Question

The pulley arrangements shown in the figure are identical, the mass of the rope being negligible. In case $(a)$ mass $m$ is lifted by attaching a mass of $2m$ to the other end of the rope. In case $(b)$ the mass $m$ is lifted by pulling the other end of the rope with a constant downward force $F=2mg$ , where $g$ is the acceleration due to gravity. The acceleration of mass $m$ in case $(a)$ is

• A. zero
• B. more than that in case $(b)$
• C. Less than that in case $(b)$
• D. Equal to that in case $(b)$

Answer 1

Answer: (C)

Let's consider, the acceleration of the blocks in case $\left(a\right)$ and in case $\left(b\right)$ is $a_1$ and $a_2$ respectively.
In case $\left(a\right)$ ,
Let's assume the Tension force in the string as $T_1$ as shown.

From the free body diagram,
$2mg-T_1=2m{a}_1$ and
$T_1-mg=m{a}_1$
From the above two equations,
$2mg-\left(mg+m{a}_1\right)=2m{a}_1$ ,
$\Rightarrow mg=3m{a}_1$
$\Rightarrow a_1=\frac{g}{3}$ .
In case $\left(b\right)$ ,
The Tension in the string, $T_2=F=2mg$

Thus, from the free body diagram,
$T_2-mg=m{a}_2$
$\Rightarrow 2mg-mg=m{a}_2$
$\Rightarrow a_2=g$ .
Therefore, the acceleration in case $\left(a\right)$ is less than that in case $\left(b\right)$ .