thanks for your help! Mass m_1 = 11.3 kg is on a horizontal surface. Mass m_2 =

Question

thanks for your help!

Mass m_1 = 11.3 kg is on a horizontal surface. Mass m_2 = 8.25 kg hangs freely on a rope which is attached to the first mass. The coefficient of static friction between m_1 and the horizontal surface is mu_s = 0.624, while the coefficient of kinetic friction is mu_k = 0.141. If the system is in motion with m_1 moving to the left, then what will be the magnitude of the system's acceleration? Consider the pulley to be massless and frictionless.______If the system is in motion with m_1 moving to the right, then what will be the magnitude of the system's acceleration?_____

Explanation / Answer

1)

The motivating force is the weight of m2 or
Wt = 8.25(9.81) = 80.93 N

The kinetic friction force of the table on m1 is

Ffk = 11.3(9.81)(0.141)
Ffk = 15.63 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 = 15.63 + 80.73
F = 96.36 N

so acceleration of the system will be
a = F/m
a = 96.36 / (11.3 + 8.25)
a = 4.93 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 = 80.93 - 15.63
F = 65.3 N

so acceleration is
a = F/m
a = 65.3 / (11.3 + 8.25)
a = 3.34 m/s² to the right

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