ is the work done by the force on the particle.
$F→$ is the applied force vector on the object.
$dr→$ is the displacement vector of the object.

Since the force is constant that does not vary with respect to time or the position or the velocity of the object. So, the value of force can be taken out from the integration since it is constant quantity.

$W=F→∫ifdr→=F→[r→]if=F→[r→f−r→i]$

Now, here the force is constant so, the work done by this force on the object in only depends upon the end points of the displace object that shows the work done is independent of the path taken by the object to displace between the end points. But the work done is independent of the path only when the force is conservative.

Conclusion:

Therefore, the constant force applied on the object is conservative in nature.

(b)

To determine

The work done by the force $F→=(3i^+4j^) N$ on the particle along each one of the three paths shown in the figure and show that the work done along the three path is identical.

(b)

Expert Solution

Answer to Problem 44P

The work done by the force $F→=(3i^+4j^) N$ on the particle along all the three path are $35 J$ and the work done by the force is identical in all the three paths.

Explanation of Solution

Section-1:

To determine: The work done by the force $F→=(3i^+4j^) N$ on the particle along the path $OA$ .

Answer: The work done by the force on the particle as it goes from O to C along the path $OA$ is $35 J$ .

Given Information:

The force acting on the particle is $F→=(3i^+4j^) N$ and the three different path of the particle moves is shown in figure (I).

Principles of Physics: A Calculus-Based Text, Hybrid (with Enhanced WebAssign Printed Access Card), Chapter 6, Problem 44P

Figure (I)

Formula to calculate the work done by the force on the particle is,

$W=∫F→⋅dr→$

Substitute $(3i^+4j^) N$ for $F→$ and $dxi^$ for $dr^$ .

$W=∫((3i^+4j^) N)⋅(dxi^) m=∫(3dx) N⋅m×1 J1 N⋅m=[∫3dx] J$

Since along the path $OA$ the particle moves from $x=0$ to $x=5.00 m$ and there is no displacement in y-direction that is $y=0$ .

Taking the limits of the integration,

$WOA=[∫05.003dx] J=(3[x]05.00) J=(3[5.00−0]) J=15 J$

Section-2:

To determine: The work done by the force $F→=(3i^+4j^) N$ on the particle along the path $AC$ .

Answer: The work done by the force on the particle as it goes from O to C along the path $AC$ is $35 J$ .

Given Information:

The force acting on the particle is $F→=(3i^+4j^) N$ and the three different path of the particle moves is shown in figure (I).

In the path $AC$ , the particle moves from $y=0$ to $y=5.00 m$ and there is a displacement of particle in x-direction that is $x=5.00 m$ .

Formula to calculate the work done by the force on the particle is,

$W=∫F→⋅dr→$

Substitute $(3i^+4j^) N$ for $F→$ and $dyj^$ for $dr^$ .

$W=∫((3i^+4j^) N)⋅(dyj^) m=∫(4dy) N⋅m×1 J1 N⋅m=[∫4dy] J$

Taking the limits of the integration,

$WAC=[∫05.004dy] J=(4[y]05.00) J=4[5.00−0]=20 J$

Section-3:

To determine: The work done by the force $F→=(3i^+4j^) N$ on the particle along the purple path.

Answer: The work done by the force on the particle as it goes from O to C along the purple path is $35 J$ .

Given Information:

The force acting on the particle is $F→=(3i^+4j^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle along the purple path $OAC$ is,

$Wpurple=WOA+WAC$

$Wpurple$ is the work done along the purple path.
$WOA$ is the work done along the path $OA$ .
$WAC$ is the work done along the path $AC$ .

Substitute $15 J$ for $WOA$ and $20 J$ for $WAC$ .

$Wpurple=15 J+20 J=35 J$

Conclusion:

Therefore, the work done by the force on the particle as it goes from O to C along the purple path is $35 J$ .

Section-4:

To determine: The work done by the force $F→=(3i^+4j^) N$ on the particle along the path $OB$ .

Answer: The work done by the force on the particle as it goes from O to C along the path $OB$ is $35 J$ .

Given Information:

The force acting on the particle is $F→=(3i^+4j^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle is,

$W=∫F→⋅dr→$

Substitute $(3i^+4j^) N$ for $F→$ and $dyj^$ for $dr^$ .

$W=∫((3i^+4j^) N)⋅(dyj^) m=∫(4dy) N⋅m×1 J1 N⋅m=[∫4dy] J$

Since along the path $OB$ the particle moves from $y=0$ to $y=5.00 m$ and there is no displacement in x-direction that is $x=0$ .

Taking the limits of the integration,

$WOB=[∫05.004dy] J=4[y]05.00=(4[5.00−0]) J=20 J$

Section-5:

To determine: The work done by the force $F→=(3i^+4j^) N$ on the particle along the path $BC$ .

Answer: The work done by the force on the particle as it goes from O to C along the path $BC$ is $35 J$ .

Given Information:

The force acting on the particle is $F→=(3i^+4j^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle is,

$W=∫F→⋅dr→$

Substitute $(3i^+4j^) N$ for $F→$ and $dxi^$ for $dr^$ .

$W=∫((3i^+4j^) N)⋅(dxi^) m=∫(3dx) N⋅m×1 J1 N⋅m=[∫3dx] J$

In the path $BC$ , the particle moves from $x=0$ to $x=5.00 m$ and there is a displacement of particle in y-direction that is $y=5.00 m$ .

Taking the limits of the integration,

$WBC=[∫05.003dx+∫004dy] J=(3[x]05.00+0) J=(3[5.00−0]) J=15 J$

Section-6:

To determine: The work done by the force $F→=(3i^+4j^) N$ on the particle along the red path.

Answer: The work done by the force on the particle as it goes from O to C along the red path is $35 J$ .

Given Information:

The force acting on the particle is $F→=(3i^+4j^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle along the red path $OBC$ is,

$Wred=WOB+WBC$

$Wred$ is the work done along the red path.
$WOB$ is the work done along the path $OB$ .
$WBC$ is the work done along the path $BC$ .

Substitute $20 J$ for $WOB$ and $15 J$ for $WBC$ .

$Wred=20+15 J=35 J$

Conclusion:

Therefore, the work done by the force on the particle as it goes from O to C along the red path is $35 J$ .

Section-7:

To determine: The work done by the force $F→=(3i^+4j^) N$ on the particle along the blue path.

Answer: The work done by the force on the particle as it goes from O to C along the blue path is $35 J$ .

Given Information:

The force acting on the particle is $F→=(3i^+4j^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle is,

$W=[∫3dx+∫4dy] J$

The path $OC$ is a straight line passes through a origin. In this blue path the particle moves in both the direction by $5.00 m$ .

Taking the limits on integration,

$Wblue=[∫05.003dx+∫05.004dx] J=(3[x]05.00+4[y]05.00) J=(3[5.00−0]+4[5.00−0]) J=35 J$

Since the work done by the force $F→=(3i^+4j^) N$ in all three purple, red and blue path is $35 J$ hence the force is conservative and the work done is identical in all the three paths.

Conclusion:

Therefore, the work done by the force on the particle as it goes from O to C along the blue path is $35 J$ and the work done by the force is same in all the three paths.

(c)

To determine

Whether the work done by the force $F→=(4xi^+3yj^) N$ on the particle along each one of the three paths shown in the figure is identical.

(c)

Expert Solution

Answer to Problem 44P

The work done by the force $F→=(4xi^+3yj^) N$ on the particle along all the three path are $35 J$ and the work done by the force is identical in all the three paths.

Explanation of Solution

Section-1:

To determine: The work done by the force $F→=(4xi^+3yj^) N$ on the particle along the path $OA$ .

Answer: The work done by the force on the particle as it goes from O to C along the path $OA$ is $35 J$ .

Given Information:

The force acting on the particle is $F→=(4xi^+3yj^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle is,

$W=∫F→⋅dr→$

Substitute $(4xi^+3yj^) N$ for $F→$ and $dxi^$ for $dr^$ .

$W=∫((4xi^+3yj^) N)⋅(dxi^) m=∫(4xdx) N⋅m×1 J1 N⋅m=[∫4xdx] J$

Since along the path $OA$ the particle moves from $x=0$ to $x=5.00 m$ and there is no displacement in y-direction that is $y=0$ .

Taking the limits of the integration,

$WOA=[∫05.004xdx] J=(4[x]05.00) J=(4[5.00−0]) J=20 J$

Section-2:

To determine: The work done by the force $F→=(4xi^+3yj^) N$ on the particle along the path $AC$ .

Answer: The work done by the force on the particle as it goes from O to C along the path $AC$ is $35 J$ .

Given Information:

The force acting on the particle is $F→=(4xi^+3yj^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle is,

$W=∫F→⋅dr→$

Substitute $(4xi^+3yj^) N$ for $F→$ and $dyj^$ for $dr^$ .

$W=∫((4xi^+3yj^) N)⋅(dyj^) m=∫(3ydy) N⋅m×1 J1 N⋅m=[∫3ydy] J$

In the path $AC$ , the particle moves from $y=0$ to $y=5.00 m$ and there is a displacement of particle in x-direction that is $x=5.00 m$ .

Taking the limits of the integration,

$WAC=[∫05.003ydy] J=(3[y]05.00) J=3[5.00−0]=15 J$

Section-3:

To determine: The work done by the force $F→=(4xi^+3yj^) N$ on the particle along the purple path.

Answer: The work done by the force on the particle as it goes from O to C along the purple path is $35 J$ .

Given Information:

The force acting on the particle is $F→=(4xi^+3yj^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle along the purple path $OAC$ is,

$Wpurple=WOA+WAC$ (III)

$Wpurple$ is the work done along the purple path.
$WOA$ is the work done along the path $OA$ .
$WAC$ is the work done along the path $AC$ .

Substitute $20 J$ for $WOA$ and $15 J$ for $WAC$ in equation (II).

$Wpurple=20 J+15 J=35 J$

Conclusion:

Therefore, the work done by the force on the particle as it goes from O to C along the purple path is $35 J$ .

Section-4:

To determine: The work done by the force $F→=(4xi^+3yj^) N$ on the particle along the path $OB$ .

Answer: The work done by the force on the particle as it goes from O to C along the path $OB$ is $35 J$ .

Given Information:

The force acting on the particle is $F→=(4xi^+3yj^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle is,

$W=∫F→⋅dr→$

Substitute $(4xi^+3yj^) N$ for $F→$ and $dyj^$ for $dr^$ .

$W=∫((4xi^+3yj^) N)⋅(dyj^) m=∫(3ydy) N⋅m×1 J1 N⋅m=[∫3ydy] J$

Since along the path $OB$ the particle moves from $y=0$ to $y=5.00 m$ and there is no displacement in x-direction that is $x=0$ .

Taking the limits of the integration,

$WOB=[∫05.003ydy] J=3[y]05.00=(3[5.00−0]) J=15 J$

Section-5:

To determine: The work done by the force $F→=(4xi^+3yj^) N$ on the particle along the path $BC$ .

Answer: The work done by the force on the particle as it goes from O to C along the path $BC$ is $20 J$ .

Given Information:

The force acting on the particle is $F→=(4xi^+3yj^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle is,

$W=∫F→⋅dr→$

Substitute $(4xi^+3yj^) N$ for $F→$ and $dxi^$ for $dr^$ .

$W=∫((4xi^+3yj^) N)⋅(dxi^) m=∫(4xdx) N⋅m×1 J1 N⋅m=[∫4xdx] J$

In the path $BC$ , the particle moves from $x=0$ to $x=5.00 m$ and there is a displacement of particle in y-direction that is $y=5.00 m$ .

Taking the limits of the integration,

$WBC=[∫05.004xdx] J=(4[x]05.00) J=(4[5.00−0]) J=20 J$

Section-6:

To determine: The work done by the force $F→=(4xi^+3yj^) N$ on the particle along the red path.

Answer: The work done by the force on the particle as it goes from O to C along the red path is $35 J$ .

Given Information:

The force acting on the particle is $F→=(4xi^+3yj^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle along the red path $OBC$ is,

$Wred=WOB+WBC$

$Wred$ is the work done along the red path.
$WOB$ is the work done along the path $OB$ .
$WBC$ is the work done along the path $BC$ .

Substitute $15 J$ for $WOB$ and $20 J$ for $WBC$ .

$Wred=15 J+20 J=35 J$

Conclusion:

Therefore, the work done by the force on the particle as it goes from O to C along the red path is $35 J$ .

Section-7:

To determine: The work done by the force $F→=(4xi^+3yj^) N$ on the particle along the blue path.

Answer: The work done by the force on the particle as it goes from O to C along the blue path is $35 J$ .

