# Pulling a box by a 10N force

1. Mar 8, 2007

### future_vet

1. The problem statement, all variables and given/known data
A 2.0 kg box is pulled along a horizontal plank by a 10N force parallel to the direction of motion. The rate of acceleration is 0.50 m/s^2. Find the coefficient of kinetic friction.

I think we use the formula: coefficient of friction = F (so 10) / N.
But I am not sure how to use the acceleration and the N variable.

2. Relevant equations
F - μk mg = ma
10 - 2(9.8)μk = 2 x 0.5

Would this be the correct way to go?

Thanks!

2. Mar 8, 2007

### PhanthomJay

Yes, but it contradicts what you said in the problem statement.

3. Mar 9, 2007

### future_vet

Yes, so I use the second equation? Is the data correctly plugged in?

4. Mar 9, 2007

### cristo

Staff Emeritus
Yes it is.

5. Mar 9, 2007

### future_vet

I am getting -0.005 as the answer... does it sound right? Where should the acceleration be used?

Thanks for all the help! :)

6. Mar 9, 2007

### PhanthomJay

you have the correct equation, with the given acceleration entered correctly, but your math is off.
$$10 -19.6\mu_k = 2(.5) = 1$$
$$-19.6\mu_k = -9$$
$$\mu_k = ?$$

7. Mar 9, 2007

### future_vet

oops, it's 0.46, right?

8. Mar 9, 2007

### PhanthomJay

yes, correct.