Posted by Mary on .
Please tell me where I am going wrong.
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)= ?

Physics drwls please help 
tu papi,
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