Question 91 of 183

A ball of mass $100g$ is dropped from a height $h=10cm$ on a platform fixed at the top of a vertical spring (as shown in figure). The ball stays on the platform and the platform is depressed by a distance $\dfrac{h}{2}$. The spring constant is _____ $N{m}^{-1}$
(Use $g=10{ms}^{-2}$)

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Answer: 120

Loss in the gravitational potential energy of the ball= gain in the potential energy by the spring
$-∆P{E}_{gravitational}=∆P{E}_{spring}$
$mg\left(h+\dfrac{h}{2}\right)=\dfrac{1}{2}k{\left(\dfrac{h}{2}\right)}^{2}$$\Rightarrow$$mg\dfrac{3h}{2}=\dfrac{k{h}^{2}}{8}$
$\Rightarrow k=\dfrac{12mg}{h}=\dfrac{12\times 0.1\times 10}{0.1}=120N{m}^{-1}$