Question

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

Answer 1

$mg-T=m_{2}a$

$T-mg=m_{1}a$

$(m_2-m_1)g=(m_1+m)a$

$a=\frac{\left(0.36\right)g}{0.36+0.72}=\frac{g}{3}m/s^2$

$T=m_1(a+g)=\frac{4m_1}{3}g=0.48g=4.8N$

work done by T on $m_1=T\left(\frac{1}{2}a{t}^2\right)$ at $t=1$sec

$=4.8\times \frac{1}{2}\times \frac{g}{3}\times 1$

$=1.6\times 5=8J$