Question

Two unequal masses of $1$ kg and $2$ 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 tension on both $1$ kg and $2$ kg blocks in $1$ s. (Take $g=10m/s^2$).

• A. $\frac{200}{9}J,-\frac{200}{9}J$
• B. $-\frac{200}{9}J,+\frac{200}{9}J$
• C. $+\frac{200}{9}J,+\frac{200}{9}J$
• D. $-\frac{200}{9}J,-\frac{200}{9}J$

Answer 1

The correct option is A $\frac{200}{9}J,-\frac{200}{9}J$

Net pulling force on the system is
$F_{net}=2g-1g=20-10=10N$
Total mass being pulled $m=(1+2)=3kg$
Therefore, acceleration of the system will be $a=\frac{F_{net}}{m}=\frac{10}{3}m/s^2$
Displacement of both the blocks in 1 s is
$S=\frac{1}{2}a{t}^2=\frac{1}{2}\left(\frac{10}{3}\right){\left(1\right)}^2=\frac{5}{3}m$
From free body diagram of 2 kg block
Using $\Sigma F=ma,$ we get
$20-T=2a=2\left(\frac{10}{3}\right)$
or $T=20-\frac{20}{3}=\frac{40}{3}N$
$\therefore$ Work done by string (tension) on 1 kg block in 1 s is
$W_1=\left(T\right)\left(S\right)\cos 0^{\circ }$
$=\left(\frac{40}{3}\right)\left(\frac{5}{3}\right)\left(1\right)=\frac{200}{9}J$
Similarly, work done by string on 2 kg block in 1 s will be
$W_2=\left(T\right)\left(S\right)\left(\cos {180}^{\circ }\right)$
$=\left(\frac{40}{3}\right)\left(\frac{5}{3}\right)(-1)=-\frac{200}{9}J$