Question

Assuming that the surface is frictionless and strings are inextensible, find the acceleration of the block of mass $2m$ shown in the figure.

• A. $\frac{g}{6}$
• B. $\frac{g}{3}$
• C. $\frac{g}{4}$
• D. $\frac{g}{2}$

Answer 1

The correct option is A $\frac{g}{6}$

Let the acceleration of pulley, block of mass $m$ and $2m$ be $a_p$, $a$ and $a_2$ respectively.

So, from the method of intercept we have
$l_1+l_2=\text{constant}$
Differentiating w.r.t time, we get equation in the form of velocity i.e
$\stackrel{.}{l_1}+\stackrel{.}{l_2}=0$
Again differentiating w.r.t time, we get equation in the form of acceleration i.e
$\Rightarrow a_p+a_p-a=0$
$\Rightarrow a=2a_p$ ........... (1)
Also,
$l_3+l_4=\text{constant}$
Double differentiating w.r.t time, we get equation in the form of acceleration i.e
$\stackrel{.}{l_3}+\stackrel{.}{l_4}=0$
$\Rightarrow -a_p+a_2=0$
$\Rightarrow a_2=a_p$...............(2)
From (1) & (2), we can say that
$a=2a_2$

Now, from the FBD of $m$, Pulley and $2m$ we have,

From the FBD of block of mass $m$,
$T-mg\sin \theta =m\left(2a_2\right)$
As $\theta ={30}^{\circ }$
$\Rightarrow T-\frac{mg}{2}=2m{a}_2$..........(3)
From FBD of pulley,
$T^{{}^{\prime }}=2T$
From the FBD of block of mass $2m$,
$2mg-T^{{}^{\prime }}=2m{a}_2\Rightarrow 2mg-2T=2m{a}_2$
$\Rightarrow mg-T=m{a}_2$...........(4)
From (3) & (4), we have
$mg-\frac{mg}{2}=3m{a}_2\Rightarrow a_2=\frac{g}{6}$
Thus, acceleration of the mass $2m$ is $\frac{g}{6}$