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JEE Main 2022 (24 Jun, Shift 1) · Spring & Elastic Energy

Spring & Elastic Energy question from JEE Main 2022 (24 Jun, Shift 1)

Q14JEE Main 2022 · 24 JUN · SHIFT 1Medium

Spring & Elastic Energy

A ball of mass 100 g is dropped from a height h = 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\frac{h}{2}. The spring constant is _____________ Nm−1{}^{-1}.

(Use g = 10 ms−2{}^{-2})

Figure

Numerical answer type. Work it out, then open the answer.

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Answer: 120

  1. The ball falls h+h2=0.15h + \frac{h}{2} = 0.15 m in all; this PE goes into the spring.
  2. mg(0.15)=12k(0.05)2mg(0.15) = \frac{1}{2}k\left(0.05\right)^2
  3. 0.1×10×0.15=0.15=0.00125k0.1 \times 10 \times 0.15 = 0.15 = 0.00125k
  4. So k=120 N m−1k = 120\ \text{N m}^{-1}.
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