In the figure below, m1 = 10.0 kg and m2 = 4.5 kg. The coefficient of static fri
ID: 2049016 • Letter: I
Question
In the figure below, m1 = 10.0 kg and m2 = 4.5 kg. The coefficient of static friction between m1 and the horizontal surface is 0.60 while the coefficient of kinetic friction is 0.30.
the picture: http://www.webassign.net/sercp/p4-30.gif
(a) If the system is released from rest, what will its acceleration be? m/s^2
(b) If the system is set in motion with m2 moving downward, what will be the acceleration of the system? m/s^2
Explanation / Answer
(a) Check first if the system would even move. To check, get the static friction. static friction = µmg = (0.60)(10kg)(9.8m/s2) = 58.8N Then compare it with the gravitational force on m2, which is the force that would move the system. F = m2g = 4.5kg(9.8m/s2) = 44.1N Since 58.8N > 44.1N, the system won't move. The gravitational force is not enough to overcome the static friction of mass 1. acceleration = 0 m/s2 (b) When it is moving downward already, the kinetic friction takes action, not static friction. net force on system = gravitational force on m2 - kinetic friction on m1 (m1 + m2)a = m2 g - µm1 g (10 + 4.5)a = (4.5)(9.8) - (0.30)(10)(9.8) a = 1.01m/s2 Share: by email on facebook on twitter