Two unequal masses of $1 \mathrm{kg}$ and $2 \mathrm{kg}$ are attached at the two ends of a light inextensible string passing over a smooth pulley as shown in Figure. If the system is released from rest, find the work done by string on both the blocks in 1 s.

(Take$g=10 \mathrm{m}/{\mathrm{s}}^2$ ).

Net pulling force on the system is         

$F_{\text{nd }}=2g-1g=20-10=10 \mathrm{N}$

Total mass being pulled $m=(1+2)=3 \mathrm{kg}$

Therefore, acceleration of the system will be

a=$\frac{F_{\mathrm{nt}}}{m}=\frac{10}{3} \mathrm{m}/{\mathrm{s}}^2$

Displacement of both the blocks in 1 s is

S=$\frac{1}{2}a{t}^2=\frac{1}{2}\left(\frac{10}{3}\right)(1{)}^2=\frac{5}{3} \mathrm{m}$

Free body diagram of $2 \mathrm{kg}$ block is shown in Fig.

Using (Sigma F=ma), we get 

$20-T=2a=2\left(\frac{10}{3}\right)$

or $T=20-\frac{20}{3}=\frac{40}{3} \mathrm{N}$

$\therefore$ Work done by string (tension) on 1 kg block in 1 s is

$W_1=\left(T\right)\left(S\right)\cos 0°$

=$\left(\frac{40}{3}\right)\left(\frac{5}{3}\right)\left(1\right)=\frac{200}{9} \mathrm{J}$

Similarly, work done by string on $2 \mathrm{kg}$ block in I s will be

$W_2=\left(T\right)\left(S\right)\left(\cos 180°\right)$

=$\left(\frac{40}{3}\right)\left(\frac{5}{3}\right)(-1)=-\frac{200}{9} \mathrm{J}$