A 55.0 N frictional force is exerted onto a box that is moving on flat ground. The box slows down at - 1.88m / (s ^ 2) before coming to a complete stop. What is the coefficient of kinetic friction ?



Answer :

Answer:

0.192

Explanation:

Friction force is equal to the normal force (N) times the coefficient of friction (μ). From Newton's second law, the net force on an object is equal to the mass times acceleration. Drawing a free body diagram, there are 3 forces on the box: weight force mg pulling down, normal force N pushing up, and friction force Nμ pushing to the left.

Sum of forces in the y direction:

∑F = ma

N − mg = 0

N = mg

Sum of forces in the x direction:

∑F = ma

-Nμ = ma

Substitute:

-mgμ = ma

-gμ = a

μ = -a / g

Plug in and solve.

μ = -(-1.88) / 9.8

μ = 0.192