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)= ?
To find the coefficient of kinetic friction (mu,k), you need to set the initial kinetic energy equal to the work done against friction.
The initial kinetic energy of the sled and person is given by (1/2)Mtotal*Vf^2, where Mtotal is the total mass of the sled and person (71.7 kg) and Vf is the final velocity of the sled and person (3.41 m/s).
The work done against friction is given by Mtotal*g*(mu,k)*X, where g is the acceleration due to gravity (9.8 m/s^2) and X is the distance the sled and person coast before coming to rest (30.0 m).
Setting the initial kinetic energy equal to the work done against friction, we have:
(1/2)Mtotal*Vf^2 = Mtotal*g*(mu,k)*X
Now, you can solve for the coefficient of kinetic friction (mu,k):
mu,k = Vf^2 / (2*g*X)
Plugging in the values you have:
mu,k = (3.41 m/s)^2 / (2 * 9.8 m/s^2 * 30.0 m)
Calculating this expression will give you the coefficient of kinetic friction.