Two monkeys with the same mass stand on a bra | vedclass

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Question

(C) According to the Work-Energy Theorem,the total work done on an object is equal to the change in its kinetic energy: $W = \Delta K = K_f - K_i$.Sin

Vedclass · Accepted Answer

$W_A = W_B, v_A = v_B$

Vedclass · Answer

(C) According to the Work-Energy Theorem,the total work done on an object is equal to the change in its kinetic energy: $W = \Delta K = K_f - K_i$.Since both monkeys start from rest at the same height $h$,their initial kinetic energy $K_i = 0$ is the same.Both monkeys hit the floor at the same final height (zero),so their final kinetic energy $K_f$ depends only on the change in potential energy,which is $\Delta U = mgh$ for both.For monkey $B$,the only force doing work is gravity,so $W_B = \Delta K_B = mgh$.For monkey $A$,the forces acting are gravity and the tension in the rope. The tension acts perpendicular to the displacement during the swing,so it does zero work. When monkey $A$ lets go,it is in free fall. Thus,the total work done on monkey $A$ is also $W_A = mgh$.Since $W_A = W_B$,both monkeys have the same change in kinetic energy,and since they start from rest,they have the same final kinetic energy $K_f = mgh$.Therefore,their final speeds $v = \sqrt{2gh}$ are also equal: $v_A = v_B$.

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