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Chapter 6, Problem 43P

(a)

To determine

The amount of work done by the force along the purple path.

(a)

Expert Solution
Check Mark

Answer to Problem 43P

The work done by the force along the purple path is 125J .

Explanation of Solution

Given info: The coordinates of point (C) is (5.00m, 5.00m) and the force acting on the body is F=2yi^+x2j^ .

Along path OA the work is done is only in x direction so, work done along OA is,

WOA=05.00Fdxi^

Here,

F is the force acting on the particle.

Substitute 2yi^+x2j^ for F in the above equation.

WOA=05.00(2yi^+x2j^)dxi^=05.00(2y)dx

The value y along path OA is 0 .

Substitute 0 for y in the above equation.

WOA=05.00(2(0))dx=0J

Thus, work done along OA is 0J .

Work done along AC is,

WAC=05.00Fdyj^

Substitute 2yi^+x2j^ for F in the above equation.

WAC=05.00(2yi^+x2j^)dyj^=05.00(x2)dy=x2[y]05.00=5x2J

The value of x along the path AC is 5.00 .

Substitute 5.00 for x in the above equation.

WAC=5(5.002)J=125J

Thus, the work done along AC is 125J .

The work done along purple path is,

Wpurple=WOA+WAC

Substitute 0J for WOA and 125J for WAC in the above equation.

Wpurple=0J+125J=125J

Conclusion:

Therefore, the work done in the purple path is 125J .

(b)

To determine

The amount of work done by the force along the red path.

(b)

Expert Solution
Check Mark

Answer to Problem 43P

The work done by the force along the red path is 50.0J .

Explanation of Solution

Given info: The coordinates of point (C) is (5.00m, 5.00m) and the force acting on the body is F=2yi^+x2j^ .

Work done along OB is,

WOB=05.00Fdyi^

Substitute 2yi^+x2j^ for F in the above equation.

WOB=05.00(2yi^+x2j^)dyj^=05.00(x2)dy

The value of x along path (OB) is zero.

Substitute 0 for x in the above equation.

WOB=05.00(0)dy=0J

Thus, the work done along the path OB is 0J .

Work done along BC is,

WBC=05.00Fdxi^

Substitute 2yi^+x2j^ for F in the above equation.

WBC=05.00(2yi^+x2j^)dxi^=05.00(2y)dx=2y[x]05.00=10yJ

Along path BC the value of y is 5.00 .

Substitute 5.00 for y in the above equation.

WBC=10(5.00)J=50.0J

Thus, the work done along the path OB is 50.0J .

The work done along red path is,

Wred=WOB+WBC

Substitute 50.0J for WBC and 0J for WOB in the above equation.

Wred=0J+50.0J=50.0J

Conclusion:

Therefore, the work done by the force along the red path is 50.0J .

(c)

To determine

The amount of work done by the force along the blue path.

(c)

Expert Solution
Check Mark

Answer to Problem 43P

The work done by the force along the blue path is 66.7J .

Explanation of Solution

Given info: The coordinates of point (C) is (5.00m, 5.00m) and the force acting on the body is F=2yi^+x2j^ .

The work done along the blue path is,

WOC=05.00F(dxi^+dyi^)

Substitute 2yi^+x2j^ for F in the above equation

WOC=05.002yi^+x2j^(dxi^+dyi^)

The value of x and y along path OC is same, so substitute x for y in the above equation.

WOC=05.002xi^+x2j^(dx)=[x2+x33]05.00=66.66J66.7J

Conclusion:

Therefore, the work done by the force along the blue path is 66.7J .

(d)

To determine

Whether F is conservative.

(d)

Expert Solution
Check Mark

Answer to Problem 43P

The force F is not a conservative force.

Explanation of Solution

Given info: The coordinates of point (C) is (5.00m, 5.00m) and the force acting on the body is F=2yi^+x2j^ .

The work done by the force is non conservative in nature as a force is a conservative force if the work done is independent of the path the particle takes.

For the given case the work done in every case is different as the path changes for each of the cases. From the above calculation the work done by the force F is different for each path taken. So, the force F is non-conservative in nature.

Conclusion:

Therefore, the work done by force along every path is different hence it is a non conservative force in nature.

(e)

To determine

The reason for the force F is not a conservative force.

(e)

Expert Solution
Check Mark

Answer to Problem 43P

The work done along each path is different.

Explanation of Solution

Given info: The coordinates of point (C) is (5.00m, 5.00m) and the force acting on the body is F=2yi^+x2j^ .

The work done by the force F along the purple path is 125J , the work done along the red path is 50.0J and the work done along the blue path is 66.7J .

The work done by the given force F is different for each path the particle traversed. The work done by a conservative force is independent of the path taken. So the force is non conservative in nature.

Conclusion:

Therefore, the force is non conservative in nature because work done along each path is different.

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Chapter 6 Solutions

Bundle: Principles of Physics: A Calculus-Based Text, 5th + WebAssign Printed Access Card for Serway/Jewett's Principles of Physics: A Calculus-Based Text, 5th Edition, Multi-Term

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