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MHT CET 2026 (13 Apr, Shift 2) · Energy in SHM

Energy in SHM question from MHT CET 2026 (13 Apr, Shift 2)

Q14MHT CET 2026 · 13 APR · SHIFT 2Medium

Energy in SHM

A block is fastened to a horizontal spring. The block is pulled to a distance x=10x = 10 cm from its equilibrium position (at x=0x = 0) on a frictionless surface from rest. The kinetic energy of the block at x=5x = 5 cm is 0.250.25 J. The spring constant of the spring is nearly (in Nm−1-1)

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Answer: (c)

  1. Released from rest at 10 cm, so A=0.1A = 0.1 m.
  2. KE=12k(A2−x2)=12k(0.01−0.0025)=0.00375kKE = \frac{1}{2}k(A^2 - x^2) = \frac{1}{2}k(0.01 - 0.0025) = 0.00375k
  3. 0.00375k=0.25⇒k=66.7≈670.00375k = 0.25 \Rightarrow k = 66.7 \approx 67 N/m
  4. So the answer is (c) 67.
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