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

Answer:

a) $\mu_s =0.40$

b) $\mu_k =0.20$

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

⇒$\mu_s$ = $\frac{200}{499.8}=0.40$

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

⇒$\mu_k$ = $\frac{100}{499.8}=0.20$