Question

A 53.6-kg person, running horizontally with a velocity of +4.72 m/s, jumps onto a 19.1-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. (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?

Answer 1

(a) Momentum conservation : m1v1 + m2v2 = m1v1' + m2v2'
Since the person jumped onto the sled, they will move as onetherefore, we can rewrite the equation as: m1v1 + m2v2 =(m1+m2)V'
Plug in values:
(53.6x4.72) + (19.1x0) = (53.6+19.1)V'
V' = 252.99/72.7
V' = 3.47 m/s

(b) Energy conservation: the lost energy is the result of work donefriction
Initial energy is kinetic energy = (1/2)mv2 =0.5x(53.6+19.1)x3.47^2 = 437.7J
Friction force = μk x (Normal force) =μk mg = 72.7x9.8μk
Work done by friction is W = FS = (friction force)x(displacement) =72.7x9.8μk x(30)

Therefore 437.7J = 72.7x9.8μk x(30)
437.7 = 21373.8μk
μk = 0.0204