Question

A block of mass $m_1$ rests on a horizontal table. A string tied to this block is passed over a frictional pulley fixed at one end of the table and another block of mass $m_2$ is hung to the other end of the string. The acceleration $(a)$ of the system is :

• A. $ \frac{m_{1}m_{2}g}{m_{1}-m_{2}} $
• B. $ \frac{m_{2}g}{m_{1}+m_{2}} $
• C. $ \frac{m_{1}g}{m_{1}+m_{2}} $
• D. $ g $

Answer 1

Answer: (B)

Force acting on block which causes acceleration is $F= ma$.
The free body diagram depicting the situation is shown.

Resultant force acting on hanging block is
$\left(m_{2} g-T\right)$.
Force acting on block $m_{2}$ which causes acceleration is $m_{2} a$
Hence, $m_{2} g-T=m_{2} a$ ... (i)
Also, $T=m_{1} a$ ... (ii)
From Eqs. (i) and (ii), we get
$a=\frac{m_{2} g}{m_{1}+m_{2}}$