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JEE Main 2022 (28 Jun, Shift 2) · Work by a Variable Force

Work by a Variable Force question from JEE Main 2022 (28 Jun, Shift 2)

Q11JEE Main 2022 · 28 JUN · SHIFT 2Medium

Work by a Variable Force

A block of mass 2 kg moving on a horizontal surface with speed of 4 ms−1{}^{-1} enters a rough surface ranging from x = 0.5 m to x = 1.5 m. The retarding force in this range of rough surface is related to distance by F = −- kx where k = 12 Nm−1{}^{-1}. The speed of the block as it just crosses the rough surface will be:

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Answer: (c)

  1. Work by the retarding force: W=−∫0.51.512x dx=−6(1.52−0.52)=−12W = -\int_{0.5}^{1.5} 12x\,dx = -6(1.5^2 - 0.5^2) = -12 J
  2. Initial KE =12(2)(4)2=16= \frac{1}{2}(2)(4)^2 = 16 J, so final KE =16−12=4= 16 - 12 = 4 J
  3. 12(2)v2=4⇒v=2\frac{1}{2}(2)v^2 = 4 \Rightarrow v = 2 m/s
  4. So the answer is (c) 2.0 ms−1^{-1}.
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