Given Information:

The force acting on the particle is $F→=(4xi^+3yj^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle is,

$W=[∫4xdx+∫3ydy] J$

The path $OC$ is a straight line passes through a origin. In this blue path the particle moves in both the direction by $5.00 m$ .

Taking the limits on integration,

$Wred=[∫05.004xdx+∫05.003ydx] J=(4[x]05.00+3[y]05.00) J=(4[5.00−0]+3[5.00−0]) J=35 J$

Since the work done by the force $F→=(3i^+4j^) N$ in all three purple, red and blue path is $35 J$ hence the force is conservative and the work done is identical in all the three paths.

Conclusion:

Therefore, the work done by the force on the particle as it goes from O to C along the blue path is $35 J$ and the work done by the force is same in all the three paths.

(d)

To determine

Whether the work done by the force $F→=(yi^−xj^) N$ on the particle along each one of the three paths shown in the figure is identical.

(d)

Expert Solution

Answer to Problem 44P

The work done by the force $F→=(yi^−xj^) N$ on the particle along all the three path are not identical.

Explanation of Solution

Section-1:

To determine: The work done by the force $F→=(yi^−xj^) N$ on the particle along the path $OA$ .

Answer: The work done by the force on the particle as it goes from O to C along the path $OA$ is $−25 J$ .

Given Information:

The force acting on the particle is $F→=(yi^−xj^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle is,

$W=∫F→⋅dr→$

Substitute $(yi^−xj^) N$ for $F→$ and $dxi^$ for $dr^$ .

$W=∫((yi^−xj^) N)⋅(dxi^) m=∫(ydx) N⋅m×1 J1 N⋅m=[∫ydx] J$

Since along the path $OA$ the particle moves from $x=0$ to $x=5.00 m$ and there is no displacement in y-direction that is $y=0$ .

Substitute $0$ for $y$ and $0$ for $dy$ .

$WOA=[∫0dx] J=0 J$

Section-2:

To determine: The work done by the force $F→=(yi^−xj^) N$ on the particle along the path $AC$ .

Answer: The work done by the force on the particle as it goes from O to C along the path $AC$ is $−25 J$ .

Given Information:

The force acting on the particle is $F→=(yi^−xj^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle is,

$W=∫F→⋅dr→$

Substitute $(yi^−xj^) N$ for $F→$ and $dyj^$ for $dr^$ .

$W=∫((yi^−xj^) N)⋅(dyj) m=−∫(xdy) N⋅m×1 J1 N⋅m=−[∫xdy] J$

In the path $AC$ , the particle moves from $y=0$ to $y=5.00 m$ and there is a displacement of particle in x-direction that is $x=5.00 m$ .

Substitute $5.00$ for $x$ and $0$ for $dx$ .

$WAC=[−∫(5.00)dy] J$

Taking the limits of the integration,

$WAC=[∫05.00(5.00)dy] J=(−5[y]05.00) J=−5[5.00−0]=−25 J$

Section-3:

To determine: The work done by the force $F→=(yi^−xj^) N$ on the particle along the path $AC$ .

Answer: The work done by the force on the particle as it goes from O to C along the path $AC$ is $−25 J$ .

Given Information:

The force acting on the particle is $F→=(yi^−xj^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle along the purple path $OAC$ is,

$Wpurple=WOA+WAC$

$Wpurple$ is the work done along the purple path.
$WOA$ is the work done along the path $OA$ .
$WAC$ is the work done along the path $AC$ .

Substitute $0$ for $WOA$ and $−25 J$ for $WAC$ .

$Wpurple=0 J−25J=−25 J$

Conclusion:

Therefore, the work done by the force on the particle as it goes from O to C along the purple path is $−25 J$ .

Section-4:

To determine: The work done by the force $F→=(yi^−xj^) N$ on the particle along the path $OB$ .

Answer: The work done by the force on the particle as it goes from O to C along the path $OB$ is $25 J$ .

Given Information:

The force acting on the particle is $F→=(yi^−xj^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle is,

$W=∫F→⋅dr→$

Substitute $(yi^−xj^) N$ for $F→$ and $dyj^$ for $dr^$ .

$W=∫((yi^−xj^) N)⋅(dyj) m=−∫(xdy) N⋅m×1 J1 N⋅m=−[∫xdy] J$

Since along the path $OB$ the particle moves from $y=0$ to $y=5.00 m$ and there is no displacement in x-direction that is $x=0$ .

Substitute $0$ for $x$ in above integration.

$WOB=[∫05.000dy] J=0$

Section-5:

To determine: The work done by the force $F→=(yi^−xj^) N$ on the particle along the path $BC$ .

Answer: The work done by the force on the particle as it goes from O to C along the path $BC$ is $25 J$ .

Given Information:

The force acting on the particle is $F→=(yi^−xj^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle is,

$W=∫F→⋅dr→$

Substitute $(yi^−xj^) N$ for $F→$ and $dxi^$ for $dr^$ .

$W=∫((yi^−xj^) N)⋅(dxi^) m=∫(ydx) N⋅m×1 J1 N⋅m=[∫ydx] J$

In the path $BC$ , the particle moves from $x=0$ to $x=5.00 m$ and there is a displacement of particle in y-direction that is $y=5.00 m$ .

Substitute $5.00$ for $y$ and $0$ for $dy$ .

Taking the limits of the integration,

$WBC=[∫05.005dx] J=(5[x]05.00) J=(5[5.00−0]) J=25 J$

Section-6:

To determine: The work done by the force $F→=(yi^−xj^) N$ on the particle along the red path.

Answer: The work done by the force on the particle as it goes from O to C along the red path is $25 J$ .

Given Information:

The force acting on the particle is $F→=(yi^−xj^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle along the red path $OBC$ is,

$Wred=WOB+WBC$ (VI)

$Wred$ is the work done along the red path.
$WOB$ is the work done along the path $OB$ .
$WBC$ is the work done along the path $BC$ .

Substitute $0$ for $WOB$ and $25 J$ for $WBC$ in equation (VI).

$Wred=0 J+25 J=25 J$

Conclusion:

Therefore, the work done by the force on the particle as it goes from O to C along the red path is $25 J$ .

Section-7:

To determine: The work done by the force $F→=(yi^−xj^) N$ on the particle along the blue path.

Answer: The work done by the force on the particle as it goes from O to C along the blue path is $0$ .

Given Information:

The force acting on the particle is $F→=(yi^−xj^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle is,

$W=∫F→⋅dr→$

Substitute $(yi^−xj^) N$ for $F→$ and $dxi^+dyj^$ for $dr^$ .

$W=∫((yi^−xj^) N)⋅(dxi^+dyj^) m=∫(ydx−xdy) N⋅m×1 J1 N⋅m=[∫ydx−xdy] J$

The path $OC$ is a straight line passes through a origin. In this blue path the particle moves in both the direction by $5.00 m$ . Hence equation of line become $y=x$ .

Substitute $x$ for $y$ and $dx$ for $dy$ .

Taking the limits on integration,

$Wblue=[∫xdx−∫xdx] J=0$

Conclusion:

Therefore, the work done by the force on the particle as it goes along the three paths is not same.

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Answer 4

Question

Chapter 6, Problem 44P

(a)

To determine

To Show: That the constant force act on the object is conservative.

(a)

Expert Solution

Answer to Problem 44P

the constant force applied on the object is conservative in nature.

Explanation of Solution

The work done by a conservative force on a particle moving between any two points is independent of the path taken by the particle. Its only depends upon the end points of the path taken by the particle to move.

Given Information:

The general definition for work done by a force is,

$W=∫ifF→⋅dr→$ .

Formula to calculate the work done by the force on the object is,

$W=∫ifF→⋅dr→$

$W$ is the work done by the force on the particle.
$F→$ is the applied force vector on the object.
$dr→$ is the displacement vector of the object.

Since the force is constant that does not vary with respect to time or the position or the velocity of the object. So, the value of force can be taken out from the integration since it is constant quantity.

$W=F→∫ifdr→=F→[r→]if=F→[r→f−r→i]$

Now, here the force is constant so, the work done by this force on the object in only depends upon the end points of the displace object that shows the work done is independent of the path taken by the object to displace between the end points. But the work done is independent of the path only when the force is conservative.

Conclusion:

Therefore, the constant force applied on the object is conservative in nature.

(b)

To determine

The work done by the force $F→=(3i^+4j^) N$ on the particle along each one of the three paths shown in the figure and show that the work done along the three path is identical.

(b)

Expert Solution

Answer to Problem 44P

The work done by the force $F→=(3i^+4j^) N$ on the particle along all the three path are $35 J$ and the work done by the force is identical in all the three paths.

Explanation of Solution

Section-1:

To determine: The work done by the force $F→=(3i^+4j^) N$ on the particle along the path $OA$ .

Answer: The work done by the force on the particle as it goes from O to C along the path $OA$ is $35 J$ .

Given Information:

The force acting on the particle is $F→=(3i^+4j^) N$ and the three different path of the particle moves is shown in figure (I).

Principles of Physics: A Calculus-Based Text, Hybrid (with Enhanced WebAssign Printed Access Card), Chapter 6, Problem 44P

Figure (I)

Formula to calculate the work done by the force on the particle is,

$W=∫F→⋅dr→$

Substitute $(3i^+4j^) N$ for $F→$ and $dxi^$ for $dr^$ .

$W=∫((3i^+4j^) N)⋅(dxi^) m=∫(3dx) N⋅m×1 J1 N⋅m=[∫3dx] J$

Since along the path $OA$ the particle moves from $x=0$ to $x=5.00 m$ and there is no displacement in y-direction that is $y=0$ .

Taking the limits of the integration,

$WOA=[∫05.003dx] J=(3[x]05.00) J=(3[5.00−0]) J=15 J$

Section-2:

To determine: The work done by the force $F→=(3i^+4j^) N$ on the particle along the path $AC$ .

Answer: The work done by the force on the particle as it goes from O to C along the path $AC$ is $35 J$ .

Given Information:

The force acting on the particle is $F→=(3i^+4j^) N$ and the three different path of the particle moves is shown in figure (I).

In the path $AC$ , the particle moves from $y=0$ to $y=5.00 m$ and there is a displacement of particle in x-direction that is $x=5.00 m$ .

Formula to calculate the work done by the force on the particle is,

$W=∫F→⋅dr→$

Substitute $(3i^+4j^) N$ for $F→$ and $dyj^$ for $dr^$ .

$W=∫((3i^+4j^) N)⋅(dyj^) m=∫(4dy) N⋅m×1 J1 N⋅m=[∫4dy] J$

Taking the limits of the integration,

$WAC=[∫05.004dy] J=(4[y]05.00) J=4[5.00−0]=20 J$

Section-3:

To determine: The work done by the force $F→=(3i^+4j^) N$ on the particle along the purple path.

Answer: The work done by the force on the particle as it goes from O to C along the purple path is $35 J$ .

Given Information:

The force acting on the particle is $F→=(3i^+4j^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle along the purple path $OAC$ is,

$Wpurple=WOA+WAC$

$Wpurple$ is the work done along the purple path.
$WOA$ is the work done along the path $OA$ .
$WAC$ is the work done along the path $AC$ .

Substitute $15 J$ for $WOA$ and $20 J$ for $WAC$ .

$Wpurple=15 J+20 J=35 J$

Conclusion:

Therefore, the work done by the force on the particle as it goes from O to C along the purple path is $35 J$ .

Section-4:

To determine: The work done by the force $F→=(3i^+4j^) N$ on the particle along the path $OB$ .

Answer: The work done by the force on the particle as it goes from O to C along the path $OB$ is $35 J$ .

Given Information:

The force acting on the particle is $F→=(3i^+4j^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle is,

$W=∫F→⋅dr→$

Substitute $(3i^+4j^) N$ for $F→$ and $dyj^$ for $dr^$ .

$W=∫((3i^+4j^) N)⋅(dyj^) m=∫(4dy) N⋅m×1 J1 N⋅m=[∫4dy] J$

Since along the path $OB$ the particle moves from $y=0$ to $y=5.00 m$ and there is no displacement in x-direction that is $x=0$ .

Taking the limits of the integration,

$WOB=[∫05.004dy] J=4[y]05.00=(4[5.00−0]) J=20 J$

Section-5:

To determine: The work done by the force $F→=(3i^+4j^) N$ on the particle along the path $BC$ .

Answer: The work done by the force on the particle as it goes from O to C along the path $BC$ is $35 J$ .

Given Information:

The force acting on the particle is $F→=(3i^+4j^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle is,

$W=∫F→⋅dr→$

Substitute $(3i^+4j^) N$ for $F→$ and $dxi^$ for $dr^$ .

