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.
(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?
v01= 3.90 m/s
m2= 12.2 kg
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 -