Physics for Scientists and Engineers
Physics for Scientists and Engineers
10th Edition
ISBN: 9781337553278
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Textbook Question
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Chapter 7, Problem 32P

(a) Suppose a constant force acts on an object. The force does not vary with time or with the position or the velocity of the object. Start with the general definition for work done by a force

W = i f F d r

and show that the force is conservative. (b) As a special case, suppose the force F = ( 3 i ^ + 4 j ^ ) N acts on a panicle that moves from O to © in Figure P7.31. Calculate the work done by F on the particle as it moves along each one of the three paths shown in the figure and show that the work done along the three paths is identical. (c) What If? Is the work done also identical along the three paths for the force F = ( 4 x i ^ + 3 y j ^ ) , where F is in newtons and x and y are in meters, from Problem 19? (d) What If? Suppose the force is given by F = ( y i ^ + x j ^ ) , where F is in newtons and x and y are in meters. Is the work done identical along the three paths for this force?

(a)

Expert Solution
Check Mark
To determine

That the constant force act on the object is conservative.

Answer to Problem 32P

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.

The general definition for work done by a force

    W=ifFdr.

Write the formula to calculate the work done by the force on the object

    W=ifFdr

Here, W is the work done by the force on the particle, F is the applied force vector on the object and 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=Fifdr=F[r]if=F[rfri]

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)

Expert Solution
Check Mark
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 to Problem 32P

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

Explanation of Solution

The given  force is F=(3i^+4j^)N.

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).

Physics for Scientists and Engineers, Chapter 7, Problem 32P

Figure (I)

Write the formula to calculate the work done by the force on the particle

    W=Fdr

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

    W=((3i^+4j^)N)(dxi^)m=(3dx)Nm×1J1Nm=[3dx]J

Since along the path OA the particle moves from x=0 to x=5.00m 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[5.000])J=15J

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

Write the formula to calculate the work done by the force on the particle

    W=Fdr

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

    W=((3i^+4j^)N)(dyj^)m=(4dy)Nm×1J1Nm=[4dy]J

Taking the limits of the integration,

    WAC=[05.004dy]J=(4[y]05.00)J=4[5.000]=20J

Write the formula to calculate the work done by the force on the particle along the purple path OAC

    Wpurple=WOA+WAC

Here, Wpurple is the work done along the purple path, WOA is the work done along the path OA and WAC is the work done along the path AC.

Substitute 15J for WOA and 20J for WAC.

    Wpurple=15J+20J=35J

Write the formula to calculate the work done by the force on the particle

    W=Fdr

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

    W=((3i^+4j^)N)(dyj^)m=(4dy)Nm×1J1Nm=[4dy]J

Since along the path OB the particle moves from y=0 to y=5.00m 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[5.000])J=20J

Write the formula to calculate the work done by the force on the particle

    W=Fdr

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

    W=((3i^+4j^)N)(dxi^)m=(3dx)Nm×1J1Nm=[3dx]J

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

Taking the limits of the integration,

    WBC=[05.003dx+004dy]J=(3[x]05.00+0)J=(3[5.000])J=15J

Write the formula to calculate the work done by the force on the particle along the red path OBC

    Wred=WOB+WBC

Here, Wred is the work done along the red path, WOB is the work done along the path OB and WBC is the work done along the path BC.

Substitute 20J for WOB and 15J for WBC.

    Wred=20+15J=35J

Write the formula to calculate the work done by the force on the particle

    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 5.00m.

Taking the limits on integration,

    Wblue=[05.003dx+05.004dx]J=(3[x]05.00+4[y]05.00)J=(3[5.000]+4[5.000])J=35J

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

(c)

Expert Solution
Check Mark
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 to Problem 32P

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

Explanation of Solution

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).

