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Mass m1 = 11.7 kg is on a horizontal surface. Mass m2 = 5.35 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 μs = 0.574, while the coefficient of kinetic friction is μk = 0.106. If the system is set in motion with m1 moving to the right, then what will be the magnitude of the system's acceleration? Consider the pulley to be massless and frictionless. Submit Answer Tries 0/12

User Datoraki
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Answer:

The magnitude of the system's acceleration is 2.362 m/s².

Step-by-step explanation:

Given that,

Mass of one m₁= 11.7 kg

Mass of another m₂=5.35 kg

Coefficient of static friction = 0.574

Coefficient of kinetic friction = 0.106

We need to calculate the acceleration

Using balance equation

For m₂,


m_(2)g-T=m_(2)a


T=m_(2)g-m_(2)a...(I)

For m₁,


T-f=m_(1)a...(II)

Put the value of T in equation (II)


m_(2)g-m_(2)a-\mu_(k)m_(1)g=m_(1)a


a=(m_(2)g-\mu m_(1)g)/(m_(1)+m_(2))

Put the value into the formula


a=(5.35*9.8-0.106*11.7*9.8)/(11.7+5.35)


a=2.362\ m/s^2

Hence, The magnitude of the system's acceleration is 2.362 m/s².

Mass m1 = 11.7 kg is on a horizontal surface. Mass m2 = 5.35 kg hangs freely on a-example-1
User Elliott Darfink
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