what is the work done by friction? and find the distance

Question

1.     A hockey puck (mass = 0.164 kg) slides down the ice rink. As it travels the length of the rink, its speed drops from 18.9 m/s to 16.6 m/s. (a) How much work is done by friction as the puck slides down the ice? 

(b) If the coefficient of friction between the puck and the ice is 0.07, how far did the puck travel down the ice?

Answer 1

For part a use work energy theorem. In words this means the change in kinetic energy over the interval is equal to the work done. In this case the work is done by friction as evidenced by the fact that the puck slows down. In equation form this amounts to

(1/2) m(v2 ^2 - v1^2 = W

Where v2 = 16.6 and v1 = 18.9. Calculating, we get W = - 6.7 J, the minus because the work is negative.

For the b part we want the distance d covered for the given coefficient of friction. To get use the equation

(mu)mgd = W

where mu is the coefficient of friction = 0.07

(mu)mg = 0,07 (0.164) (9.8) = friction force with product of g and m is normal force

Solve for d and we get 58 meters.