Question

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

• A. 4 J
• B. 2 J
• C. 8 J
• D. 10 J

Answer 1

Answer: (C)

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