In the figure below m1 95 kg and m2 35 kg The coefficient

In the figure below, m_1 = 9.5 kg and m_2 = 3.5 kg. The coefficient of static friction between m_1 and the horizontal surface is 0.60, and the coefficient of kinetic friction is 0.30. If the system is released from rest, what will its acceleration be? If the system is set in motion with m_2 moving downward, what will be the acceleration of the system?

Solution

The total mass accelerated = 3.5 + 9.5 = 13kg.
The net force = 3.5*9.8 - 9.5*9.8*0.6 = -21.56N which means the block won\'t slide since the friction force of m1 on the table exceeds the weight of m2
When the block is moving, the net force
= 3.5*9.8 - 9.5*9.8*0.3 =6.37N
a = F/(total mass) = 6.37/13 = 0.49m/s²