Q73 / 100 · BITSAT 2024 (Memory Based Paper 3) · Physics

A force of $F - 0.5 \mathrm{~N}$ is applied on lower block as shown in figure. The work done by lower block on upper block for a displacement of 3 m of the upper block with respect to ground is (Take, $\left.g=10 \mathrm{~m} / \mathrm{s}^2\right)$ Figure

  • A
    -0.5 J
  • Correct answer
    B
    0.5 J
  • C
    2 J
  • D
    -2 J

Solution

Maximum acceleration of 1 kg block may be $a_{\max }=\mu g=1 \mathrm{~m} / \mathrm{s}^2$ Common acceleration without relative motion between two blocks may be, $a=\frac{0.5}{3} \mathrm{~m} / \mathrm{s}^2$ Since, $a \lt a_{\max }$ There will be no relative motion and blocks will move with acceleration $\frac{0.5}{3} \mathrm{~m} / \mathrm{s}^2$. Force of friction by lower block on upper block, $\begin{aligned} f & =m a=(1)\left(\frac{0.5}{3}\right)=\frac{1}{6} \mathrm{~N}(\text { towards right }) \\ \therefore \quad W & =f \times s \\ & =\frac{1}{6} \times 3=0.5 \mathrm{~J} \end{aligned}$

Hence, the correct answer is option: 'B'.

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