Question

A block of mass m1 rests on a horizontal table. A string tied to the block is passed on a frictionless pulley fixed at the end of the table and to the other end of string is hung another block of mass m2. The acceleration of the system is

Answer 1

Answer: The acceleration of the system is $a = g (\mu m_1 - m_2 ) / M$

Given data states that the block $m_1$ is on horizontal table which will experience force of friction $F_1 = \mu m_1 g$ and the second block attached with string to $m_1$ is $m_2$ which will experience Force of gravity $F_2 = m_2 g.$

Therefore, the net force be $F = F_1 - F_2$ in order for the motion to happen with an acceleration a.

                         $=> F = \mu m_1 g - m_2 g \\=> m a = g (\mu m_1 - m_2 )\\ => a = g (\mu m_1 - m_2 ) / M\\ where M = m_1 + m_2.$

Answer 2

Hey !

Mass of block resting on the horizontal table m₁

Mass of the block hanging from the string

              = m₂ = T

Let tension in the string and acceleration of the blocks = a

Resultant force acting on the hanging block

                  = m₂g - T

Force acting on the body which causes acceleration = m₂a

Hence, m₂g - T = m₂a  --------------> (1)

Similarly, resultant force on the block resting on the table = T

Therefore,           T = m₁a    ---------------> (2)

Solving equation (1) and (2), we get

                 a = m₂g/m₁ + m₂

Good luck !