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March 26, 2015

March 26, 2015

Posted by **Mary** on Saturday, March 3, 2007 at 1:44am.

A 59.5 kg person, running horizontally with a velocity of +3.90 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.

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?

m1= 59.5kg

v01= 3.90 m/s

m2= 12.2 kg

v02= 0

vf= (59.5kg x 3.90m/s)+ (12.2m/s x 0)/ (59.5kg + 12.2 kg)= 244.25m/s /71.7= 3.41 m/s

It looks like you calculated Vf correctly.

For the coefficient of friction, set the initial kinetic energy equal to the work done against fraction.

(1/2)Mtotal*Vf^2 = Mtotal*g*(mu,k)*X

mu,k = Vf^2/(2 g X)= ?

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