Answer:
a) [tex]\mu_s =0.40[/tex]
b) [tex]\mu_k =0.20[/tex]
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 ([tex]\mu_s[/tex])× Normal reaction(N)
N = mg
where, g = acceleration due to gravity
⇒N = 51 kg × 9.8 m/s²
⇒N = 499.8 N
thus, 200N = [tex]\mu_s[/tex] × 499.8 N
⇒[tex]\mu_s[/tex] = [tex]\frac{200}{499.8}=0.40[/tex]
b) Applied force = 100 N
since the cabinet is moving, thus the coefficient of kinetic([tex]\mu_k[/tex]) friction will come into action
Now the force required to move the from the state of rest (F) = coefficient of kinetic friction ([tex]\mu_k[/tex])× Normal reaction(N)
N = mg
where, g = acceleration due to gravity
⇒N = 51 kg × 9.8 m/s²
⇒N = 499.8 N
thus, 100N = [tex]\mu_k[/tex] × 499.8 N
⇒[tex]\mu_k[/tex] = [tex]\frac{100}{499.8}=0.20[/tex]
