Question

In the figure shown, the pulleys and strings are massless. The acceleration of the block of mass 4m just after the system is released from rest is $(\theta =si{n}^{-1}\frac{3}{5})$

• A. $\frac{2g}{5}$ downwards
• B. $\frac{2g}{5}$ upwards
• C. $\frac{5g}{11}$ upwards
• D. $\frac{5g}{11}$ downwards.

Answer 1

The correct option is D $\frac{5g}{11}$ downwards.

The FBD of blocks is as shown


From Newton's second law
$4mg-2Tcos\theta =4mA....\left(1\right)$
and T - mg = ma.....(2)
$cos\theta =\frac{4}{5}$ and from constraint we get $a=Acos\theta ......\left(3\right)$
Solving equation (1), (2) and (3)
we get acceleration of block of mass 4m , $A=\frac{5g}{11}$ downwards.