Question

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?

Answer 1

Applied force = 89.7 N

Mass = 22.5 kg

Acceleration = 1.18 m/s​​​​​​2

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)

Answer 2

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)