# Coefficient of Kinetic Friction

1. Sep 26, 2008

### kblue!1

SOLVED

1. The problem statement, all variables and given/known data
Suppose the coefficient of kinetic friction between m1 and the plane in the figure is $$\mu$$k= 0.24
m1 = m2 = 2.5kg

As m2 moves down, determine the magnitude of the acceleration of m1 and m2 given $$\Theta$$ = 26.

2. Relevant equations
F=ma

3. The attempt at a solution
m2g = (2.5kg)(9.8m/s2)= 24.5N
24.5 - T = 2.5a

(2.5kg)(9.8m/s2)sin26 = 18.6827N
(2.5kg)(9.8m/s2)cos26 = 15.8495N
(.24)(15.8495N) = 3.80388N

T - 18.6827 - 3.80388 = 2.5a
24.5 - T = 2.5a
(-18.6827-3.80388+24.5)= (2.5a+2.5a)

2.01342 = 5a
a=.402 m/s2

did I do any miscalculations?

Last edited: Sep 26, 2008
2. Sep 26, 2008