Question

To move a 51kg cabinet across a floor requires a force of 200N to start it moving, but then only 100N to keep it moving a steady speed. (a) Find the maximum coefficient of static friction between the cabinet and the floor. (b) Find the coefficient of kinetic friction between the cabinet and the floor.

Answer 1

a)
`\mu_s =0.40`

b)
`\mu_k =0.20`

Step-by-step explanation:

Given:

Mass of the cabinet, m = 51 kg

a) Applied force = 200 N

Now the force required to move the from the state of rest (F) = coefficient of static friction (
`\mu_s`)× Normal reaction(N)

N = mg

where, g = acceleration due to gravity

⇒N = 51 kg × 9.8 m/s²

⇒N = 499.8 N

thus, 200N =
`\mu_s` × 499.8 N

⇒
`\mu_s` =
`(200)/(499.8)=0.40`

b) Applied force = 100 N

since the cabinet is moving, thus the coefficient of kinetic(
`\mu_k`) friction will come into action

Now the force required to move the from the state of rest (F) = coefficient of kinetic friction (
`\mu_k`)× Normal reaction(N)

N = mg

where, g = acceleration due to gravity

⇒N = 51 kg × 9.8 m/s²

⇒N = 499.8 N

thus, 100N =
`\mu_k` × 499.8 N

⇒
`\mu_k` =
`(100)/(499.8)=0.20`