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JEE Main 2017 · Work-Energy Theorem

Work-Energy Theorem question from JEE Main 2017

Q22JEE Main 2017Medium

Work-Energy Theorem

A time dependent force F = 6t acts on a particle of mass 1 kg. If the particle starts from rest, the work done by the force during the first 1 sec. will be:

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Answer: (b)

  1. a=Fm=6ta = \frac{F}{m} = 6t, so v=∫0t6t dt=3t2v = \int_0^t 6t\,dt = 3t^2.
  2. v(1)=3 m/sv(1) = 3\text{ m/s}
  3. W=ΔK=12(1)(32)=4.5 JW = \Delta K = \frac{1}{2}(1)(3^2) = 4.5\text{ J}
  4. So the answer is (b) 4.5 J.
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