Question

In the shown pulley-block system, strings are light. Pulleys are massless and smooth, and the system is released from rest. In <!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> $0.6\text{s}$, what is the work done by gravitation on <!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> $20\text{kg}$ and <!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> $40\text{kg}$ block, respectively ?
Take [$g=10\text{m}/{\text{s}}^2$]

• A. <!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> $-120\text{J}$ , $480\text{J}$
• B. $120\text{J}$ , $240\text{J}$
• C. $-240\text{J}$ , $120\text{J}$
• D. $-240\text{J}$ , $480\text{J}$

Answer 1

The correct option is A <!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> $-120\text{J}$ , $480\text{J}$

Using constraints relation, if acceleration of $20\text{kg}$ kg is $a\&$ upwards, then acceleration of $40\text{kg}$ is $2a\&$downwards,

FBD for both bodies as shown,


Writing equation of motion for the two blocks respectively,

$2T-20g=20a..........\left(1\right)$

$40g-T=80a..........\left(2\right)$

From eq (1) and (2) we get,

$60g=180a$

$\therefore a=\frac{g}{3}$

Displacement of $40\text{kg}$ block in $0.6\text{s}$ is,

$S=ut+\frac{1}{2}{t}^2=\frac{1}{2}\left(2a\right)t^2=\frac{g}{3}{\left(0.6\right)}^2=1.2\text{m}$

$\therefore$ Displacement of $20\text{kg}$ will be,

$S=\frac{1.2}{2}=0.6\text{m}$

$\therefore$ Work done on $40\text{kg}$ block by gravity is,

$W=mgh=40\times 10\times 1.2=480\text{J}$

$\therefore$ Work done on $20\text{kg}$ block by gravity is,

$W=mgh=20\times 10\times 0.6=-120\text{J}$

The negative sign shows that, $20\text{kg}$ block is moving against gravity.

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, option $\left(a\right)$ is correct.

Key concept: When force is constant the work done is $W$=$\overrightarrow{F.}$$\overrightarrow{S}$