Mass m1 11.4 kg is on a horizontal surface. Mass m25.10 kg hangs freely on a rop
ID: 1658363 • Letter: M
Question
Mass m1 11.4 kg is on a horizontal surface. Mass m25.10 kg hangs freely on a rope which is attached to the first mass. The coefficient of static friction between m1 and the horizontal surface is Hs = 0.450, while the coefficient of kinetic friction is 0.183 If the system is in motion with m1 moving to the left, then what will be the magnitude of the system's acceleration? Consider the pulley to be massless and frictionless. Submit Answer Tries 0/20 If the system is in motion with m1 moving to the right, then what will be the magnitude of the system's acceleration? Submit Answer Tries 0/20Explanation / Answer
1)
The motivating force is the weight of m2
Wt = 5.10(9.81) = 50.03 N
The kinetic friction force of the table on m1 is
Ffk = 11.4(9.81)(0.183)
Ffk = 20.47 N
with m1 moving to the left, both friction and the motivating force are acting to the right so the net force to the right will
F = Ffk + Wt
F = 20.47 + 50.3
F = 70.5 N
so acceleration of the system will be
a = F/m
a = 70.5 / (11.4 + 5.10)
a = 4.273 m/s^2 to the right
2)
if m1 is moving to the right the friction force acts to the left so
net force to the right is
F = 50.03 - 20.47
F = 29.56 N
so acceleration is
a = F/m
a = 29.56 / (11.4 + 5.10)
a = 1.792 m/s^2 to the right