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JEE Mains · Physics · STD 11 - Work,energy,power and collision
  • JEE Mains 2022
  • Medium
  • Numerical

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

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Answer & Solution

Correct Answer

(D) \(120\)

Step-by-step Solution

Detailed explanation

By energy conservation

\(PE = KE\)

\(mg \left( H +\frac{ H }{2}\right)=\frac{1}{2} kx ^{2}\left( x =\frac{ H }{2}\right)\)

\(0.100 \times 10 \times \frac{3}{2}(0.10)=\frac{1}{2} k (0.05 \times 0.05)\)

\(k =\frac{3 \times 0.10}{0.05 \times 0.05}\)

\(=\frac{3 \times 1000}{25}=120\, N / m\)

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