In the figure below, m 1 = 9.5 kg and m 2 = 4.0 kg. The coefficient of static fr
ID: 2132855 • Letter: I
Question
In the figure below, m1 = 9.5 kg and m2 = 4.0 kg. The coefficient of static friction between m1 and the horizontal surface is 0.50, and the coefficient of kinetic friction is 0.30.
In the figure below, m1 = 9.5 kg and m2 = 4.0 kg. The coefficient of static friction between m1 and the horizontal surface is 0.50, 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 m2 moving downward, what will be the acceleration of the system?Explanation / Answer
a). when the system is released from rest, the only external forces acting on the whole system will be (m2*g) and frictional force on block m1. so, net force acting on the system = m2*g - 0.5*m1*g = -7.36 N Since the frictional force is greater than the driving force responsible for motion of the system, the system will remain at rest. b). now the frictional force acting on block m1 will be calculated from coefficient of kinetic friction. frictional force acting = 0.3*9.5*9.81 = 27.96 N net force acting on system = m2*g - 27.96 = 11.28 N acceleration = F/(m1 + m2) = 11.28/(9.5+4) = 0.84 m/s^2