What am I doing worng with this physics problem?

Question

A 65.0 kg person, running horizontally with a velocity of +3.68 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.
I got this right with 3.1 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?

I've gotten different answes... .016, .16, .02, .4.... all those are incorrect.

What am I doing wrong in this problem? I cannot figure it out and it's frustrating me.

According to my homework website, 3.1 is correct, and if the answer was 3.37, then .019 would be correct.... however, 3.1 is the correct answer and should be used when finding the coefficient of kinetic friction... however, if you did it the same way that you figured it out, it would come out to .016 but the website is telling me that that is wrong.

Hmmm... during the 1st part, we ignored the effects of friction... what would the 1st part be if we included friction?

Answer 1

Kinetic Energy, KE=1/2mv^2, where m is the mass of the person+sled equal to 77.2kg. This you got right. v is their initial velocity of 3.1m/s which you got right, too. Now substitute known values:

KE=1/2*77.2*3.1^2
=370.9Joules

They came to rest because of the work done by the force of friction. The formula for work is F*s, where F is the force we are interested in and s is the displacement given as equal to 30m.

Therefore, Work =F*30

Based on the principle of conservation of energy, kinetic energy=work.

370.9=30F
F=370.9/30
=12.4N

Coefficient of friction=F/R where R is the total weight normal to snow which you must already have computed correctly as well.

Coefficient of friction=12.4/(77.2*9.8)
=0.016

Let's skin the cat another way using the formula,

v^2-u^2=2as, where v is the final velocity of 0, u is the initital velocity which you correctly got , equal to 3.1m/s, s the displacement of 30m, and a the acceleration. Now substitute known values:

0-3.1^2=2*a*30
-9.61=60a
a=-9.61/60
=-0.16m/s^2 it's negative because the person and sled were decelerating.

Now solve for the force of friction using the formula, F=Ma.

F=77.2*0.16
=12.4N Same as above so coefficient will also be the same, i.e. .016, viz:

Coefficient of kinetic friction=12.4/756.6
=0.016


I also cannot figure out what I did wrong either, if you say the above solutions and answer are incorrect.

Answer 2

Okay, now you got initial velocity wrong it should be 3.37 m/s, then you have final velocity, and you have the distance traveled.... find the acceleration, which should come out to -.189 m/s^2... now you know the equation that "mu" or friction should be the resistive force of friction divided by perpendicular force pushing the two objects together.... then you should get .0193...

Answer 3

the kinetic energy when the man jumps onto a sled is 0.5 * 65.0 * 3.68**2 =440.128
this kinetic energy remains the same after the man jumped onto the sled that is 0.5 * (65+12.2) * v**2 = 38.6 * v**2
So, 38.6 * v**2 = 440.128
V=3.377 m/sec

Let Friction force = F
work done by friction = F*30

F*30 = 440.128
F= 14.67 N
Coef of kinetic = 14.67/440.128 = 0.019