Homework Help??

Question

A hockey puck is hit on a frozen lake and starts moving with a speed of 11.2 m/s. Five seconds later, its speed is 5.50 m/s.
What is the average value of the coefficient of kinetic friction between puck and ice?

Answer 1

a=( change in velocity)/(change in time)
=(Vfinal - Vinitial)/2
= (5.5-11.2)/5
=--1.14 m/s2

Force of Friction = mxa

since we don't know the mass of the puck
let m=M

Force of Friction = mxa = 1.14*M

Coefficient of Friction = Force of Fiction / Weight

Weight = mxg again we don't know puck mass m=M

Weight = 9.81x M

Plug this into

Coefficient of Friction = Force of Fiction / Weight

=1.14xM /(9.81xM) , the M's cancel out
=1.14/9.81
=0.116

Answer 2

Acceleration is the rate of change of velocity. The average acceleration is (5.50 - 11.2) / 5 = -1.14 m/s^2.

Force is mass times acceleration. The force is -1.14M newtons, where M is the mass of the puck.

The restraining force is due to friction, which equals the weight of the puck times the coefficient of kinetic friction. The weight of the puck is Mg = 9.8M, so

-1.14M = -9.8M mu

where mu, the coefficient, equals 1.14 / 9.8 = 0.116.

That's your answer.