A29 kg chair initially at rest on a horizontal floor requires a 171 n horizontal force to set it in motion. once the chair is in motion, a 135 n horizontal force keeps it moving at a constant velocity. the acceleration of gravity is 9.81 m/s 2 . a) what is the coefficient of static friction between the chair and the floor? 017 (part 2 of 2) 10.0 points b) what is the coefficient of kinetic friction between the chair and the floor?

a)   μ = 0.602   and b)   μ = 0.475

Explanation:

a) Let's use Newton's second law for this exercise. The reference system the x axis is parallel to the floor and the axis and perpendicular.

Y Axis

      N-W = 0

      N = W

X axis

    F-fr = ma

The friction force equation is

    fr = μN

At the point where the movement begins the acceleration is zero

Let's replace

     F = μ (mg) = 0

    F = μ mg

    μ = F / mg

    μ = 171/29 9.8

    μ = 0.602

b) in this case we have the same formula, but the friction is kinetic

     μ = F2 / mg

     μ = 135/29 9.8

     μ = 0.475