Mass m1 = 12.8 kg is on a horizontal surface. Mass m2 = 5.21 kg hangs freely on
ID: 1412103 • Letter: M
Question
Mass m1 = 12.8 kg is on a horizontal surface. Mass m2 = 5.21 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 s = 0.501, while the coefficient of kinetic friction is k = 0.196.
1) 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.
2) If the system is in motion with m1 moving to the right, then what will be the magnitude of the system's acceleration? If you could show steps as well thatd be great!
Explanation / Answer
1)
The motivating force is the weight of m2 or
Wt = 5.210(9.81) = 51 N
The kinetic friction force of the table on m1 is
Ffk = 12.8(9.81)(0.196)
Ffk = 24.6 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 be
F = Ffk + Wt
F = 24.6 + 51
F = 75.6 N
so acceleration of the system will be
a = F/m
a = 75.6 / (12.8 + 5.210)
a = 4.196 m/s² 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 = 75.6 - 24.6
F = 51N
so acceleration is
a = F/m
a = 51 / (12.8 + 5.21)
a = 2.67 m/s² to the right