$W=∫((3i^+4j^) N)⋅(dxi^) m=∫(3dx) N⋅m×1 J1 N⋅m=[∫3dx] J$

In the path $BC$ , the particle moves from $x=0$ to $x=5.00 m$ and there is a displacement of particle in y-direction that is $y=5.00 m$ .

Taking the limits of the integration,

$WBC=[∫05.003dx+∫004dy] J=(3[x]05.00+0) J=(3[5.00−0]) J=15 J$

Section-6:

To determine: The work done by the force $F→=(3i^+4j^) N$ on the particle along the red path.

Answer: The work done by the force on the particle as it goes from O to C along the red path is $35 J$ .

Given Information:

The force acting on the particle is $F→=(3i^+4j^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle along the red path $OBC$ is,

$Wred=WOB+WBC$

$Wred$ is the work done along the red path.
$WOB$ is the work done along the path $OB$ .
$WBC$ is the work done along the path $BC$ .

Substitute $20 J$ for $WOB$ and $15 J$ for $WBC$ .

$Wred=20+15 J=35 J$

Conclusion:

Therefore, the work done by the force on the particle as it goes from O to C along the red path is $35 J$ .

Section-7:

To determine: The work done by the force $F→=(3i^+4j^) N$ on the particle along the blue path.

Answer: The work done by the force on the particle as it goes from O to C along the blue path is $35 J$ .

Given Information:

The force acting on the particle is $F→=(3i^+4j^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle is,

$W=[∫3dx+∫4dy] J$

The path $OC$ is a straight line passes through a origin. In this blue path the particle moves in both the direction by $5.00 m$ .

Taking the limits on integration,

$Wblue=[∫05.003dx+∫05.004dx] J=(3[x]05.00+4[y]05.00) J=(3[5.00−0]+4[5.00−0]) J=35 J$

Since the work done by the force $F→=(3i^+4j^) N$ in all three purple, red and blue path is $35 J$ hence the force is conservative and the work done is identical in all the three paths.

Conclusion:

Therefore, the work done by the force on the particle as it goes from O to C along the blue path is $35 J$ and the work done by the force is same in all the three paths.

(c)

To determine

Whether the work done by the force $F→=(4xi^+3yj^) N$ on the particle along each one of the three paths shown in the figure is identical.

(c)

Expert Solution

Answer to Problem 44P

The work done by the force $F→=(4xi^+3yj^) N$ on the particle along all the three path are $35 J$ and the work done by the force is identical in all the three paths.

Explanation of Solution

Section-1:

To determine: The work done by the force $F→=(4xi^+3yj^) N$ on the particle along the path $OA$ .

Answer: The work done by the force on the particle as it goes from O to C along the path $OA$ is $35 J$ .

Given Information:

The force acting on the particle is $F→=(4xi^+3yj^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle is,

$W=∫F→⋅dr→$

Substitute $(4xi^+3yj^) N$ for $F→$ and $dxi^$ for $dr^$ .

$W=∫((4xi^+3yj^) N)⋅(dxi^) m=∫(4xdx) N⋅m×1 J1 N⋅m=[∫4xdx] J$

Since along the path $OA$ the particle moves from $x=0$ to $x=5.00 m$ and there is no displacement in y-direction that is $y=0$ .

Taking the limits of the integration,

$WOA=[∫05.004xdx] J=(4[x]05.00) J=(4[5.00−0]) J=20 J$

Section-2:

To determine: The work done by the force $F→=(4xi^+3yj^) N$ on the particle along the path $AC$ .

Answer: The work done by the force on the particle as it goes from O to C along the path $AC$ is $35 J$ .

Given Information:

The force acting on the particle is $F→=(4xi^+3yj^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle is,

$W=∫F→⋅dr→$

Substitute $(4xi^+3yj^) N$ for $F→$ and $dyj^$ for $dr^$ .

$W=∫((4xi^+3yj^) N)⋅(dyj^) m=∫(3ydy) N⋅m×1 J1 N⋅m=[∫3ydy] J$

In the path $AC$ , the particle moves from $y=0$ to $y=5.00 m$ and there is a displacement of particle in x-direction that is $x=5.00 m$ .

Taking the limits of the integration,

$WAC=[∫05.003ydy] J=(3[y]05.00) J=3[5.00−0]=15 J$

Section-3:

To determine: The work done by the force $F→=(4xi^+3yj^) N$ on the particle along the purple path.

Answer: The work done by the force on the particle as it goes from O to C along the purple path is $35 J$ .

Given Information:

The force acting on the particle is $F→=(4xi^+3yj^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle along the purple path $OAC$ is,

$Wpurple=WOA+WAC$ (III)

$Wpurple$ is the work done along the purple path.
$WOA$ is the work done along the path $OA$ .
$WAC$ is the work done along the path $AC$ .

Substitute $20 J$ for $WOA$ and $15 J$ for $WAC$ in equation (II).

$Wpurple=20 J+15 J=35 J$

Conclusion:

Therefore, the work done by the force on the particle as it goes from O to C along the purple path is $35 J$ .

Section-4:

To determine: The work done by the force $F→=(4xi^+3yj^) N$ on the particle along the path $OB$ .

Answer: The work done by the force on the particle as it goes from O to C along the path $OB$ is $35 J$ .

Given Information:

The force acting on the particle is $F→=(4xi^+3yj^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle is,

$W=∫F→⋅dr→$

Substitute $(4xi^+3yj^) N$ for $F→$ and $dyj^$ for $dr^$ .

$W=∫((4xi^+3yj^) N)⋅(dyj^) m=∫(3ydy) N⋅m×1 J1 N⋅m=[∫3ydy] J$

Since along the path $OB$ the particle moves from $y=0$ to $y=5.00 m$ and there is no displacement in x-direction that is $x=0$ .

Taking the limits of the integration,

$WOB=[∫05.003ydy] J=3[y]05.00=(3[5.00−0]) J=15 J$

Section-5:

To determine: The work done by the force $F→=(4xi^+3yj^) N$ on the particle along the path $BC$ .

Answer: The work done by the force on the particle as it goes from O to C along the path $BC$ is $20 J$ .

Given Information:

The force acting on the particle is $F→=(4xi^+3yj^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle is,

$W=∫F→⋅dr→$

Substitute $(4xi^+3yj^) N$ for $F→$ and $dxi^$ for $dr^$ .

$W=∫((4xi^+3yj^) N)⋅(dxi^) m=∫(4xdx) N⋅m×1 J1 N⋅m=[∫4xdx] J$

In the path $BC$ , the particle moves from $x=0$ to $x=5.00 m$ and there is a displacement of particle in y-direction that is $y=5.00 m$ .

Taking the limits of the integration,

$WBC=[∫05.004xdx] J=(4[x]05.00) J=(4[5.00−0]) J=20 J$

Section-6:

To determine: The work done by the force $F→=(4xi^+3yj^) N$ on the particle along the red path.

Answer: The work done by the force on the particle as it goes from O to C along the red path is $35 J$ .

Given Information:

The force acting on the particle is $F→=(4xi^+3yj^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle along the red path $OBC$ is,

$Wred=WOB+WBC$

$Wred$ is the work done along the red path.
$WOB$ is the work done along the path $OB$ .
$WBC$ is the work done along the path $BC$ .

Substitute $15 J$ for $WOB$ and $20 J$ for $WBC$ .

$Wred=15 J+20 J=35 J$

Conclusion:

Therefore, the work done by the force on the particle as it goes from O to C along the red path is $35 J$ .

Section-7:

To determine: The work done by the force $F→=(4xi^+3yj^) N$ on the particle along the blue path.

Answer: The work done by the force on the particle as it goes from O to C along the blue path is $35 J$ .

Given Information:

The force acting on the particle is $F→=(4xi^+3yj^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle is,

$W=[∫4xdx+∫3ydy] J$

The path $OC$ is a straight line passes through a origin. In this blue path the particle moves in both the direction by $5.00 m$ .

Taking the limits on integration,

$Wred=[∫05.004xdx+∫05.003ydx] J=(4[x]05.00+3[y]05.00) J=(4[5.00−0]+3[5.00−0]) J=35 J$

Since the work done by the force $F→=(3i^+4j^) N$ in all three purple, red and blue path is $35 J$ hence the force is conservative and the work done is identical in all the three paths.

Conclusion:

Therefore, the work done by the force on the particle as it goes from O to C along the blue path is $35 J$ and the work done by the force is same in all the three paths.

(d)

To determine

Whether the work done by the force $F→=(yi^−xj^) N$ on the particle along each one of the three paths shown in the figure is identical.

(d)

Expert Solution

Answer to Problem 44P

The work done by the force $F→=(yi^−xj^) N$ on the particle along all the three path are not identical.

Explanation of Solution

Section-1:

To determine: The work done by the force $F→=(yi^−xj^) N$ on the particle along the path $OA$ .

Answer: The work done by the force on the particle as it goes from O to C along the path $OA$ is $−25 J$ .

Given Information:

The force acting on the particle is $F→=(yi^−xj^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle is,

$W=∫F→⋅dr→$

Substitute $(yi^−xj^) N$ for $F→$ and $dxi^$ for $dr^$ .

$W=∫((yi^−xj^) N)⋅(dxi^) m=∫(ydx) N⋅m×1 J1 N⋅m=[∫ydx] J$

Since along the path $OA$ the particle moves from $x=0$ to $x=5.00 m$ and there is no displacement in y-direction that is $y=0$ .

Substitute $0$ for $y$ and $0$ for $dy$ .

$WOA=[∫0dx] J=0 J$

Section-2:

To determine: The work done by the force $F→=(yi^−xj^) N$ on the particle along the path $AC$ .

Answer: The work done by the force on the particle as it goes from O to C along the path $AC$ is $−25 J$ .

Given Information:

The force acting on the particle is $F→=(yi^−xj^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle is,

$W=∫F→⋅dr→$

Substitute $(yi^−xj^) N$ for $F→$ and $dyj^$ for $dr^$ .

$W=∫((yi^−xj^) N)⋅(dyj) m=−∫(xdy) N⋅m×1 J1 N⋅m=−[∫xdy] J$

In the path $AC$ , the particle moves from $y=0$ to $y=5.00 m$ and there is a displacement of particle in x-direction that is $x=5.00 m$ .

Substitute $5.00$ for $x$ and $0$ for $dx$ .

$WAC=[−∫(5.00)dy] J$

Taking the limits of the integration,

$WAC=[∫05.00(5.00)dy] J=(−5[y]05.00) J=−5[5.00−0]=−25 J$

Section-3:

To determine: The work done by the force $F→=(yi^−xj^) N$ on the particle along the path $AC$ .

Answer: The work done by the force on the particle as it goes from O to C along the path $AC$ is $−25 J$ .

Given Information:

The force acting on the particle is $F→=(yi^−xj^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle along the purple path $OAC$ is,

$Wpurple=WOA+WAC$

$Wpurple$ is the work done along the purple path.
$WOA$ is the work done along the path $OA$ .
$WAC$ is the work done along the path $AC$ .

Substitute $0$ for $WOA$ and $−25 J$ for $WAC$ .

$Wpurple=0 J−25J=−25 J$

Conclusion:

Therefore, the work done by the force on the particle as it goes from O to C along the purple path is $−25 J$ .

Section-4:

To determine: The work done by the force $F→=(yi^−xj^) N$ on the particle along the path $OB$ .

Answer: The work done by the force on the particle as it goes from O to C along the path $OB$ is $25 J$ .

Given Information:

The force acting on the particle is $F→=(yi^−xj^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle is,

$W=∫F→⋅dr→$

Substitute $(yi^−xj^) N$ for $F→$ and $dyj^$ for $dr^$ .

$W=∫((yi^−xj^) N)⋅(dyj) m=−∫(xdy) N⋅m×1 J1 N⋅m=−[∫xdy] J$

Since along the path $OB$ the particle moves from $y=0$ to $y=5.00 m$ and there is no displacement in x-direction that is $x=0$ .

Substitute $0$ for $x$ in above integration.

$WOB=[∫05.000dy] J=0$

Section-5:

To determine: The work done by the force $F→=(yi^−xj^) N$ on the particle along the path $BC$ .

Answer: The work done by the force on the particle as it goes from O to C along the path $BC$ is $25 J$ .

Given Information:

The force acting on the particle is $F→=(yi^−xj^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle is,

$W=∫F→⋅dr→$

Substitute $(yi^−xj^) N$ for $F→$ and $dxi^$ for $dr^$ .

