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(C) According to the Law of Conservation of Mechanical Energy $(COME)$,the total mechanical energy of the system remains constant.Let the initial posi

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$\frac{2mg(h + x)}{x^2}$

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(C) According to the Law of Conservation of Mechanical Energy $(COME)$,the total mechanical energy of the system remains constant.Let the initial position of the ball be the reference level for gravitational potential energy $(PE = 0)$.Initial state: The ball is at height $h$ above the platform. The spring is at its natural length. Initial kinetic energy $(KE_i)$ is $0$,and initial potential energy $(PE_i)$ is $0$ (if we measure from the release point).Final state: The ball and the platform have moved down by a distance $x$. The ball is now at a vertical position $- (h + x)$ relative to the release point. The spring is compressed by $x$. Final kinetic energy $(KE_f)$ is $0$ (at the point of maximum compression).Applying the principle of conservation of energy:$PE_i + KE_i = PE_f + KE_f$$0 + 0 = -mg(h + x) + \frac{1}{2}kx^2$Rearranging the equation to solve for the spring constant $k$:$mg(h + x) = \frac{1}{2}kx^2$$k = \frac{2mg(h + x)}{x^2}$

$A$ ball of mass $m$ is dropped from a height $h$ on a platform fixed at the top of a vertical spring,as shown in the figure. The platform is depressed by a distance $x$. Then the spring constant is

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