# Coefficient of kinetic friction, need today

1. Apr 30, 2010

### an9890

coefficient of kinetic friction, URGENT need today

The system shown in the figure below has an acceleration of magnitude 0.51 m/s2, where m1 =
4.30 kg and m2 = 6.00 kg. Assume that the coefficient of kinetic
friction between block and incline is the same for both inclines.

http://img9.imageshack.us/img9/3437/figureyh.jpg [Broken]
m1 is box on the left, m2 is box on the right. both angles are 35 degrees.

a) find the coefficient of kinetic friction

Here's what I tried:
coefficient of kinetic friction = (gsinΘ - a) / gcos Θ
and got .63668

when i input my answer in the system said "your answer differs from the correct answer by orders of magnitude."

b) find the tension in the string
T= m1a + m1gsinΘ
= 26.36 N

system said "your answer differs by 10% of correct answer..."

Last edited by a moderator: May 4, 2017
2. Apr 30, 2010

### hotvette

Re: coefficient of kinetic friction, URGENT need today

Let's start with the basics. What equation of motion (i.e. $$\Sigma$$F = ma) did you use for each mass?

3. Apr 30, 2010

### an9890

Re: coefficient of kinetic friction, URGENT need today

^

(a)
∑ Fx = ma --> mgsinΘ - fk = ma
∑ Fy = 0 --> n - mgcosΘ = 0
n = mgcosΘ
and fk = mukn

so muk = (gsinΘ -a) / gcos Θ

4. Apr 30, 2010

### hotvette

Re: coefficient of kinetic friction, URGENT need today

I presume that's for m2. But, you forgot the force of the tension. Also, you need a similiar equation for m1. You'll then have two equations in two unknowns (T and mu) that can be solved by substitution.

Last edited: May 1, 2010