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 the system is equal to the change in kinetic energy of the ball.Let the initial position

Vedclass · Accepted Answer

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

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(A) According to the work-energy theorem,the net work done on the system is equal to the change in kinetic energy of the ball.Let the initial position of the ball be at height $h$ above the spring. The final position is when the spring is compressed by distance $d$.The total vertical displacement of the ball is $(h + d)$.The forces acting on the ball are gravity (downward) and the spring force (upward).The work done by gravity is $W_g = mg(h + d)$.The work done by the spring force is $W_s = -\int_0^d kx \, dx = -\frac{1}{2}kd^2$.The net work done on the ball is $W_{net} = W_g + W_s = mg(h + d) - \frac{1}{2}kd^2$.Since the ball starts from rest and comes to rest momentarily at the maximum compression $d$,the change in kinetic energy is zero,which implies $W_{net} = 0$ for the entire process. However,the question asks for the net work done by the external forces (gravity and spring) during the displacement $d$,which is $mg(h + d) - \frac{1}{2}kd^2$.

$A$ 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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