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Question: In the above question find out work done by gravity from A to B and B to C....

In the above question find out work done by gravity from A to B and B to C.

Answer

Work done by gravity from A to B is mgL(1−cos⁡θ)mgL(1 - \cos \theta) and from B to C is −mgL(1−cos⁡θ)-mgL(1 - \cos \theta)

Explanation

Solution

Work done by gravity is calculated as the negative of the change in gravitational potential energy.

Wg=−ΔUg=−(Ufinal−Uinitial)=Uinitial−UfinalW_g = -\Delta U_g = -(U_{final} - U_{initial}) = U_{initial} - U_{final}.

Set the gravitational potential energy at the lowest point B to be zero, UB=0U_B = 0.

The height of points A and C above B is h=L(1−cos⁡θ)h = L(1 - \cos \theta).

The potential energy at A and C is UA=UC=mgh=mgL(1−cos⁡θ)U_A = U_C = mgh = mgL(1 - \cos \theta).

Work done from A to B: WA→B=UA−UB=mgL(1−cos⁡θ)−0=mgL(1−cos⁡θ)W_{A \to B} = U_A - U_B = mgL(1 - \cos \theta) - 0 = mgL(1 - \cos \theta).

Work done from B to C: WB→C=UB−UC=0−mgL(1−cos⁡θ)=−mgL(1−cos⁡θ)W_{B \to C} = U_B - U_C = 0 - mgL(1 - \cos \theta) = -mgL(1 - \cos \theta).