A spring of force constant k is fixed vertica | vedclass

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(B) According to the work-energy theorem,the total work done by all forces (gravity and spring force) on the ball is equal to the change in its kineti

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) According to the work-energy theorem,the total work done by all forces (gravity and spring force) on the ball is equal to the change in its kinetic energy.The ball starts from rest at height $h$ above the spring and comes to rest momentarily after compressing the spring by distance $d$. Thus,the initial and final kinetic energy of the ball is $0$.Change in kinetic energy $\Delta K = K_f - K_i = 0 - 0 = 0$.Work done by gravity $W_g = mg(h + d)$.Work done by the spring force $W_s = -\frac{1}{2}kd^2$.According to the work-energy theorem,$W_{total} = W_g + W_s = \Delta K$.$mg(h + d) - \frac{1}{2}kd^2 = 0$.

$A$ spring of force constant $k$ is fixed vertically on a table. $A$ ball of mass $m$ is dropped from a height $h$ onto the free end of the spring,causing the spring to compress by a distance $d$. What is the total work done in this process?

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