# Coefficient of Kinetic Friction

1. Mar 16, 2009

### yb1013

1. The problem statement, all variables and given/known data

A 5.20 kg block is set into motion up an inclined plane with an initial speed of v0 = 7.70 m/s. The block comes to rest after traveling 3.00 m along the plane, which is inclined at an angle of 30.0° to the horizontal.

(a) For this motion, determine the change in the block's kinetic energy.

I got -154.154 (correct)

(b) For this motion, determine the change in potential energy of the block-Earth system.

I got 75.6 (correct)

(c) Determine the frictional force exerted on the block (assumed to be constant).

I got 26.18 (correct)

(d) What is the coefficient of kinetic friction?

However, I could not figure this out for the life of me. Please help with the kinetic friction?

2. Relevant equations

I think the coeff. can be found with f=μmgCosθ , however my answer keeps coming out wrong...

3. The attempt at a solution

after using the previous formula, I came out with a value of μ=3.327 (26.18=u(5.2)(9.81)(cos(30))

2. Mar 16, 2009

### brentwoodbc

3. Mar 16, 2009

### brentwoodbc

I dont understand how you get 3.327... coefficiant of friction isnt going to be more than 1

if you calculate what you have

(26.18=u(5.2)(9.81)(cos(30))

u get

26.18=44.13u

divide by 44.13
=.59

4. Mar 16, 2009

### yb1013

oooooo haha, thank you sir, my calculator somehow reverted itself back to radians?

Thanks again!

5. Mar 16, 2009

### brentwoodbc

no prob. you did all the hard work haha