Question #95733

A 28.0-kg block is initially at rest on a horizontal surface. A horizontal force of 80.0 N is required to set the block in motion, after which a horizontal force of 57.0 N is required to keep the block moving with constant speed.
(a) Find the coefficient of static friction between the block and the surface.
(b) Find the coefficient of kinetic friction between the block and the surface.

Expert's answer

Let Fa be force applied to block, m be mass of the block, Fk and Fs be kinetic and static friction respectively, kk and ks be coefficients of kinetic and static friction.

a)To move object from the rest force applied to block must overcome static friction force,

therefore

F_a=F_s=k_s\cdot m\cdot d,\

ks=Fam⋅g=8028⋅9.8≈0.29k_s=\frac{F_a}{m\cdot g}=\frac{80}{28\cdot 9.8}\approx0.29

b) Since acceleration of the block is zero

Fa=Fk=kk⋅m⋅g,F_a=F_k=k_k\cdot m\cdot g,

kk=Fam⋅g=5728⋅9.8≈0.2k_k=\frac{F_a}{m\cdot g}=\frac{57}{28\cdot 9.8}\approx0.2

Answer: a)0.29, b) 0.2.


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