Question

Two blocks A and B of equal masses $m\text{kg}$ are suspended with the help of ideal pulleys and string arrangement as shown in figure. Then acceleration $\left({\text{in m/s}}^2\right)$ of mass B will be:

• A. $\frac{g}{3}{\text{m/s}}^2$
• B. $\frac{5g}{3}{\text{m/s}}^2$
• C. $\frac{4g}{3}{\text{m/s}}^2$
• D. $\frac{2g}{5}{\text{m/s}}^2$

Answer 1

The correct option is D $\frac{2g}{5}{\text{m/s}}^2$


Assume block A is moving upwards with acceleration $a_A$ and block B is moving downwards with acceleration $a_B$.

Suppose $a_P$ is the acceleration of the pulley.
Then, using string constraint relations, $a_P=a_A$ and
$a_B-a_P-a_P=a_B-2a_P=0\Rightarrow a_B=2a_P$
i.e $a_B=2a_A$

Differentiating w.r.t time twice ,we get relation between accelerations:
$\therefore a_B=2a_A=2a$ (taking $a_A=a$)

Applying Newton's $2^{nd}$ on blocks A and B respectively in direction of acceleration.

$2T-mg=ma$.................(i)
$mg-T=m\times \left(2a\right)=2ma$...(ii)

$\text{From Eq.(i) and (ii)}$:
$a=\frac{g}{5}{\text{m/s}}^2$
$\therefore a_B=2a=\frac{2g}{5}{\text{m/s}}^2$