A 26.0-kg block is initially at rest on a horizontal surface. A horizontal force of 72.0 N is required to set the block in motion, after which a horizontal force of 58.0 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.

a)  μ = 0.283 , b)    μ = 0.228

Explanation:

For this exercise we must use the translational equilibrium conditions in each exercise

Y Axis  

       N- W = 0

       N = W = mg

X axis

       F -fr = 0

       F = fr

friction force has an expression

        Fr = μ N

 we substitute

       F = μ mg

        μ = F / mg

  let's calculate

a) μ = 72.0 / 26.0 9.8      static

      μ = 0.283

b) μ = 58/26 9.8             dinamicic

      μ = 0.228

the tires are rotating at 0.434 rpmlinear speed=45 m/sdiameter=0.550 mradius=0.275 mnow v= rωω= v/rω=45/0.275ω=163.6 rad/sω=163.6 rad/s (1rev/2π rad) (60s/1m)ω=1563 rpm