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JEE Main 2019 (10 Jan, Shift 1) · Work by a Constant Force

Work by a Constant Force question from JEE Main 2019 (10 Jan, Shift 1)

Q17JEE Main 2019 · 10 JAN · SHIFT 1Medium

Work by a Constant Force

A block of mass m is kept on a platform which starts from rest with constant acceleration g/2 upward, as shown in figure. Work done by normal reaction on block in time t is -

Figure

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Answer: (b)

  1. On the block: N−mg=mg2⇒N=3mg2N - mg = m\frac{g}{2} \Rightarrow N = \frac{3mg}{2}
  2. Displacement from rest: s=12⋅g2t2=gt24s = \frac{1}{2}\cdot\frac{g}{2}t^2 = \frac{gt^2}{4} (upward, along NN)
  3. WN=Ns=3mg2⋅gt24=3mg2t28W_N = Ns = \frac{3mg}{2}\cdot\frac{gt^2}{4} = \frac{3mg^2t^2}{8}
  4. So the answer is (b) 3mg2t28\frac{3mg^2t^2}{8}.
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