$W=∫((yi^−xj^) N)⋅(dxi^) m=∫(ydx) N⋅m×1 J1 N⋅m=[∫ydx] J$

In the path $BC$ , the particle moves from $x=0$ to $x=5.00 m$ and there is a displacement of particle in y-direction that is $y=5.00 m$ .

Substitute $5.00$ for $y$ and $0$ for $dy$ .

Taking the limits of the integration,

$WBC=[∫05.005dx] J=(5[x]05.00) J=(5[5.00−0]) J=25 J$

Section-6:

To determine: The work done by the force $F→=(yi^−xj^) N$ on the particle along the red path.

Answer: The work done by the force on the particle as it goes from O to C along the red path is $25 J$ .

Given Information:

The force acting on the particle is $F→=(yi^−xj^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle along the red path $OBC$ is,

$Wred=WOB+WBC$ (VI)

$Wred$ is the work done along the red path.
$WOB$ is the work done along the path $OB$ .
$WBC$ is the work done along the path $BC$ .

Substitute $0$ for $WOB$ and $25 J$ for $WBC$ in equation (VI).

$Wred=0 J+25 J=25 J$

Conclusion:

Therefore, the work done by the force on the particle as it goes from O to C along the red path is $25 J$ .

Section-7:

To determine: The work done by the force $F→=(yi^−xj^) N$ on the particle along the blue path.

Answer: The work done by the force on the particle as it goes from O to C along the blue path is $0$ .

Given Information:

The force acting on the particle is $F→=(yi^−xj^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle is,

$W=∫F→⋅dr→$

Substitute $(yi^−xj^) N$ for $F→$ and $dxi^+dyj^$ for $dr^$ .

$W=∫((yi^−xj^) N)⋅(dxi^+dyj^) m=∫(ydx−xdy) N⋅m×1 J1 N⋅m=[∫ydx−xdy] J$

The path $OC$ is a straight line passes through a origin. In this blue path the particle moves in both the direction by $5.00 m$ . Hence equation of line become $y=x$ .

Substitute $x$ for $y$ and $dx$ for $dy$ .

Taking the limits on integration,

$Wblue=[∫xdx−∫xdx] J=0$

Conclusion:

Therefore, the work done by the force on the particle as it goes along the three paths is not same.

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Answer 5

Chapter 6, Problem 44P

(a)

To determine

To Show: That the constant force act on the object is conservative.

(a)

Expert Solution

Answer to Problem 44P

the constant force applied on the object is conservative in nature.

Explanation of Solution

The work done by a conservative force on a particle moving between any two points is independent of the path taken by the particle. Its only depends upon the end points of the path taken by the particle to move.

Given Information:

The general definition for work done by a force is,

$W=∫ifF→⋅dr→$ .

Formula to calculate the work done by the force on the object is,

$W=∫ifF→⋅dr→$

$W$ is the work done by the force on the particle.
$F→$ is the applied force vector on the object.
$dr→$ is the displacement vector of the object.

Since the force is constant that does not vary with respect to time or the position or the velocity of the object. So, the value of force can be taken out from the integration since it is constant quantity.

$W=F→∫ifdr→=F→[r→]if=F→[r→f−r→i]$

Now, here the force is constant so, the work done by this force on the object in only depends upon the end points of the displace object that shows the work done is independent of the path taken by the object to displace between the end points. But the work done is independent of the path only when the force is conservative.

Conclusion:

Therefore, the constant force applied on the object is conservative in nature.

(b)

To determine

The work done by the force $F→=(3i^+4j^) N$ on the particle along each one of the three paths shown in the figure and show that the work done along the three path is identical.

(b)

Expert Solution

Answer to Problem 44P

The work done by the force $F→=(3i^+4j^) N$ on the particle along all the three path are $35 J$ and the work done by the force is identical in all the three paths.

Explanation of Solution

Section-1:

To determine: The work done by the force $F→=(3i^+4j^) N$ on the particle along the path $OA$ .

Answer: The work done by the force on the particle as it goes from O to C along the path $OA$ is $35 J$ .

Given Information:

The force acting on the particle is $F→=(3i^+4j^) N$ and the three different path of the particle moves is shown in figure (I).

Principles of Physics: A Calculus-Based Text, Hybrid (with Enhanced WebAssign Printed Access Card), Chapter 6, Problem 44P

Figure (I)

Formula to calculate the work done by the force on the particle is,

$W=∫F→⋅dr→$

Substitute $(3i^+4j^) N$ for $F→$ and $dxi^$ for $dr^$ .

$W=∫((3i^+4j^) N)⋅(dxi^) m=∫(3dx) N⋅m×1 J1 N⋅m=[∫3dx] J$

Since along the path $OA$ the particle moves from $x=0$ to $x=5.00 m$ and there is no displacement in y-direction that is $y=0$ .

Taking the limits of the integration,

$WOA=[∫05.003dx] J=(3[x]05.00) J=(3[5.00−0]) J=15 J$

Section-2:

To determine: The work done by the force $F→=(3i^+4j^) N$ on the particle along the path $AC$ .

Answer: The work done by the force on the particle as it goes from O to C along the path $AC$ is $35 J$ .

Given Information:

The force acting on the particle is $F→=(3i^+4j^) N$ and the three different path of the particle moves is shown in figure (I).

In the path $AC$ , the particle moves from $y=0$ to $y=5.00 m$ and there is a displacement of particle in x-direction that is $x=5.00 m$ .

Formula to calculate the work done by the force on the particle is,

$W=∫F→⋅dr→$

Substitute $(3i^+4j^) N$ for $F→$ and $dyj^$ for $dr^$ .

$W=∫((3i^+4j^) N)⋅(dyj^) m=∫(4dy) N⋅m×1 J1 N⋅m=[∫4dy] J$

Taking the limits of the integration,

$WAC=[∫05.004dy] J=(4[y]05.00) J=4[5.00−0]=20 J$

Section-3:

To determine: The work done by the force $F→=(3i^+4j^) N$ on the particle along the purple path.

Answer: The work done by the force on the particle as it goes from O to C along the purple path is $35 J$ .

Given Information:

The force acting on the particle is $F→=(3i^+4j^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle along the purple path $OAC$ is,

$Wpurple=WOA+WAC$

$Wpurple$ is the work done along the purple path.
$WOA$ is the work done along the path $OA$ .
$WAC$ is the work done along the path $AC$ .

Substitute $15 J$ for $WOA$ and $20 J$ for $WAC$ .

$Wpurple=15 J+20 J=35 J$

Conclusion:

Therefore, the work done by the force on the particle as it goes from O to C along the purple path is $35 J$ .

Section-4:

To determine: The work done by the force $F→=(3i^+4j^) N$ on the particle along the path $OB$ .

Answer: The work done by the force on the particle as it goes from O to C along the path $OB$ is $35 J$ .

Given Information:

The force acting on the particle is $F→=(3i^+4j^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle is,

$W=∫F→⋅dr→$

Substitute $(3i^+4j^) N$ for $F→$ and $dyj^$ for $dr^$ .

$W=∫((3i^+4j^) N)⋅(dyj^) m=∫(4dy) N⋅m×1 J1 N⋅m=[∫4dy] J$

Since along the path $OB$ the particle moves from $y=0$ to $y=5.00 m$ and there is no displacement in x-direction that is $x=0$ .

Taking the limits of the integration,

$WOB=[∫05.004dy] J=4[y]05.00=(4[5.00−0]) J=20 J$

Section-5:

To determine: The work done by the force $F→=(3i^+4j^) N$ on the particle along the path $BC$ .

Answer: The work done by the force on the particle as it goes from O to C along the path $BC$ is $35 J$ .

Given Information:

The force acting on the particle is $F→=(3i^+4j^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle is,

$W=∫F→⋅dr→$

Substitute $(3i^+4j^) N$ for $F→$ and $dxi^$ for $dr^$ .

$W=∫((3i^+4j^) N)⋅(dxi^) m=∫(3dx) N⋅m×1 J1 N⋅m=[∫3dx] J$

In the path $BC$ , the particle moves from $x=0$ to $x=5.00 m$ and there is a displacement of particle in y-direction that is $y=5.00 m$ .

Taking the limits of the integration,

$WBC=[∫05.003dx+∫004dy] J=(3[x]05.00+0) J=(3[5.00−0]) J=15 J$

Section-6:

To determine: The work done by the force $F→=(3i^+4j^) N$ on the particle along the red path.

Answer: The work done by the force on the particle as it goes from O to C along the red path is $35 J$ .

Given Information:

The force acting on the particle is $F→=(3i^+4j^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle along the red path $OBC$ is,

$Wred=WOB+WBC$

$Wred$ is the work done along the red path.
$WOB$ is the work done along the path $OB$ .
$WBC$ is the work done along the path $BC$ .

Substitute $20 J$ for $WOB$ and $15 J$ for $WBC$ .

$Wred=20+15 J=35 J$

Conclusion:

Therefore, the work done by the force on the particle as it goes from O to C along the red path is $35 J$ .

Section-7:

To determine: The work done by the force $F→=(3i^+4j^) N$ on the particle along the blue path.

Answer: The work done by the force on the particle as it goes from O to C along the blue path is $35 J$ .

Given Information:

The force acting on the particle is $F→=(3i^+4j^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle is,

$W=[∫3dx+∫4dy] J$

The path $OC$ is a straight line passes through a origin. In this blue path the particle moves in both the direction by $5.00 m$ .

Taking the limits on integration,

$Wblue=[∫05.003dx+∫05.004dx] J=(3[x]05.00+4[y]05.00) J=(3[5.00−0]+4[5.00−0]) J=35 J$

Since the work done by the force $F→=(3i^+4j^) N$ in all three purple, red and blue path is $35 J$ hence the force is conservative and the work done is identical in all the three paths.

Conclusion:

Therefore, the work done by the force on the particle as it goes from O to C along the blue path is $35 J$ and the work done by the force is same in all the three paths.

(c)

To determine

Whether the work done by the force $F→=(4xi^+3yj^) N$ on the particle along each one of the three paths shown in the figure is identical.

(c)

Expert Solution

Answer to Problem 44P

The work done by the force $F→=(4xi^+3yj^) N$ on the particle along all the three path are $35 J$ and the work done by the force is identical in all the three paths.

Explanation of Solution

Section-1:

To determine: The work done by the force $F→=(4xi^+3yj^) N$ on the particle along the path $OA$ .

Answer: The work done by the force on the particle as it goes from O to C along the path $OA$ is $35 J$ .

Given Information:

The force acting on the particle is $F→=(4xi^+3yj^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle is,

$W=∫F→⋅dr→$

Substitute $(4xi^+3yj^) N$ for $F→$ and $dxi^$ for $dr^$ .

$W=∫((4xi^+3yj^) N)⋅(dxi^) m=∫(4xdx) N⋅m×1 J1 N⋅m=[∫4xdx] J$

Since along the path $OA$ the particle moves from $x=0$ to $x=5.00 m$ and there is no displacement in y-direction that is $y=0$ .

Taking the limits of the integration,

$WOA=[∫05.004xdx] J=(4[x]05.00) J=(4[5.00−0]) J=20 J$

Section-2:

To determine: The work done by the force $F→=(4xi^+3yj^) N$ on the particle along the path $AC$ .

Answer: The work done by the force on the particle as it goes from O to C along the path $AC$ is $35 J$ .

Given Information:

The force acting on the particle is $F→=(4xi^+3yj^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle is,

$W=∫F→⋅dr→$

Substitute $(4xi^+3yj^) N$ for $F→$ and $dyj^$ for $dr^$ .

$W=∫((4xi^+3yj^) N)⋅(dyj^) m=∫(3ydy) N⋅m×1 J1 N⋅m=[∫3ydy] J$

In the path $AC$ , the particle moves from $y=0$ to $y=5.00 m$ and there is a displacement of particle in x-direction that is $x=5.00 m$ .

Taking the limits of the integration,

$WAC=[∫05.003ydy] J=(3[y]05.00) J=3[5.00−0]=15 J$

Section-3:

To determine: The work done by the force $F→=(4xi^+3yj^) N$ on the particle along the purple path.

Answer: The work done by the force on the particle as it goes from O to C along the purple path is $35 J$ .

Given Information:

The force acting on the particle is $F→=(4xi^+3yj^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle along the purple path $OAC$ is,

$Wpurple=WOA+WAC$ (III)

$Wpurple$ is the work done along the purple path.
$WOA$ is the work done along the path $OA$ .
$WAC$ is the work done along the path $AC$ .

Substitute $20 J$ for $WOA$ and $15 J$ for $WAC$ in equation (II).

