# Calculate the coefficient of kinetic friction for a sliding box

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1. Oct 21, 2015

### Sace Ver

• Mentor's note: Thread title changed to be decriptive of the problem
1. The problem statement, all variables and given/known data
An electric motor is used to pull a 125 kg box across a floor using a long cable. The tension in the cable is 350N and the box accelerates at 1.2m/s2 (forward) for 5.0s. The cable breaks and the box slows down and stops. Calculating the coefficient of kinetic force?

Given:
m=125 kg
a= 1.2 m/s2
t= 5.0s
Ft=350 N

2. Relevant equations
uk=Fk / FN

3. The attempt at a solution
Fnet = ma
Fnet = (125kg)(1.2m/s2)
Fnet = 150

Thought Fk would be 350N because that's the Ft.

And then Uk would equal 350N divided by 150 but I got 2.3 when the answer is 0.16 not sure how to get answer?

2. Oct 22, 2015

### SteamKing

Staff Emeritus
Ya mean, calculate the coefficient of kinetic friction?

3. Oct 22, 2015

### haruspex

You correctly got Fnet=150N, what forces is that the net of?

4. Oct 22, 2015

### Sace Ver

I thought that since the object is 125 kg it would have a force of gravity of 1225N because 125kg•9.8m/s2
so that'd mean that the normal force would also be 1225N and I know that the force of tension would be 350N. I think that Fg and FN would cancel out leaving me with only FT?

5. Oct 22, 2015

### haruspex

6. Oct 22, 2015

### Sace Ver

friction would be 350N as well I'm assuming?

7. Oct 22, 2015

### haruspex

Why would you assume that?

8. Oct 22, 2015

### Sace Ver

Because the force of tension is 350N so the opposite side is 350N as well?

9. Oct 22, 2015

### haruspex

That would make the net horizontal force zero, but you have already calculated that the net horizontal force is not zero.

Do you understand what the equation $\Sigma F_x = ma_x$ means?

10. Oct 22, 2015

### Sace Ver

Fnet = mass • acceleration that's all I know

11. Oct 22, 2015

### haruspex

OK, and you used that to calculate Fnet, which is to say the sum of the horizontal forces. The value was not zero. So why did you suggest in post #8 that the horizontal forces should balance, which would give a zero Fnet? Do you understand what net force means?

12. Oct 22, 2015

### Sace Ver

Net force is the sum of all forces acting on an object

13. Oct 22, 2015

### haruspex

Right.
As you said, the vertical forces balance, so it just leaves the horizontal forces: tension and friction.
Write out the $\Sigma F_x=F_{net}$ equation using the known horizontal forces and net force.

14. Oct 22, 2015

### Sace Ver

(350N) Fx = 150N
Fx = 150N - 350N
Fx = 200N [backwards]

15. Oct 22, 2015

### haruspex

I have no idea where you get that from.
That is not quite what I meant, but it's close. "Fx" was meant to stand for a generic horizontal force. Let's use Ff for frictional force:
$150N=ma_x=F_{x net}=\Sigma F_x=$Tension+Friction$=350N+F_f$.
So, yes, Ff has magnitude 200N.
Now deduce the coefficient of friction.

16. Oct 23, 2015

### Sace Ver

Do I use that entire formula and sub in all my information?

17. Oct 23, 2015

### haruspex

Do you mean the whole of this?
No, that expresses the chain of logic. Do you understand each of the steps in there?
It leads to result that $150N=350N+F_f$, which I assume is what you meant by your equation in post #14.
It remains to express $F_f$ in some way involving the coefficient of friction.

18. Oct 23, 2015

### Sace Ver

150N = 350N + Ff is what I meant!

19. Oct 23, 2015

### Sace Ver

Just not sure where to go after finding out that the magnitude of Ff equals 200N.

20. Oct 23, 2015