Write the formula to calculate the work done by the force on the particle

    W=Fdr

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

    W=((4xi^+3yj^)N)(dxi^)m=(4xdx)Nm×1J1Nm=[4xdx]J

Since along the path OA the particle moves from x=0 to x=5.00m 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[5.000])J=20J

Write the formula to calculate the work done by the force on the particle

    W=Fdr

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

    W=((4xi^+3yj^)N)(dyj^)m=(3ydy)Nm×1J1Nm=[3ydy]J

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

Taking the limits of the integration,

    WAC=[05.003ydy]J=(3[y]05.00)J=3[5.000]=15J

Write the formula to calculate the work done by the force on the particle along the purple path OAC

    Wpurple=WOA+WAC                                 (III)

Here, Wpurple is the work done along the purple path, WOA is the work done along the path OA and WAC is the work done along the path AC.

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

    Wpurple=20J+15J=35J

Write the formula to calculate the work done by the force on the particle

    W=Fdr

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

    W=((4xi^+3yj^)N)(dyj^)m=(3ydy)Nm×1J1Nm=[3ydy]J

Since along the path OB the particle moves from y=0 to y=5.00m 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[5.000])J=15J

Write the formula to calculate the work done by the force on the particle

    W=Fdr

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

    W=((4xi^+3yj^)N)(dxi^)m=(4xdx)Nm×1J1Nm=[4xdx]J

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

Taking the limits of the integration,

    WBC=[05.004xdx]J=(4[x]05.00)J=(4[5.000])J=20J

Write the formula to calculate the work done by the force on the particle along the red path OBC

    Wred=WOB+WBC

Here, Wred is the work done along the red path, WOB is the work done along the path OB and WBC is the work done along the path BC.

Substitute 15J for WOB and 20J for WBC.

    Wred=15J+20J=35J

Write the formula to calculate the work done by the force on the particle

    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 5.00m.

Taking the limits on integration,

    Wred=[05.004xdx+05.003ydx]J=(4[x]05.00+3[y]05.00)J=(4[5.000]+3[5.000])J=35J

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

(d)

Expert Solution
Check Mark
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 to Problem 32P

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

Explanation of Solution

Write the formula to calculate the work done by the force on the particle

    W=Fdr

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

    W=((yi^xj^)N)(dxi^)m=(ydx)Nm×1J1Nm=[ydx]J

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

Substitute 0 for y and 0 for dy.

    WOA=[0dx]J=0J

Write the formula to calculate the work done by the force on the particle

    W=Fdr

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

    W=((yi^xj^)N)(dyj)m=(xdy)Nm×1J1Nm=[xdy]J

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

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[5.000]=25J

Write the formula to calculate the work done by the force on the particle along the purple path OAC

    Wpurple=WOA+WAC

Here, Wpurple is the work done along the purple path, WOA is the work done along the path OA and WAC is the work done along the path AC.

Substitute 0 for WOA and 25J for WAC.

    Wpurple=0J25J=25J

Write the formula to calculate the work done by the force on the particle

    W=Fdr

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

    W=((yi^xj^)N)(dyj)m=(xdy)Nm×1J1Nm=[xdy]J

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

Substitute 0 for x in above integration.

    WOB=[05.000dy]J=0

Write the formula to calculate the work done by the force on the particle

    W=Fdr

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

    W=((yi^xj^)N)(dxi^)m=(ydx)Nm×1J1Nm=[ydx]J

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

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[5.000])J=25J

Write the formula to calculate the work done by the force on the particle along the red path OBC

    Wred=WOB+WBC                                 (VI)

Here, Wred is the work done along the red path, WOB is the work done along the path OB and WBC is the work done along the path BC.

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

    Wred=0J+25J=25J

Write the formula to calculate the work done by the force on the particle

    W=Fdr

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

    W=((yi^xj^)N)(dxi^+dyj^)m=(ydxxdy)Nm×1J1Nm=[ydxxdy]J

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

Substitute x for y and dx for dy.

Taking the limits on integration,

    Wblue=[xdxxdx]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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Chapter 7 Solutions

Physics for Scientists and Engineers

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