The change in potential for a closed path in a conservative force field is zero.Does this imply that the force on an object going along the path is also zero? Explain with an example.

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ishpiro | College Teacher | (Level 1) Educator

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This does not imply that the force along the path is zero, but it does imply that the WORK of the force along the closed path is zero.

The work along the closed path is zero is because the work of a conservative force along any path depends only on the beginning and the end point of the path, that is, it is only depends on the orginal and final position of the object, but not on the shape of the path.

As an example, consider the work of the gravity, which is a conservative force, on the object of mass m falling from height H to height h. The work done by gravity is the force times distance, H - h:

W = mg(H - h) = mgH - mgh.

If instead the object slid down the slanted ramp of lenght L from height H to height h, the work would be 

`W=mgLcos(theta)`

where `theta`

is the angle between the vertical and the ramp. But, `Lcos(theta) = H-h`

from the right triangle formed by the ramp, the ground and the vertical. Thus the work still only depends on the original and final height:

W = mgH - mgh.

Along closed path, the original and final position are the same, so the work performed must be 0.

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