Your answer is partially correct. Try again. A 75.6-kg person, running horizonta

Question

Your answer is partially correct. Try again. A 75.6-kg person, running horizontally with a velocity of +2.66 m/s, jumps onto a 18.6-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? Number 2.13 JUnits m/s + Number Units No units Click if you would like to sow Work for this question: Open Show Work

Explanation / Answer

Part B.

Using Work-energy theorem

Net Work = Change in kinetic energy

W = dKE

Net Work = Work-done by friction

Wf = Ff*d

Ff = uk*N = uk*m*g

dKE = KEf - KEi

Vi = initial Velocity = 0

Vf = final velocity = 2.13 m/sec (from part A)

So,

uk*m*g*d = 0.5*m*Vf^2 - 0.5*m*Vi^2

uk*m*g*d = 0.5*m*Vf^2 - 0

uk = 0.5*Vf^2/(g*d)

uk = 0.5*2.13^2/(9.81*30)

uk = 0.0077

Please Upvote. Let me know if you have any doubt.

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