A light inextensible string that goes over a smooth fixed pulley as shown in the figure connects two blocks of masses $0.36 \mathrm{~kg}$ and $0.72 \mathrm{~kg}$. Taking $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2},$ find the work done $(\mathrm{in}$ joules) by the string on the block of mass $0.36 \mathrm{~kg}$ during the first second after the system is released from rest.

• A

  $4 \mathrm{~J}$
• B

  $2 \mathrm{~J}$
• Correct answer
  C

  $8 \mathrm{~J}$
• D

  $10 \mathrm{~J}$

Solution

Given $m=0.36 \mathrm{~kg}, M=0.72 \mathrm{~kg}$. The figure shows the forces on $m$ and $M$. When the system is released, let the acceleration be $a$. Then $T-m g=m a$ $M g-T=M a$ $$ a=\frac{(M-m) g}{M+m}=g / 3 $$ and $T=4 \mathrm{mg} / 3$ For block m: $$ u=0, a=g / 3, t=1, s=? $$
$$ s=u t+\frac{1}{2} a t^{2}=0+\frac{1}{2} \times \frac{g}{3} \times 1^{2}=g / 6 $$ Work done by the string on $m$ is $$ \bar{T} \bar{s}=T s=4 \frac{m g}{3} \times \frac{g}{6}=\frac{4 \times 0.36 \times 10 \times 10}{3 \times 6}=8 \mathrm{~J} $$

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