# Kinetic Friction Coefficient Problem

1. Jun 25, 2009

### iamtrojan3

1. The problem statement, all variables and given/known data
A block is at rest on a plank whose angle can be varied. The angle is gradually increased from 0 deg. At 26.5°, the block starts to slide down the incline, traveling 3.10 m down the incline in 2.00 s. Calculate the kinetic coefficent of friction between the block and the plank.

2. Relevant equations
Uk = friction Coefficient
D = Vot + 1/2at^2
F= ma
kinetic Friction = cos(theta)*m*g*Uk

3. The attempt at a solution
First i used the kinematic equation to solve for the acceleration,
3.1 = 0 + 1/2a(2)^2
a = 1.55 m/s^2

Then with the acceleration i got here pluged it in the F= ma to get,
F = m (1.55)
F = sin(theta)*m*g - cos(theta)*m*g*Uk
Notice the m cancels out so i get

Sin(theta)g-Cos(theta)*g*Uk= a
g = 9.8m/s^2

I plugged everything in, theta = 26.5, a = 1.55, g = 9.8 and solved for Uk

Uk turned out to be -.352 which is wrong and makes no sense... but nevertheless, .352 is also wrong.

Any idea on what i did wrong?
thanks for the help!

2. Jun 25, 2009

### LowlyPion

Maybe a calculation mistake?

Using
Sin 26.5 = .446
Cosine 26.5 = .895