A horizontal force of 89.7 N is applied to a 22.5 kg crate on a rough, level surface. If the crate accelerates at 1.18 m/s2, what is the magnitude of the force of kinetic friction (in N) acting on the crate?
Applied force = 89.7 N
Mass = 22.5 kg
Acceleration = 1.18 m/s2
Newton's second law - The summation of all the forces acting on a system is equal to mass times the acceleration of the system
So,
Applied force - Frictional force = mass X acceleration
==) 89.7 - Frictional force = 22.5 X 1.18
Frictional force = 89.7 -26.55 = 63.15 N
Also,
Frictional force = kinetic friction coefficient X mass X acceleration due to gravity
63.15 = kinetic friction coefficient X 22.5 X 9.8
So,
Kinetic Friction coefficient = 0.286 (Answer)
SOLUTION :
N = m g - 22.5*9.8 = 220.5 (N)
P - F = m a
=> F = P - m a = 89.7 - 22.5*1.18 = 63.15 (N)
F = µk * N
=> µk = F/N = 63.15/220.5 = 0.286 (ANSWER)
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