# Coefficient of kinetic friction question

1. Jan 19, 2012

### Meco

1. The problem statement, all variables and given/known data
A 65.0kg person, running horizontally with a velocity of +4.25 m/s, jumps onto a 12.0-kg sled that is initially at rest.
1) Ignoring the effects of friction during the collision, find the velocity of the sled and the person as they move away.
2) The sled and the person coast 35.0 m on a level snow before coming to a rest. What is the coefficient of kinetic friction between the sled and the snow?

2. Relevant equations
M1V1 + M2V2 = MV'
F=Ma

3. The attempt at a solution
1) I did 65(4.25)+ 12(0)= 77 V'
That equaled 3.5876 m/s
2) This is what I had trouble with. I am not really sure where to go next. I know the velocity is 3.5876, the mass is 77kg, the distance traveled is 35m. How do I add friction into it? F=Ma does not seem like the right equation to use,do I not have the right equation?

2. Jan 19, 2012

### Angry Citizen

Not entirely. Consider the acceleration necessary to stop an object traveling at the speed in question in order for it to travel the distance in question. Since you know the masses of the two objects, the force of the combined system will necessarily result using F=ma. Then use your equation for kinetic friction: force of friction=(coefficient of kinetic friction)*(weight of object).

This is commonly expressed as F(friction)=μk*N, where μk is pronounced mew-kay and is the coefficient of kinetic friction, and N is the normal force (AKA weight) exhibited by the object and the plane on which it is sliding.

Note that I'm assuming you are accurate with regard to the inelastic collision stuff. It's been so long since I've seen it...

3. Jan 19, 2012

### Spinnor

x(t) = x_0 + v_o*t + at^2/2

v(t) = v_o + a*t

F = ma = μmg

?

4. Jan 19, 2012

### Meco

Spinnor I don't really understand what that equation is, maybe I was just taught with different variables. Angry, I am trying to follow what you are trying to tell me and am struggling. The mass of the two objects combined is 77kgs. The force of the combined system is what? The starting velocity of the two objects combined is 3.5876 m/s. So in 35m the velocity will reach zero. How can I figure that out without knowing how long it takes to reach zero? Sorry if it is obvious and I just do not understand, but I really am trying!

5. Jan 19, 2012

### Angry Citizen

Try (velocity_final - velocity_initial)/(2*distance traveled)=acceleration. Then use F=ma using that a value.

6. Jan 19, 2012

### tannerbk

You have the initial velocity of the system, which you calculated. You have the the distance (x_f - x_o) is 35 meters. The final velocity is zero. As Spinnor showed, you have two equations and two variables: acceleration and time (you don't need time but you can solve for it if you want anyways). Once you have acceleration all you have to do is set up a net force equation. The only acceleration comes from friction, and results in slowing the sled down. Using Newton's Second Law, Net Force = ma. That will be equal to the force of friction, the equation is given by the coefficient of kinetic friction * the normal force (equal to m(total)*g). Drawing a freebody diagram once the person is in the sled should help you visualize these equations. You have mass, acceleration, and g is a constant, so you can solve for the coefficient of kinetic friction.

7. Jan 19, 2012

### Meco

Ok so
Vf=0
Vi=3.5876
Distance traveled=35
So (0-3.5876)/(70)= -.05125
F=77(-.05125)
F=-3.94625N (Newtons right?)
So now that I know force I put it into F(friction)=μk*N
So -3.94625= μk* 77 <--- the Normal force is just the mass right? It is in Kg here, that feels wrong for some reason
77/-3.94625= μk
μk= -19.512???? That seems way too high, what did I do wrong now? -.-

8. Jan 19, 2012

### Angry Citizen

No. Mass is not a force. The normal force is the weight of the object. m*g.

A discrepancy exists between the first equality used and the second equality used. Notably, the second equality 77/-3.94625 does not equal μk, it equals 1/μk (well, it doesn't, because you forgot to multiply the mass by the acceleration due to gravity). You messed up the division here.

9. Jan 19, 2012

### Meco

So I tried to work it out like you said and came up with this:
77kg = mass
77*9.8=754.6 = the Normal force
-.05125= the acceleration
77*-.05125 = net force = μmg
so 77*-.05125 = -3.946= μ77(9.8)
-3.946/754.6= μ
-.0052279= μ

10. Jan 19, 2012

### Angry Citizen

Well, the first thing that comes to mind is that you can drop the negative sign. The purpose of the negative sign was merely to indicate the vector quantity of the acceleration, which holds no meaning for mew-kay.

The second thing is that your value doesn't make sense. The coefficient of friction between two surfaces coated in teflon is about 0.04. Yours is an order of magnitude lower than one of the smoothest surfaces known to mankind. Hrm.

After about five minutes, I discovered the problem. I gave you (mildly) incorrect information - but luckily, we knew what an answer 'should' look like by comparing it to extreme values. This is often a good check of your answer, and I think is one of the greatest lessons one can learn in introductory physics: learning to know when you're egregiously wrong due to a simple error.

So, what caused the problem was, the equation used to calculate acceleration was wrong. I told you it was (v_final - v_initial)/(2*d); in fact, it's ((v_final)^2 - (v_initial)^2)/(2*d). My sincerest apologies; I just transcribed it wrong.

11. Jan 19, 2012

### Meco

So when you do
(0-3.833^2)/70
you get -2.0988=a
ma=F
77*2.0988=161.6076=net force = μmg
161.6076= μ(754.6)
161.6076/754.6=μ
μ=.21416 <--- looks a lot more reasonable :D
What units is μ in?

12. Jan 19, 2012

### Angry Citizen

Much more reasonable! Anyway, mu is dimensionless. Force=mu*force means mu must be dimensionless.

13. Jan 19, 2012

### Meco

Ok thanks again, hurray! One problem down!