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JEE Main 2018 (16 Apr, Shift 1) · Work-Energy Theorem

Work-Energy Theorem question from JEE Main 2018 (16 Apr, Shift 1)

Q20JEE Main 2018 · 16 APR · SHIFT 1Medium

Work-Energy Theorem

A body of mass m starts moving from rest along x-axis so that its velocity varies as v=asv = a\sqrt{s} where a is a constant and s is the distance covered by the body. The total work done by all the forces acting on the body in the first t seconds after the start of the motion is:

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Answer: (a)

  1. v2=a2sv^2 = a^2 s; compare with v2=2acsv^2 = 2a_c s: constant acceleration ac=a22a_c = \frac{a^2}{2}.
  2. From rest, v=act=a2t2v = a_c t = \frac{a^2 t}{2}
  3. W=ΔK=12m(a2t2)2=18ma4t2W = \Delta K = \frac{1}{2}m\left(\frac{a^2 t}{2}\right)^2 = \frac{1}{8}ma^4t^2
  4. So the answer is (a) 18ma4t2\frac{1}{8}ma^4t^2.
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