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JEE Main 2026 (6 Apr, Shift 2) · Work-Energy Theorem

Work-Energy Theorem question from JEE Main 2026 (6 Apr, Shift 2)

Q2JEE Main 2026 · 6 APR · SHIFT 2Medium

Work-Energy Theorem

A body of mass 1 kg moves along a straight line with a velocity v=2x2v = 2x^2. The work done by the body during displacement from x=0x = 0 to 5 m is ______ J.

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Answer: (c)

  1. Net work == change in KE.
  2. v(0)=0v(0) = 0, v(5)=2(25)=50 m/sv(5) = 2(25) = 50\text{ m/s}
  3. W=12(1)(502)−0=1250 JW = \frac{1}{2}(1)(50^2) - 0 = 1250\text{ J}
  4. So the answer is (c) 1250.
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