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JEE Main 2023 (1 Feb, Shift 2) · Spring & Elastic Energy

Spring & Elastic Energy question from JEE Main 2023 (1 Feb, Shift 2)

Q8JEE Main 2023 · 1 FEB · SHIFT 2Easy

Spring & Elastic Energy

A block is fastened to a horizontal spring. The block is pulled to a distance x=10 cmx = 10\text{ cm} from its equilibrium position (at x=0x = 0) on a frictionless surface from rest. The energy of the block at x=5 cmx = 5\text{ cm} is 0.25 J0.25\text{ J}. The spring constant of the spring is ___________ Nm−1\text{Nm}^{-1}

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

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Answer: 67

  1. Released from rest at x=10x = 10 cm; at x=5x = 5 cm the kinetic energy equals the loss in spring PE.
  2. 12k(0.102−0.052)=0.25\frac{1}{2}k(0.10^2 - 0.05^2) = 0.25
  3. 12k(0.0075)=0.25⇒k=66.7≈67 N m−1\frac{1}{2}k(0.0075) = 0.25 \Rightarrow k = 66.7 \approx 67\ \text{N m}^{-1}
  4. So k≈67 N m−1k \approx 67\ \text{N m}^{-1}.
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