# Kinetic friction and inclined planes

1. Feb 25, 2012

### iJamJL

1. The problem statement, all variables and given/known data
A block of ice of mass m slides down an incline that makes an angle θ = 40.7° with the horizontal. In trial 1 there is no friction; the block starts at rest and takes time t to reach the bottom of the incline. In trial 2 there is friction, and the the block slides down the incline in time 3.36t. Find μk, the coefficient of kinetic friction between the ice and the incline in trial 2.

2. Relevant equations
Frictional force = μk * Normal
Components
d= Vi*t + [(a*t^2)/2]

3. The attempt at a solution
This is what I've done so far, but I'm not sure if it's correct (I always have trouble with these problems that don't give all the numbers..). I've drawn the two inclined planes, each for a trial. The force that the block is going down the plane is mg*sin(40.7). The Normal= mg*cos(40.7).

To begin this problem, I think we need to find the acceleration. What I've done is in the first trial, the force that causes the mass to slide down the incline is F=mg*sin(40.7). What I did to find the acceleration is:

mg*sin(40.7) = ma
9.81*sin(40.7) = a
a = 6.397

Since the distances in both trials are equal, I set:

[(a-first)*t^2]/2 = [(a-second)*t^2]/2

6.397*t^2 = a-second* (3.36t)^2
6.397*t^2 = a-second* 11.2896 t^2
6.397/11.2896 = a-second
a-second = .5666

In the second trial, the force that makes the mass go down the incline is:

F=ma
F=m*(.5666)
Frictional force = μk * mg*cos(40.7)

Fnet = mg*sin(40.7) - (μk * mg*cos(40.7)) = m(.5666)
6.397 - 7.4373(μk) = .5666
μk=(.78394)

Is this correct?

2. Feb 25, 2012

### iJamJL

Just wondering if anyone looked this over..