$Wpurple=20 J+15 J=35 J$

Conclusion:

Therefore, the work done by the force on the particle as it goes from O to C along the purple path is $35 J$ .

Section-4:

To determine: The work done by the force $F→=(4xi^+3yj^) N$ on the particle along the path $OB$ .

Answer: The work done by the force on the particle as it goes from O to C along the path $OB$ is $35 J$ .

Given Information:

The force acting on the particle is $F→=(4xi^+3yj^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle is,

$W=∫F→⋅dr→$

Substitute $(4xi^+3yj^) N$ for $F→$ and $dyj^$ for $dr^$ .

$W=∫((4xi^+3yj^) N)⋅(dyj^) m=∫(3ydy) N⋅m×1 J1 N⋅m=[∫3ydy] J$

Since along the path $OB$ the particle moves from $y=0$ to $y=5.00 m$ and there is no displacement in x-direction that is $x=0$ .

Taking the limits of the integration,

$WOB=[∫05.003ydy] J=3[y]05.00=(3[5.00−0]) J=15 J$

Section-5:

To determine: The work done by the force $F→=(4xi^+3yj^) N$ on the particle along the path $BC$ .

Answer: The work done by the force on the particle as it goes from O to C along the path $BC$ is $20 J$ .

Given Information:

The force acting on the particle is $F→=(4xi^+3yj^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle is,

$W=∫F→⋅dr→$

Substitute $(4xi^+3yj^) N$ for $F→$ and $dxi^$ for $dr^$ .

$W=∫((4xi^+3yj^) N)⋅(dxi^) m=∫(4xdx) N⋅m×1 J1 N⋅m=[∫4xdx] J$

In the path $BC$ , the particle moves from $x=0$ to $x=5.00 m$ and there is a displacement of particle in y-direction that is $y=5.00 m$ .

Taking the limits of the integration,

$WBC=[∫05.004xdx] J=(4[x]05.00) J=(4[5.00−0]) J=20 J$

Section-6:

To determine: The work done by the force $F→=(4xi^+3yj^) N$ on the particle along the red path.

Answer: The work done by the force on the particle as it goes from O to C along the red path is $35 J$ .

Given Information:

The force acting on the particle is $F→=(4xi^+3yj^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle along the red path $OBC$ is,

$Wred=WOB+WBC$

$Wred$ is the work done along the red path.
$WOB$ is the work done along the path $OB$ .
$WBC$ is the work done along the path $BC$ .

Substitute $15 J$ for $WOB$ and $20 J$ for $WBC$ .

$Wred=15 J+20 J=35 J$

Conclusion:

Therefore, the work done by the force on the particle as it goes from O to C along the red path is $35 J$ .

Section-7:

To determine: The work done by the force $F→=(4xi^+3yj^) N$ on the particle along the blue path.

Answer: The work done by the force on the particle as it goes from O to C along the blue path is $35 J$ .

Given Information:

The force acting on the particle is $F→=(4xi^+3yj^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle is,

$W=[∫4xdx+∫3ydy] J$

The path $OC$ is a straight line passes through a origin. In this blue path the particle moves in both the direction by $5.00 m$ .

Taking the limits on integration,

$Wred=[∫05.004xdx+∫05.003ydx] J=(4[x]05.00+3[y]05.00) J=(4[5.00−0]+3[5.00−0]) J=35 J$

Since the work done by the force $F→=(3i^+4j^) N$ in all three purple, red and blue path is $35 J$ hence the force is conservative and the work done is identical in all the three paths.

Conclusion:

Therefore, the work done by the force on the particle as it goes from O to C along the blue path is $35 J$ and the work done by the force is same in all the three paths.

(d)

To determine

Whether the work done by the force $F→=(yi^−xj^) N$ on the particle along each one of the three paths shown in the figure is identical.

(d)

Expert Solution

Answer to Problem 44P

The work done by the force $F→=(yi^−xj^) N$ on the particle along all the three path are not identical.

Explanation of Solution

Section-1:

To determine: The work done by the force $F→=(yi^−xj^) N$ on the particle along the path $OA$ .

Answer: The work done by the force on the particle as it goes from O to C along the path $OA$ is $−25 J$ .

Given Information:

The force acting on the particle is $F→=(yi^−xj^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle is,

$W=∫F→⋅dr→$

Substitute $(yi^−xj^) N$ for $F→$ and $dxi^$ for $dr^$ .

$W=∫((yi^−xj^) N)⋅(dxi^) m=∫(ydx) N⋅m×1 J1 N⋅m=[∫ydx] J$

Since along the path $OA$ the particle moves from $x=0$ to $x=5.00 m$ and there is no displacement in y-direction that is $y=0$ .

Substitute $0$ for $y$ and $0$ for $dy$ .

$WOA=[∫0dx] J=0 J$

Section-2:

To determine: The work done by the force $F→=(yi^−xj^) N$ on the particle along the path $AC$ .

Answer: The work done by the force on the particle as it goes from O to C along the path $AC$ is $−25 J$ .

Given Information:

The force acting on the particle is $F→=(yi^−xj^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle is,

$W=∫F→⋅dr→$

Substitute $(yi^−xj^) N$ for $F→$ and $dyj^$ for $dr^$ .

$W=∫((yi^−xj^) N)⋅(dyj) m=−∫(xdy) N⋅m×1 J1 N⋅m=−[∫xdy] J$

In the path $AC$ , the particle moves from $y=0$ to $y=5.00 m$ and there is a displacement of particle in x-direction that is $x=5.00 m$ .

Substitute $5.00$ for $x$ and $0$ for $dx$ .

$WAC=[−∫(5.00)dy] J$

Taking the limits of the integration,

$WAC=[∫05.00(5.00)dy] J=(−5[y]05.00) J=−5[5.00−0]=−25 J$

Section-3:

To determine: The work done by the force $F→=(yi^−xj^) N$ on the particle along the path $AC$ .

Answer: The work done by the force on the particle as it goes from O to C along the path $AC$ is $−25 J$ .

Given Information:

The force acting on the particle is $F→=(yi^−xj^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle along the purple path $OAC$ is,

$Wpurple=WOA+WAC$

$Wpurple$ is the work done along the purple path.
$WOA$ is the work done along the path $OA$ .
$WAC$ is the work done along the path $AC$ .

Substitute $0$ for $WOA$ and $−25 J$ for $WAC$ .

$Wpurple=0 J−25J=−25 J$

Conclusion:

Therefore, the work done by the force on the particle as it goes from O to C along the purple path is $−25 J$ .

Section-4:

To determine: The work done by the force $F→=(yi^−xj^) N$ on the particle along the path $OB$ .

Answer: The work done by the force on the particle as it goes from O to C along the path $OB$ is $25 J$ .

Given Information:

The force acting on the particle is $F→=(yi^−xj^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle is,

$W=∫F→⋅dr→$

Substitute $(yi^−xj^) N$ for $F→$ and $dyj^$ for $dr^$ .

$W=∫((yi^−xj^) N)⋅(dyj) m=−∫(xdy) N⋅m×1 J1 N⋅m=−[∫xdy] J$

Since along the path $OB$ the particle moves from $y=0$ to $y=5.00 m$ and there is no displacement in x-direction that is $x=0$ .

Substitute $0$ for $x$ in above integration.

$WOB=[∫05.000dy] J=0$

Section-5:

To determine: The work done by the force $F→=(yi^−xj^) N$ on the particle along the path $BC$ .

Answer: The work done by the force on the particle as it goes from O to C along the path $BC$ is $25 J$ .

Given Information:

The force acting on the particle is $F→=(yi^−xj^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle is,

$W=∫F→⋅dr→$

Substitute $(yi^−xj^) N$ for $F→$ and $dxi^$ for $dr^$ .

$W=∫((yi^−xj^) N)⋅(dxi^) m=∫(ydx) N⋅m×1 J1 N⋅m=[∫ydx] J$

In the path $BC$ , the particle moves from $x=0$ to $x=5.00 m$ and there is a displacement of particle in y-direction that is $y=5.00 m$ .

Substitute $5.00$ for $y$ and $0$ for $dy$ .

Taking the limits of the integration,

$WBC=[∫05.005dx] J=(5[x]05.00) J=(5[5.00−0]) J=25 J$

Section-6:

To determine: The work done by the force $F→=(yi^−xj^) N$ on the particle along the red path.

Answer: The work done by the force on the particle as it goes from O to C along the red path is $25 J$ .

Given Information:

The force acting on the particle is $F→=(yi^−xj^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle along the red path $OBC$ is,

$Wred=WOB+WBC$ (VI)

$Wred$ is the work done along the red path.
$WOB$ is the work done along the path $OB$ .
$WBC$ is the work done along the path $BC$ .

Substitute $0$ for $WOB$ and $25 J$ for $WBC$ in equation (VI).

$Wred=0 J+25 J=25 J$

Conclusion:

Therefore, the work done by the force on the particle as it goes from O to C along the red path is $25 J$ .

Section-7:

To determine: The work done by the force $F→=(yi^−xj^) N$ on the particle along the blue path.

Answer: The work done by the force on the particle as it goes from O to C along the blue path is $0$ .

Given Information:

The force acting on the particle is $F→=(yi^−xj^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle is,

$W=∫F→⋅dr→$

Substitute $(yi^−xj^) N$ for $F→$ and $dxi^+dyj^$ for $dr^$ .

$W=∫((yi^−xj^) N)⋅(dxi^+dyj^) m=∫(ydx−xdy) N⋅m×1 J1 N⋅m=[∫ydx−xdy] J$

The path $OC$ is a straight line passes through a origin. In this blue path the particle moves in both the direction by $5.00 m$ . Hence equation of line become $y=x$ .

Substitute $x$ for $y$ and $dx$ for $dy$ .

Taking the limits on integration,

$Wblue=[∫xdx−∫xdx] J=0$

Conclusion:

Therefore, the work done by the force on the particle as it goes along the three paths is not same.

Answer 6

Chapter 6, Problem 44P

(a)

To determine

To Show: That the constant force act on the object is conservative.

(a)

Expert Solution

Answer to Problem 44P

the constant force applied on the object is conservative in nature.

Explanation of Solution

The work done by a conservative force on a particle moving between any two points is independent of the path taken by the particle. Its only depends upon the end points of the path taken by the particle to move.

Given Information:

The general definition for work done by a force is,

$W=∫ifF→⋅dr→$ .

Formula to calculate the work done by the force on the object is,

$W=∫ifF→⋅dr→$

$W$ is the work done by the force on the particle.
$F→$ is the applied force vector on the object.
$dr→$ is the displacement vector of the object.

Since the force is constant that does not vary with respect to time or the position or the velocity of the object. So, the value of force can be taken out from the integration since it is constant quantity.

$W=F→∫ifdr→=F→[r→]if=F→[r→f−r→i]$

Now, here the force is constant so, the work done by this force on the object in only depends upon the end points of the displace object that shows the work done is independent of the path taken by the object to displace between the end points. But the work done is independent of the path only when the force is conservative.

Conclusion:

Therefore, the constant force applied on the object is conservative in nature.

Answer 7

Chapter 6, Problem 44P

(a)

To determine

To Show: That the constant force act on the object is conservative.

Answer 8

(a)

Expert Solution

Answer to Problem 44P

the constant force applied on the object is conservative in nature.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

The work done by a conservative force on a particle moving between any two points is independent of the path taken by the particle. Its only depends upon the end points of the path taken by the particle to move.

Given Information:

The general definition for work done by a force is,

$W=∫ifF→⋅dr→$ .

Formula to calculate the work done by the force on the object is,

$W=∫ifF→⋅dr→$

$W$ is the work done by the force on the particle.
$F→$ is the applied force vector on the object.
$dr→$ is the displacement vector of the object.

Since the force is constant that does not vary with respect to time or the position or the velocity of the object. So, the value of force can be taken out from the integration since it is constant quantity.

$W=F→∫ifdr→=F→[r→]if=F→[r→f−r→i]$

Now, here the force is constant so, the work done by this force on the object in only depends upon the end points of the displace object that shows the work done is independent of the path taken by the object to displace between the end points. But the work done is independent of the path only when the force is conservative.

Conclusion:

Therefore, the constant force applied on the object is conservative in nature.

Answer 13

(b)

To determine

The work done by the force $F→=(3i^+4j^) N$ on the particle along each one of the three paths shown in the figure and show that the work done along the three path is identical.

(b)

Expert Solution

Answer to Problem 44P

The work done by the force $F→=(3i^+4j^) N$ on the particle along all the three path are $35 J$ and the work done by the force is identical in all the three paths.

Explanation of Solution

Section-1:

To determine: The work done by the force $F→=(3i^+4j^) N$ on the particle along the path $OA$ .

