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JEE Main 2022 (29 Jun, Shift 1) · Work-Energy Theorem

Work-Energy Theorem question from JEE Main 2022 (29 Jun, Shift 1)

Q13JEE Main 2022 · 29 JUN · SHIFT 1Medium

Work-Energy Theorem

A particle of mass 500 gm is moving in a straight line with velocity v = b x5/2. The work done by the net force during its displacement from x = 0 to x = 4 m is: (Take b = 0.25 m−3{}^{-3}/2 s−1{}^{-1}).

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Answer: (d)

  1. Net work == change in KE.
  2. v(0)=0v(0) = 0, v(4)=0.25×45/2=0.25×32=8 m/sv(4) = 0.25 \times 4^{5/2} = 0.25 \times 32 = 8\text{ m/s}
  3. W=12(0.5)(82)=16 JW = \frac{1}{2}(0.5)(8^2) = 16\text{ J}
  4. So the answer is (d) 16 J.
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