# Pulling a box by a 10N force

• future_vet
In summary, the coefficient of kinetic friction for a 2.0 kg box being pulled along a horizontal plank by a 10N force with a rate of acceleration of 0.50 m/s^2 is 0.46.

## Homework Statement

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.

## Homework Equations

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

Would this be the correct way to go?

Thanks!

future_vet said:

## Homework Statement

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.

## Homework Equations

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

Would this be the correct way to go?

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

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

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

Yes it is.

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

Thanks for all the help! :)

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

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

oops, it's 0.46, right?

future_vet said:
oops, it's 0.46, right?
yes, correct.