Answer: The work done by the force on the particle as it goes from O to C along the path $OA$ is $35 J$ .

Given Information:

The force acting on the particle is $F→=(3i^+4j^) N$ and the three different path of the particle moves is shown in figure (I).

Principles of Physics: A Calculus-Based Text, Hybrid (with Enhanced WebAssign Printed Access Card), Chapter 6, Problem 44P

Figure (I)

Formula to calculate the work done by the force on the particle is,

$W=∫F→⋅dr→$

Substitute $(3i^+4j^) N$ for $F→$ and $dxi^$ for $dr^$ .

$W=∫((3i^+4j^) N)⋅(dxi^) m=∫(3dx) N⋅m×1 J1 N⋅m=[∫3dx] J$

Since along the path $OA$ the particle moves from $x=0$ to $x=5.00 m$ and there is no displacement in y-direction that is $y=0$ .

Taking the limits of the integration,

$WOA=[∫05.003dx] J=(3[x]05.00) J=(3[5.00−0]) J=15 J$

Section-2:

To determine: The work done by the force $F→=(3i^+4j^) N$ on the particle along the path $AC$ .

Answer: The work done by the force on the particle as it goes from O to C along the path $AC$ is $35 J$ .

Given Information:

The force acting on the particle is $F→=(3i^+4j^) N$ and the three different path of the particle moves is shown in figure (I).

In the path $AC$ , the particle moves from $y=0$ to $y=5.00 m$ and there is a displacement of particle in x-direction that is $x=5.00 m$ .

Formula to calculate the work done by the force on the particle is,

$W=∫F→⋅dr→$

Substitute $(3i^+4j^) N$ for $F→$ and $dyj^$ for $dr^$ .

$W=∫((3i^+4j^) N)⋅(dyj^) m=∫(4dy) N⋅m×1 J1 N⋅m=[∫4dy] J$

Taking the limits of the integration,

$WAC=[∫05.004dy] J=(4[y]05.00) J=4[5.00−0]=20 J$

Section-3:

To determine: The work done by the force $F→=(3i^+4j^) N$ on the particle along the purple path.

Answer: The work done by the force on the particle as it goes from O to C along the purple path is $35 J$ .

Given Information:

The force acting on the particle is $F→=(3i^+4j^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle along the purple path $OAC$ is,

$Wpurple=WOA+WAC$

$Wpurple$ is the work done along the purple path.
$WOA$ is the work done along the path $OA$ .
$WAC$ is the work done along the path $AC$ .

Substitute $15 J$ for $WOA$ and $20 J$ for $WAC$ .

$Wpurple=15 J+20 J=35 J$

Conclusion:

Therefore, the work done by the force on the particle as it goes from O to C along the purple path is $35 J$ .

Section-4:

To determine: The work done by the force $F→=(3i^+4j^) N$ on the particle along the path $OB$ .

Answer: The work done by the force on the particle as it goes from O to C along the path $OB$ is $35 J$ .

Given Information:

The force acting on the particle is $F→=(3i^+4j^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle is,

$W=∫F→⋅dr→$

Substitute $(3i^+4j^) N$ for $F→$ and $dyj^$ for $dr^$ .

$W=∫((3i^+4j^) N)⋅(dyj^) m=∫(4dy) N⋅m×1 J1 N⋅m=[∫4dy] J$

Since along the path $OB$ the particle moves from $y=0$ to $y=5.00 m$ and there is no displacement in x-direction that is $x=0$ .

Taking the limits of the integration,

$WOB=[∫05.004dy] J=4[y]05.00=(4[5.00−0]) J=20 J$

Section-5:

To determine: The work done by the force $F→=(3i^+4j^) N$ on the particle along the path $BC$ .

Answer: The work done by the force on the particle as it goes from O to C along the path $BC$ is $35 J$ .

Given Information:

The force acting on the particle is $F→=(3i^+4j^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle is,

$W=∫F→⋅dr→$

Substitute $(3i^+4j^) N$ for $F→$ and $dxi^$ for $dr^$ .

$W=∫((3i^+4j^) N)⋅(dxi^) m=∫(3dx) N⋅m×1 J1 N⋅m=[∫3dx] J$

In the path $BC$ , the particle moves from $x=0$ to $x=5.00 m$ and there is a displacement of particle in y-direction that is $y=5.00 m$ .

Taking the limits of the integration,

$WBC=[∫05.003dx+∫004dy] J=(3[x]05.00+0) J=(3[5.00−0]) J=15 J$

Section-6:

To determine: The work done by the force $F→=(3i^+4j^) N$ on the particle along the red path.

Answer: The work done by the force on the particle as it goes from O to C along the red path is $35 J$ .

Given Information:

The force acting on the particle is $F→=(3i^+4j^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle along the red path $OBC$ is,

$Wred=WOB+WBC$

$Wred$ is the work done along the red path.
$WOB$ is the work done along the path $OB$ .
$WBC$ is the work done along the path $BC$ .

Substitute $20 J$ for $WOB$ and $15 J$ for $WBC$ .

$Wred=20+15 J=35 J$

Conclusion:

Therefore, the work done by the force on the particle as it goes from O to C along the red path is $35 J$ .

Section-7:

To determine: The work done by the force $F→=(3i^+4j^) N$ on the particle along the blue path.

Answer: The work done by the force on the particle as it goes from O to C along the blue path is $35 J$ .

Given Information:

The force acting on the particle is $F→=(3i^+4j^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle is,

$W=[∫3dx+∫4dy] J$

The path $OC$ is a straight line passes through a origin. In this blue path the particle moves in both the direction by $5.00 m$ .

Taking the limits on integration,

$Wblue=[∫05.003dx+∫05.004dx] J=(3[x]05.00+4[y]05.00) J=(3[5.00−0]+4[5.00−0]) J=35 J$

Since the work done by the force $F→=(3i^+4j^) N$ in all three purple, red and blue path is $35 J$ hence the force is conservative and the work done is identical in all the three paths.

Conclusion:

Therefore, the work done by the force on the particle as it goes from O to C along the blue path is $35 J$ and the work done by the force is same in all the three paths.

Answer 14

(b)

To determine

The work done by the force $F→=(3i^+4j^) N$ on the particle along each one of the three paths shown in the figure and show that the work done along the three path is identical.

Answer 15

(b)

Expert Solution

Answer to Problem 44P

The work done by the force $F→=(3i^+4j^) N$ on the particle along all the three path are $35 J$ and the work done by the force is identical in all the three paths.

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Section-1:

To determine: The work done by the force $F→=(3i^+4j^) N$ on the particle along the path $OA$ .

Answer: The work done by the force on the particle as it goes from O to C along the path $OA$ is $35 J$ .

Given Information:

The force acting on the particle is $F→=(3i^+4j^) N$ and the three different path of the particle moves is shown in figure (I).

Principles of Physics: A Calculus-Based Text, Hybrid (with Enhanced WebAssign Printed Access Card), Chapter 6, Problem 44P

Figure (I)

Formula to calculate the work done by the force on the particle is,

$W=∫F→⋅dr→$

Substitute $(3i^+4j^) N$ for $F→$ and $dxi^$ for $dr^$ .

$W=∫((3i^+4j^) N)⋅(dxi^) m=∫(3dx) N⋅m×1 J1 N⋅m=[∫3dx] J$

Since along the path $OA$ the particle moves from $x=0$ to $x=5.00 m$ and there is no displacement in y-direction that is $y=0$ .

Taking the limits of the integration,

$WOA=[∫05.003dx] J=(3[x]05.00) J=(3[5.00−0]) J=15 J$

Section-2:

To determine: The work done by the force $F→=(3i^+4j^) N$ on the particle along the path $AC$ .

Answer: The work done by the force on the particle as it goes from O to C along the path $AC$ is $35 J$ .

Given Information:

The force acting on the particle is $F→=(3i^+4j^) N$ and the three different path of the particle moves is shown in figure (I).

In the path $AC$ , the particle moves from $y=0$ to $y=5.00 m$ and there is a displacement of particle in x-direction that is $x=5.00 m$ .

Formula to calculate the work done by the force on the particle is,

$W=∫F→⋅dr→$

Substitute $(3i^+4j^) N$ for $F→$ and $dyj^$ for $dr^$ .

$W=∫((3i^+4j^) N)⋅(dyj^) m=∫(4dy) N⋅m×1 J1 N⋅m=[∫4dy] J$

Taking the limits of the integration,

$WAC=[∫05.004dy] J=(4[y]05.00) J=4[5.00−0]=20 J$

Section-3:

To determine: The work done by the force $F→=(3i^+4j^) N$ on the particle along the purple path.

Answer: The work done by the force on the particle as it goes from O to C along the purple path is $35 J$ .

Given Information:

The force acting on the particle is $F→=(3i^+4j^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle along the purple path $OAC$ is,

$Wpurple=WOA+WAC$

$Wpurple$ is the work done along the purple path.
$WOA$ is the work done along the path $OA$ .
$WAC$ is the work done along the path $AC$ .

Substitute $15 J$ for $WOA$ and $20 J$ for $WAC$ .

$Wpurple=15 J+20 J=35 J$

Conclusion:

Therefore, the work done by the force on the particle as it goes from O to C along the purple path is $35 J$ .

Section-4:

To determine: The work done by the force $F→=(3i^+4j^) N$ on the particle along the path $OB$ .

Answer: The work done by the force on the particle as it goes from O to C along the path $OB$ is $35 J$ .

Given Information:

The force acting on the particle is $F→=(3i^+4j^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle is,

$W=∫F→⋅dr→$

Substitute $(3i^+4j^) N$ for $F→$ and $dyj^$ for $dr^$ .

$W=∫((3i^+4j^) N)⋅(dyj^) m=∫(4dy) N⋅m×1 J1 N⋅m=[∫4dy] J$

Since along the path $OB$ the particle moves from $y=0$ to $y=5.00 m$ and there is no displacement in x-direction that is $x=0$ .

Taking the limits of the integration,

$WOB=[∫05.004dy] J=4[y]05.00=(4[5.00−0]) J=20 J$

Section-5:

To determine: The work done by the force $F→=(3i^+4j^) N$ on the particle along the path $BC$ .

Answer: The work done by the force on the particle as it goes from O to C along the path $BC$ is $35 J$ .

Given Information:

The force acting on the particle is $F→=(3i^+4j^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle is,

$W=∫F→⋅dr→$

Substitute $(3i^+4j^) N$ for $F→$ and $dxi^$ for $dr^$ .

$W=∫((3i^+4j^) N)⋅(dxi^) m=∫(3dx) N⋅m×1 J1 N⋅m=[∫3dx] J$

In the path $BC$ , the particle moves from $x=0$ to $x=5.00 m$ and there is a displacement of particle in y-direction that is $y=5.00 m$ .

Taking the limits of the integration,

$WBC=[∫05.003dx+∫004dy] J=(3[x]05.00+0) J=(3[5.00−0]) J=15 J$

Section-6:

To determine: The work done by the force $F→=(3i^+4j^) N$ on the particle along the red path.

Answer: The work done by the force on the particle as it goes from O to C along the red path is $35 J$ .

Given Information:

The force acting on the particle is $F→=(3i^+4j^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle along the red path $OBC$ is,

$Wred=WOB+WBC$

$Wred$ is the work done along the red path.
$WOB$ is the work done along the path $OB$ .
$WBC$ is the work done along the path $BC$ .

Substitute $20 J$ for $WOB$ and $15 J$ for $WBC$ .

$Wred=20+15 J=35 J$

Conclusion:

Therefore, the work done by the force on the particle as it goes from O to C along the red path is $35 J$ .

Section-7:

To determine: The work done by the force $F→=(3i^+4j^) N$ on the particle along the blue path.

Answer: The work done by the force on the particle as it goes from O to C along the blue path is $35 J$ .

Given Information:

The force acting on the particle is $F→=(3i^+4j^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle is,

$W=[∫3dx+∫4dy] J$

The path $OC$ is a straight line passes through a origin. In this blue path the particle moves in both the direction by $5.00 m$ .

Taking the limits on integration,

$Wblue=[∫05.003dx+∫05.004dx] J=(3[x]05.00+4[y]05.00) J=(3[5.00−0]+4[5.00−0]) J=35 J$

Since the work done by the force $F→=(3i^+4j^) N$ in all three purple, red and blue path is $35 J$ hence the force is conservative and the work done is identical in all the three paths.

Conclusion:

Therefore, the work done by the force on the particle as it goes from O to C along the blue path is $35 J$ and the work done by the force is same in all the three paths.

