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WBJEE · Physics · Laws of Motion

A block of mass \(m(=0.1 \mathrm{kg})\) is hanging over a frictionless light fixed pulley by an inextensible string of negligible mass. The other end of the string is pulled by a constant force \(F\) in the vertically downward direction. The linear momentum of the block increases by \(2 \mathrm{kgms}^{-1}\) in \(1 \mathrm{s}\) after the block starts from rest. Then, (given \(g=10 \mathrm{ms}^{-2}\) )

  1. A The tension in the string is \(F\)
  2. B The tension in the string is \(3 \mathrm{N}\)
  3. C The work done by the tension on the block is \(20 \mathrm{J}\) during this 1 s
  4. D The work done against the force of gravity is \(10 \mathrm{J}\)
Verified Solution

Answer & Solution

Correct Answer

(D) The work done against the force of gravity is \(10 \mathrm{J}\)

Step-by-step Solution

Detailed explanation

The free body diagram \(\Rightarrow \quad F=2+m g=3 N\) also, \(a=\frac{\text { unbalanced force }}{\text { mass }}=\frac{2}{0.1}=20 \mathrm{m} / \mathrm{s}^{2}\) \(S=\frac{1}{2} a t^{2}=\frac{1}{2} \times 20 \times 1=10 \mathrm{m}\) Honce, work done by tension…