What about mu?

Question

A 65.0 kg person, running horizontally with a velocity of +3.68 m/s, jumps onto a 12.2 kg sled that is initially at rest.

(a) Ignoring the effects of friction during the collision, find the velocity of the sled and person as they move away.
(this is right with 3.1 m/s)
(b) The sled and person coast 30.0 m on level snow before coming to rest. What is the coefficient of kinetic friction between the sled and the snow?
How do I figure out the coefficient of friction.

Answer 1

a) m1*V1 + m2*V2= (m1+m2)*Vf where v2 =0
Vf = 3.1 m/s

b) If they coast for s = 30.0 m, the deceleration, a, is given by
the equation
V^2 = Vo^2 +2*a*s
Where V = 0 and Vo = Vf from part a). Solve for a, deceleration.
Then use a in F = m*a. F is the force of friction given by mu*N. Since they are on level snow the normal force is the weight of the sled and person.

Ohh! Watch out behind you.
Happy Halloween