Physics review question

Question

A 5 kg hockey puck is initially traveling wit a velocity of 10m/s and comes to a stop after traveling 75 m. Calculate the coefficient of kinetic friction between the ice and the puck.

Answer 1

Always start with a free body diagram:

The sum of the forces in the y-direction is zero because there is no motion in that direction.

∑Fy = N + W = 0 where N is the normal force and W is the weight. W = mg = 5•9.8 = 49 N (downward)

N - 49 = 0

N = 49 N

The sum of the forces in the x-direction equals mass time acceleration.

The acceleration is calculated from the kinematic equation:

v_f² = v_i² + 2a(x_f - x_i) or

a = (v_f² - v_i²)/(2(x_f - x_i)) = (0² - 10²)/(2•75) = -2/3 m/s²

So in the x-direction:

∑Fx = ma

F_F = ma

F_F = (5)(-2/3)

F_F = -10/3 N

But we also know that F_F = μN so μ = F_F/N = (10/3)/49 = 0.068 or, with 1 sig fig, μ = 0.07