NEET Physics Question - Work Energy and Power - Work Done by Constant and Variable Forces

An object of mass m is tied to string of length L and a variable horizontal force is applied on it which starts at zero and gradually increases (it is pulled extremely slowly so that equilibrium exists at all times) until the string makes an angle θ with the vertical. Work done by the gravitational force mg is : Image related to

mgL (1 – cosθ)

- mgL (1 – cosθ)

mgL (1 – sinθ)

-mgL (1 – sin θ)

Solution:

Be definition of Work, Work Done = F.S. cos θ

Taking F and cos θ together like

W= (F. cos θ) . s = Component of Force along displacement × displacement

Here gravitational force mg is acting downwards and displacement in upward direction is (L- L cos θ), So

Work done = mg.(L-Lcosθ)cos (180°)= -mgL(1-cosθ)

Rankers Physics

Solution:

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