Question

A body of mass m was slowly pulled up the hill by a force F which at each point was directed along the tangent of the trajectory. All surfaces are smooth. Find work done by this force

Answer 1

Work done by force "F" is mgh.

Explanation:

Applying work energy theorem

Change in kinetic energy = Work done by all forces

=  W(mg) +  W(ext)  

• Work done by Force "F" is W(ext).
• Change in Kinetic energy = 0

So W(ext) - mgh = 0

W(ext) = mgh

Thus work done by force "F" is mgh.

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Two forces F1 and F2 of magnitude 5N each inclined to each other at 60 degree,act on the body . Find the resultant force acting on the body ?

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Answer 2

Explanation:

applying work energy theorem change in kinetic

energy= work done by all forces

=Wmg+Wext

work by f is Wext

change in kinetic energy =0

so Wext - mgh =0

Wext = mgh

so work by force F is mgh.