science math

Question

A 160-g hockey puck is at rest on the ice. A player then strikes the puck to accelerate it to a speed of 30 m/s. The puck slides on the ice, decelerating at a constant rate due to friction with the ice. It travels in a straight line until it comes to a rest 3.0 seconds after it began moving.
With is the force of kinetic friction between the ice and puck?

Answer 1

The force of friction is:

 

     F = μ·F_n

 

where:

• μ = the coefficient of kinetic friction
• F_n = the normal force on the puck = gravity = mg
• m = the mass of the puck = 0.16 kg
• g = acceleration due to gravity = 9.8 m/s²

 

By Newton's 3rd Law we have:

 

     μmg = m·(Δv/Δt)

 

where:

• Δv = change in velocity = 30 m/s - 0 m/s = 30 m/s
• Δt = change in time = 3 seconds

 

Solve for μ then plug in the numbers.