Answer 20

(c)

To determine

Whether the work done by the force $F→=(4xi^+3yj^) N$ on the particle along each one of the three paths shown in the figure is identical.

(c)

Expert Solution

Answer to Problem 44P

The work done by the force $F→=(4xi^+3yj^) N$ on the particle along all the three path are $35 J$ and the work done by the force is identical in all the three paths.

Explanation of Solution

Section-1:

To determine: The work done by the force $F→=(4xi^+3yj^) N$ on the particle along the path $OA$ .

Answer: The work done by the force on the particle as it goes from O to C along the path $OA$ is $35 J$ .

Given Information:

The force acting on the particle is $F→=(4xi^+3yj^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle is,

$W=∫F→⋅dr→$

Substitute $(4xi^+3yj^) N$ for $F→$ and $dxi^$ for $dr^$ .

$W=∫((4xi^+3yj^) N)⋅(dxi^) m=∫(4xdx) N⋅m×1 J1 N⋅m=[∫4xdx] J$

Since along the path $OA$ the particle moves from $x=0$ to $x=5.00 m$ and there is no displacement in y-direction that is $y=0$ .

Taking the limits of the integration,

$WOA=[∫05.004xdx] J=(4[x]05.00) J=(4[5.00−0]) J=20 J$

Section-2:

To determine: The work done by the force $F→=(4xi^+3yj^) N$ on the particle along the path $AC$ .

Answer: The work done by the force on the particle as it goes from O to C along the path $AC$ is $35 J$ .

Given Information:

The force acting on the particle is $F→=(4xi^+3yj^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle is,

$W=∫F→⋅dr→$

Substitute $(4xi^+3yj^) N$ for $F→$ and $dyj^$ for $dr^$ .

$W=∫((4xi^+3yj^) N)⋅(dyj^) m=∫(3ydy) N⋅m×1 J1 N⋅m=[∫3ydy] J$

In the path $AC$ , the particle moves from $y=0$ to $y=5.00 m$ and there is a displacement of particle in x-direction that is $x=5.00 m$ .

Taking the limits of the integration,

$WAC=[∫05.003ydy] J=(3[y]05.00) J=3[5.00−0]=15 J$

Section-3:

To determine: The work done by the force $F→=(4xi^+3yj^) N$ on the particle along the purple path.

Answer: The work done by the force on the particle as it goes from O to C along the purple path is $35 J$ .

Given Information:

The force acting on the particle is $F→=(4xi^+3yj^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle along the purple path $OAC$ is,

$Wpurple=WOA+WAC$ (III)

$Wpurple$ is the work done along the purple path.
$WOA$ is the work done along the path $OA$ .
$WAC$ is the work done along the path $AC$ .

Substitute $20 J$ for $WOA$ and $15 J$ for $WAC$ in equation (II).

$Wpurple=20 J+15 J=35 J$

Conclusion:

Therefore, the work done by the force on the particle as it goes from O to C along the purple path is $35 J$ .

Section-4:

To determine: The work done by the force $F→=(4xi^+3yj^) N$ on the particle along the path $OB$ .

Answer: The work done by the force on the particle as it goes from O to C along the path $OB$ is $35 J$ .

Given Information:

The force acting on the particle is $F→=(4xi^+3yj^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle is,

$W=∫F→⋅dr→$

Substitute $(4xi^+3yj^) N$ for $F→$ and $dyj^$ for $dr^$ .

$W=∫((4xi^+3yj^) N)⋅(dyj^) m=∫(3ydy) N⋅m×1 J1 N⋅m=[∫3ydy] J$

Since along the path $OB$ the particle moves from $y=0$ to $y=5.00 m$ and there is no displacement in x-direction that is $x=0$ .

Taking the limits of the integration,

$WOB=[∫05.003ydy] J=3[y]05.00=(3[5.00−0]) J=15 J$

Section-5:

To determine: The work done by the force $F→=(4xi^+3yj^) N$ on the particle along the path $BC$ .

Answer: The work done by the force on the particle as it goes from O to C along the path $BC$ is $20 J$ .

Given Information:

The force acting on the particle is $F→=(4xi^+3yj^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle is,

$W=∫F→⋅dr→$

Substitute $(4xi^+3yj^) N$ for $F→$ and $dxi^$ for $dr^$ .

$W=∫((4xi^+3yj^) N)⋅(dxi^) m=∫(4xdx) N⋅m×1 J1 N⋅m=[∫4xdx] J$

In the path $BC$ , the particle moves from $x=0$ to $x=5.00 m$ and there is a displacement of particle in y-direction that is $y=5.00 m$ .

Taking the limits of the integration,

$WBC=[∫05.004xdx] J=(4[x]05.00) J=(4[5.00−0]) J=20 J$

Section-6:

To determine: The work done by the force $F→=(4xi^+3yj^) N$ on the particle along the red path.

Answer: The work done by the force on the particle as it goes from O to C along the red path is $35 J$ .

Given Information:

The force acting on the particle is $F→=(4xi^+3yj^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle along the red path $OBC$ is,

$Wred=WOB+WBC$

$Wred$ is the work done along the red path.
$WOB$ is the work done along the path $OB$ .
$WBC$ is the work done along the path $BC$ .

Substitute $15 J$ for $WOB$ and $20 J$ for $WBC$ .

$Wred=15 J+20 J=35 J$

Conclusion:

Therefore, the work done by the force on the particle as it goes from O to C along the red path is $35 J$ .

Section-7:

To determine: The work done by the force $F→=(4xi^+3yj^) N$ on the particle along the blue path.

Answer: The work done by the force on the particle as it goes from O to C along the blue path is $35 J$ .

Given Information:

The force acting on the particle is $F→=(4xi^+3yj^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle is,

$W=[∫4xdx+∫3ydy] J$

The path $OC$ is a straight line passes through a origin. In this blue path the particle moves in both the direction by $5.00 m$ .

Taking the limits on integration,

$Wred=[∫05.004xdx+∫05.003ydx] J=(4[x]05.00+3[y]05.00) J=(4[5.00−0]+3[5.00−0]) J=35 J$

Since the work done by the force $F→=(3i^+4j^) N$ in all three purple, red and blue path is $35 J$ hence the force is conservative and the work done is identical in all the three paths.

Conclusion:

Therefore, the work done by the force on the particle as it goes from O to C along the blue path is $35 J$ and the work done by the force is same in all the three paths.

Answer 21

(c)

To determine

Whether the work done by the force $F→=(4xi^+3yj^) N$ on the particle along each one of the three paths shown in the figure is identical.

Answer 22

(c)

Expert Solution

Answer to Problem 44P

The work done by the force $F→=(4xi^+3yj^) N$ on the particle along all the three path are $35 J$ and the work done by the force is identical in all the three paths.

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Section-1:

To determine: The work done by the force $F→=(4xi^+3yj^) N$ on the particle along the path $OA$ .

Answer: The work done by the force on the particle as it goes from O to C along the path $OA$ is $35 J$ .

Given Information:

The force acting on the particle is $F→=(4xi^+3yj^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle is,

$W=∫F→⋅dr→$

Substitute $(4xi^+3yj^) N$ for $F→$ and $dxi^$ for $dr^$ .

$W=∫((4xi^+3yj^) N)⋅(dxi^) m=∫(4xdx) N⋅m×1 J1 N⋅m=[∫4xdx] J$

Since along the path $OA$ the particle moves from $x=0$ to $x=5.00 m$ and there is no displacement in y-direction that is $y=0$ .

Taking the limits of the integration,

$WOA=[∫05.004xdx] J=(4[x]05.00) J=(4[5.00−0]) J=20 J$

Section-2:

To determine: The work done by the force $F→=(4xi^+3yj^) N$ on the particle along the path $AC$ .

Answer: The work done by the force on the particle as it goes from O to C along the path $AC$ is $35 J$ .

Given Information:

The force acting on the particle is $F→=(4xi^+3yj^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle is,

$W=∫F→⋅dr→$

Substitute $(4xi^+3yj^) N$ for $F→$ and $dyj^$ for $dr^$ .

$W=∫((4xi^+3yj^) N)⋅(dyj^) m=∫(3ydy) N⋅m×1 J1 N⋅m=[∫3ydy] J$

In the path $AC$ , the particle moves from $y=0$ to $y=5.00 m$ and there is a displacement of particle in x-direction that is $x=5.00 m$ .

Taking the limits of the integration,

$WAC=[∫05.003ydy] J=(3[y]05.00) J=3[5.00−0]=15 J$

Section-3:

To determine: The work done by the force $F→=(4xi^+3yj^) N$ on the particle along the purple path.

Answer: The work done by the force on the particle as it goes from O to C along the purple path is $35 J$ .

Given Information:

The force acting on the particle is $F→=(4xi^+3yj^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle along the purple path $OAC$ is,

$Wpurple=WOA+WAC$ (III)

$Wpurple$ is the work done along the purple path.
$WOA$ is the work done along the path $OA$ .
$WAC$ is the work done along the path $AC$ .

Substitute $20 J$ for $WOA$ and $15 J$ for $WAC$ in equation (II).

$Wpurple=20 J+15 J=35 J$

Conclusion:

Therefore, the work done by the force on the particle as it goes from O to C along the purple path is $35 J$ .

Section-4:

To determine: The work done by the force $F→=(4xi^+3yj^) N$ on the particle along the path $OB$ .

Answer: The work done by the force on the particle as it goes from O to C along the path $OB$ is $35 J$ .

Given Information:

The force acting on the particle is $F→=(4xi^+3yj^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle is,

$W=∫F→⋅dr→$

Substitute $(4xi^+3yj^) N$ for $F→$ and $dyj^$ for $dr^$ .

$W=∫((4xi^+3yj^) N)⋅(dyj^) m=∫(3ydy) N⋅m×1 J1 N⋅m=[∫3ydy] J$

Since along the path $OB$ the particle moves from $y=0$ to $y=5.00 m$ and there is no displacement in x-direction that is $x=0$ .

Taking the limits of the integration,

$WOB=[∫05.003ydy] J=3[y]05.00=(3[5.00−0]) J=15 J$

Section-5:

To determine: The work done by the force $F→=(4xi^+3yj^) N$ on the particle along the path $BC$ .

Answer: The work done by the force on the particle as it goes from O to C along the path $BC$ is $20 J$ .

Given Information:

The force acting on the particle is $F→=(4xi^+3yj^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle is,

$W=∫F→⋅dr→$

Substitute $(4xi^+3yj^) N$ for $F→$ and $dxi^$ for $dr^$ .

$W=∫((4xi^+3yj^) N)⋅(dxi^) m=∫(4xdx) N⋅m×1 J1 N⋅m=[∫4xdx] J$

In the path $BC$ , the particle moves from $x=0$ to $x=5.00 m$ and there is a displacement of particle in y-direction that is $y=5.00 m$ .

Taking the limits of the integration,

$WBC=[∫05.004xdx] J=(4[x]05.00) J=(4[5.00−0]) J=20 J$

Section-6:

To determine: The work done by the force $F→=(4xi^+3yj^) N$ on the particle along the red path.

Answer: The work done by the force on the particle as it goes from O to C along the red path is $35 J$ .

Given Information:

The force acting on the particle is $F→=(4xi^+3yj^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle along the red path $OBC$ is,

$Wred=WOB+WBC$

$Wred$ is the work done along the red path.
$WOB$ is the work done along the path $OB$ .
$WBC$ is the work done along the path $BC$ .

Substitute $15 J$ for $WOB$ and $20 J$ for $WBC$ .

$Wred=15 J+20 J=35 J$

Conclusion:

Therefore, the work done by the force on the particle as it goes from O to C along the red path is $35 J$ .

Section-7:

To determine: The work done by the force $F→=(4xi^+3yj^) N$ on the particle along the blue path.

Answer: The work done by the force on the particle as it goes from O to C along the blue path is $35 J$ .

Given Information:

The force acting on the particle is $F→=(4xi^+3yj^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle is,

$W=[∫4xdx+∫3ydy] J$

The path $OC$ is a straight line passes through a origin. In this blue path the particle moves in both the direction by $5.00 m$ .

Taking the limits on integration,

$Wred=[∫05.004xdx+∫05.003ydx] J=(4[x]05.00+3[y]05.00) J=(4[5.00−0]+3[5.00−0]) J=35 J$

Since the work done by the force $F→=(3i^+4j^) N$ in all three purple, red and blue path is $35 J$ hence the force is conservative and the work done is identical in all the three paths.

Conclusion:

Therefore, the work done by the force on the particle as it goes from O to C along the blue path is $35 J$ and the work done by the force is same in all the three paths.

