A boy coasts down a hill on a sled, reaching a level surface at the bottom with a speed of 5.9 m/s. if the coefficient of friction between the sled's runners and snow is 0.045 and the boy and sled together weigh 590 n, how far does the sled travel on the level surface before coming to rest?

Frictional force = vertical force x coeff of friction = 590N x 0.045 = 26.55N

Mass of boy and sled = 590N / g = 590N / 9.8m/s^2 = 60.20 kg

Deceleration due to friction = 26.55N / 60.20kg = 0.44 m/s^2.

For constant acceleration we have:

v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and ti is time. In this case v = 0, u = 5.9 m/s, a = - 0.44 m/s^2. So we have

0 = 5.9 - 0.44t

which gives t = 5.9 / 0.44 = 13.409 s.

Distance traveled in this time d = ut + 0.5at^2 = 5.9 x 13.409 - 0.5 x 0.44 x 13.409^2 = 39.56 m

Answer: 40 meters