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JEE Main 2022 · 24 Jun · Shift 1, Physics question 58

A ball of mass 100 g is dropped from a height h=10h=10 cm 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 h2\dfrac h2. The spring constant is ______ Nm−1\mathrm{Nm^{-1}}.

(Use g=10 ms−2g=10\ \mathrm{ms^{-2}})

A ball held a height h above a platform that rests on top of a vertical spring fixed to the ground.
Answer and worked solution

Official answer: 120

Energy. The ball falls hh and then a further h2\frac h2 while compressing the spring by h2\frac h2, coming momentarily to rest:
mg(h+h2)=12k(h2)2.mg\left(h+\frac h2\right)=\frac12k\left(\frac h2\right)^2.

Numbers. 0.1×10×0.15=12k(0.05)20.1\times10\times0.15=\frac12k(0.05)^2, so 0.15=0.00125k0.15=0.00125k and k=120k=120 N m−1^{-1}.

Answer: 120

Topics:Spring potential energyConservation of mechanical energy
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