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JEE Main 2026 (21 Jan, Shift 2) · Work-Energy Theorem

Work-Energy Theorem question from JEE Main 2026 (21 Jan, Shift 2)

Q3JEE Main 2026 · 21 JAN · SHIFT 2Medium

Work-Energy Theorem

A body of mass 2 kg is moving along x-direction such that its displacement as function of time is given by x(t)=αt2+βt+γx(t) = \alpha t^2 + \beta t + \gamma m, where α=1 m/s2,β=1 m/s\alpha = 1\text{ m/s}^2, \beta = 1\text{ m/s} and γ=1 m\gamma = 1\text{ m}. The work done on the body during the time interval t=2 st = 2\text{ s} to t=3 st = 3\text{ s}, is ________ J.

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Answer: (b)

  1. v=dxdt=2αt+β=2t+1v = \frac{dx}{dt} = 2\alpha t + \beta = 2t + 1
  2. v(2)=5 m/sv(2) = 5\text{ m/s}, v(3)=7 m/sv(3) = 7\text{ m/s}
  3. W=12(2)(72−52)=49−25=24 JW = \frac{1}{2}(2)(7^2 - 5^2) = 49 - 25 = 24\text{ J}
  4. So the answer is (b) 24.
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