Answer 27

(d)

To determine

Whether the work done by the force $F→=(yi^−xj^) N$ on the particle along each one of the three paths shown in the figure is identical.

(d)

Expert Solution

Answer to Problem 44P

The work done by the force $F→=(yi^−xj^) N$ on the particle along all the three path are not identical.

Explanation of Solution

Section-1:

To determine: The work done by the force $F→=(yi^−xj^) N$ on the particle along the path $OA$ .

Answer: The work done by the force on the particle as it goes from O to C along the path $OA$ is $−25 J$ .

Given Information:

The force acting on the particle is $F→=(yi^−xj^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle is,

$W=∫F→⋅dr→$

Substitute $(yi^−xj^) N$ for $F→$ and $dxi^$ for $dr^$ .

$W=∫((yi^−xj^) N)⋅(dxi^) m=∫(ydx) N⋅m×1 J1 N⋅m=[∫ydx] J$

Since along the path $OA$ the particle moves from $x=0$ to $x=5.00 m$ and there is no displacement in y-direction that is $y=0$ .

Substitute $0$ for $y$ and $0$ for $dy$ .

$WOA=[∫0dx] J=0 J$

Section-2:

To determine: The work done by the force $F→=(yi^−xj^) N$ on the particle along the path $AC$ .

Answer: The work done by the force on the particle as it goes from O to C along the path $AC$ is $−25 J$ .

Given Information:

The force acting on the particle is $F→=(yi^−xj^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle is,

$W=∫F→⋅dr→$

Substitute $(yi^−xj^) N$ for $F→$ and $dyj^$ for $dr^$ .

$W=∫((yi^−xj^) N)⋅(dyj) m=−∫(xdy) N⋅m×1 J1 N⋅m=−[∫xdy] J$

In the path $AC$ , the particle moves from $y=0$ to $y=5.00 m$ and there is a displacement of particle in x-direction that is $x=5.00 m$ .

Substitute $5.00$ for $x$ and $0$ for $dx$ .

$WAC=[−∫(5.00)dy] J$

Taking the limits of the integration,

$WAC=[∫05.00(5.00)dy] J=(−5[y]05.00) J=−5[5.00−0]=−25 J$

Section-3:

To determine: The work done by the force $F→=(yi^−xj^) N$ on the particle along the path $AC$ .

Answer: The work done by the force on the particle as it goes from O to C along the path $AC$ is $−25 J$ .

Given Information:

The force acting on the particle is $F→=(yi^−xj^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle along the purple path $OAC$ is,

$Wpurple=WOA+WAC$

$Wpurple$ is the work done along the purple path.
$WOA$ is the work done along the path $OA$ .
$WAC$ is the work done along the path $AC$ .

Substitute $0$ for $WOA$ and $−25 J$ for $WAC$ .

$Wpurple=0 J−25J=−25 J$

Conclusion:

Therefore, the work done by the force on the particle as it goes from O to C along the purple path is $−25 J$ .

Section-4:

To determine: The work done by the force $F→=(yi^−xj^) N$ on the particle along the path $OB$ .

Answer: The work done by the force on the particle as it goes from O to C along the path $OB$ is $25 J$ .

Given Information:

The force acting on the particle is $F→=(yi^−xj^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle is,

$W=∫F→⋅dr→$

Substitute $(yi^−xj^) N$ for $F→$ and $dyj^$ for $dr^$ .

$W=∫((yi^−xj^) N)⋅(dyj) m=−∫(xdy) N⋅m×1 J1 N⋅m=−[∫xdy] J$

Since along the path $OB$ the particle moves from $y=0$ to $y=5.00 m$ and there is no displacement in x-direction that is $x=0$ .

Substitute $0$ for $x$ in above integration.

$WOB=[∫05.000dy] J=0$

Section-5:

To determine: The work done by the force $F→=(yi^−xj^) N$ on the particle along the path $BC$ .

Answer: The work done by the force on the particle as it goes from O to C along the path $BC$ is $25 J$ .

Given Information:

The force acting on the particle is $F→=(yi^−xj^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle is,

$W=∫F→⋅dr→$

Substitute $(yi^−xj^) N$ for $F→$ and $dxi^$ for $dr^$ .

$W=∫((yi^−xj^) N)⋅(dxi^) m=∫(ydx) N⋅m×1 J1 N⋅m=[∫ydx] J$

In the path $BC$ , the particle moves from $x=0$ to $x=5.00 m$ and there is a displacement of particle in y-direction that is $y=5.00 m$ .

Substitute $5.00$ for $y$ and $0$ for $dy$ .

Taking the limits of the integration,

$WBC=[∫05.005dx] J=(5[x]05.00) J=(5[5.00−0]) J=25 J$

Section-6:

To determine: The work done by the force $F→=(yi^−xj^) N$ on the particle along the red path.

Answer: The work done by the force on the particle as it goes from O to C along the red path is $25 J$ .

Given Information:

The force acting on the particle is $F→=(yi^−xj^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle along the red path $OBC$ is,

$Wred=WOB+WBC$ (VI)

$Wred$ is the work done along the red path.
$WOB$ is the work done along the path $OB$ .
$WBC$ is the work done along the path $BC$ .

Substitute $0$ for $WOB$ and $25 J$ for $WBC$ in equation (VI).

$Wred=0 J+25 J=25 J$

Conclusion:

Therefore, the work done by the force on the particle as it goes from O to C along the red path is $25 J$ .

Section-7:

To determine: The work done by the force $F→=(yi^−xj^) N$ on the particle along the blue path.

Answer: The work done by the force on the particle as it goes from O to C along the blue path is $0$ .

Given Information:

The force acting on the particle is $F→=(yi^−xj^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle is,

$W=∫F→⋅dr→$

Substitute $(yi^−xj^) N$ for $F→$ and $dxi^+dyj^$ for $dr^$ .

$W=∫((yi^−xj^) N)⋅(dxi^+dyj^) m=∫(ydx−xdy) N⋅m×1 J1 N⋅m=[∫ydx−xdy] J$

The path $OC$ is a straight line passes through a origin. In this blue path the particle moves in both the direction by $5.00 m$ . Hence equation of line become $y=x$ .

Substitute $x$ for $y$ and $dx$ for $dy$ .

Taking the limits on integration,

$Wblue=[∫xdx−∫xdx] J=0$

Conclusion:

Therefore, the work done by the force on the particle as it goes along the three paths is not same.

Answer 28

(d)

To determine

Whether the work done by the force $F→=(yi^−xj^) N$ on the particle along each one of the three paths shown in the figure is identical.

Answer 29

(d)

Expert Solution

Answer to Problem 44P

The work done by the force $F→=(yi^−xj^) N$ on the particle along all the three path are not identical.

Answer 30

(d)

Expert Solution

Answer 31

(d)

Expert Solution

Answer 32

Expert Solution

Answer 33

Explanation of Solution

Section-1:

To determine: The work done by the force $F→=(yi^−xj^) N$ on the particle along the path $OA$ .

Answer: The work done by the force on the particle as it goes from O to C along the path $OA$ is $−25 J$ .

Given Information:

The force acting on the particle is $F→=(yi^−xj^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle is,

$W=∫F→⋅dr→$

Substitute $(yi^−xj^) N$ for $F→$ and $dxi^$ for $dr^$ .

$W=∫((yi^−xj^) N)⋅(dxi^) m=∫(ydx) N⋅m×1 J1 N⋅m=[∫ydx] J$

Since along the path $OA$ the particle moves from $x=0$ to $x=5.00 m$ and there is no displacement in y-direction that is $y=0$ .

Substitute $0$ for $y$ and $0$ for $dy$ .

$WOA=[∫0dx] J=0 J$

Section-2:

To determine: The work done by the force $F→=(yi^−xj^) N$ on the particle along the path $AC$ .

Answer: The work done by the force on the particle as it goes from O to C along the path $AC$ is $−25 J$ .

Given Information:

The force acting on the particle is $F→=(yi^−xj^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle is,

$W=∫F→⋅dr→$

Substitute $(yi^−xj^) N$ for $F→$ and $dyj^$ for $dr^$ .

$W=∫((yi^−xj^) N)⋅(dyj) m=−∫(xdy) N⋅m×1 J1 N⋅m=−[∫xdy] J$

In the path $AC$ , the particle moves from $y=0$ to $y=5.00 m$ and there is a displacement of particle in x-direction that is $x=5.00 m$ .

Substitute $5.00$ for $x$ and $0$ for $dx$ .

$WAC=[−∫(5.00)dy] J$

Taking the limits of the integration,

$WAC=[∫05.00(5.00)dy] J=(−5[y]05.00) J=−5[5.00−0]=−25 J$

Section-3:

To determine: The work done by the force $F→=(yi^−xj^) N$ on the particle along the path $AC$ .

Answer: The work done by the force on the particle as it goes from O to C along the path $AC$ is $−25 J$ .

Given Information:

The force acting on the particle is $F→=(yi^−xj^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle along the purple path $OAC$ is,

$Wpurple=WOA+WAC$

$Wpurple$ is the work done along the purple path.
$WOA$ is the work done along the path $OA$ .
$WAC$ is the work done along the path $AC$ .

Substitute $0$ for $WOA$ and $−25 J$ for $WAC$ .

$Wpurple=0 J−25J=−25 J$

Conclusion:

Therefore, the work done by the force on the particle as it goes from O to C along the purple path is $−25 J$ .

Section-4:

To determine: The work done by the force $F→=(yi^−xj^) N$ on the particle along the path $OB$ .

Answer: The work done by the force on the particle as it goes from O to C along the path $OB$ is $25 J$ .

Given Information:

The force acting on the particle is $F→=(yi^−xj^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle is,

$W=∫F→⋅dr→$

Substitute $(yi^−xj^) N$ for $F→$ and $dyj^$ for $dr^$ .

$W=∫((yi^−xj^) N)⋅(dyj) m=−∫(xdy) N⋅m×1 J1 N⋅m=−[∫xdy] J$

Since along the path $OB$ the particle moves from $y=0$ to $y=5.00 m$ and there is no displacement in x-direction that is $x=0$ .

Substitute $0$ for $x$ in above integration.

$WOB=[∫05.000dy] J=0$

Section-5:

To determine: The work done by the force $F→=(yi^−xj^) N$ on the particle along the path $BC$ .

Answer: The work done by the force on the particle as it goes from O to C along the path $BC$ is $25 J$ .

Given Information:

The force acting on the particle is $F→=(yi^−xj^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle is,

$W=∫F→⋅dr→$

Substitute $(yi^−xj^) N$ for $F→$ and $dxi^$ for $dr^$ .

$W=∫((yi^−xj^) N)⋅(dxi^) m=∫(ydx) N⋅m×1 J1 N⋅m=[∫ydx] J$

In the path $BC$ , the particle moves from $x=0$ to $x=5.00 m$ and there is a displacement of particle in y-direction that is $y=5.00 m$ .

Substitute $5.00$ for $y$ and $0$ for $dy$ .

Taking the limits of the integration,

$WBC=[∫05.005dx] J=(5[x]05.00) J=(5[5.00−0]) J=25 J$

Section-6:

To determine: The work done by the force $F→=(yi^−xj^) N$ on the particle along the red path.

Answer: The work done by the force on the particle as it goes from O to C along the red path is $25 J$ .

Given Information:

The force acting on the particle is $F→=(yi^−xj^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle along the red path $OBC$ is,

$Wred=WOB+WBC$ (VI)

$Wred$ is the work done along the red path.
$WOB$ is the work done along the path $OB$ .
$WBC$ is the work done along the path $BC$ .

Substitute $0$ for $WOB$ and $25 J$ for $WBC$ in equation (VI).

$Wred=0 J+25 J=25 J$

Conclusion:

Therefore, the work done by the force on the particle as it goes from O to C along the red path is $25 J$ .

Section-7:

To determine: The work done by the force $F→=(yi^−xj^) N$ on the particle along the blue path.

Answer: The work done by the force on the particle as it goes from O to C along the blue path is $0$ .

Given Information:

The force acting on the particle is $F→=(yi^−xj^) N$ and the three different path of the particle moves is shown in figure (I).

Formula to calculate the work done by the force on the particle is,

$W=∫F→⋅dr→$

Substitute $(yi^−xj^) N$ for $F→$ and $dxi^+dyj^$ for $dr^$ .

$W=∫((yi^−xj^) N)⋅(dxi^+dyj^) m=∫(ydx−xdy) N⋅m×1 J1 N⋅m=[∫ydx−xdy] J$

The path $OC$ is a straight line passes through a origin. In this blue path the particle moves in both the direction by $5.00 m$ . Hence equation of line become $y=x$ .