A vertical spring with force constant k is fi | vedclass

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Question

(A) According to the work-energy theorem,the net work done on an object is equal to the change in its kinetic energy.In this process,the ball starts f

Vedclass · Accepted Answer

$mg(h + d) - \frac{1}{2}kd^2$

Vedclass · Answer

(A) According to the work-energy theorem,the net work done on an object is equal to the change in its kinetic energy.In this process,the ball starts from rest at height $h$ above the spring and ends at rest after compressing the spring by distance $d$. Thus,the initial kinetic energy $K_i = 0$ and the final kinetic energy $K_f = 0$.Therefore,the net work done $W_{net} = K_f - K_i = 0 - 0 = 0$.Alternatively,considering the forces acting on the ball: the gravitational force does work $mg(h + d)$ and the spring force does work $-\frac{1}{2}kd^2$. The total work done is $W_{net} = mg(h + d) - \frac{1}{2}kd^2 = 0$.

$A$ vertical spring with force constant $k$ is fixed on a table. $A$ ball of mass $m$ at a height $h$ above the free upper end of the spring falls vertically on the spring so that the spring is compressed by a distance $d$. The net work done in the process is

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