A child slides

1. Oct 26, 2010

wondermoose

A child slides....

1. The problem statement, all variables and given/known data
A child slides (starting at rest) down a 4.8 m long slide in 1.8 seconds. if the slide is inclined 28 degrees with the horizontal, what is the coefficient of friction between the child's pants and the slide?

2. Relevant equations
m a = u mg => u =a/g

3. The attempt at a solution
Beats the heck out of me. I tried solving for the x and y components, but I'm really just not entirely comfortable with the question in general. I don't even know if the formula I listed is on the right path. This is my first time posting, so I'm not really sure what the rules of solving (obvious) homework problems, but I was really hoping for a run-through of the formulas (minus values), and maybe an explanation to go along with it. Thanks!

Last edited: Oct 26, 2010
2. Oct 26, 2010

Mamooie312

Re: A child slides....

You forgot gravity. F = mgsin(28) - umgcos(28) = ma
Solve for acceleration and then use kinematics.

Proof:
Friction = uN Normal force is gravity going into slide = umgcos(28)
All movement is along the slide. The gravity component in this direction is mgsin(28)
Try to treat gravity and the normal force as hypotenuses of triangles and see if you come up with this too.

3. Oct 26, 2010

wondermoose

Re: A child slides....

Ok, so the mass cancels out of that equation, so:

a=gsin(28)-ugcos(28)
-u=a/[(gsin(28))(gcos(28))]

If the initial velocity is 0, the inital position is 0, the final position is 4.8, and the delta t is 1.8, then my kinematics equation looks like this:

sf=si+vi(t)+.5a(t^2)
4.8=.5a1.8t^2)
9.8/(1.8^2)=a
a=2.96

Put that back into the original equation....

-u=2.96/[(gsin(28))(gcos(28))]

-u=0.074
u=-0.074

How does that look? THanks!