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 cdsnig Feb26-12 05:17 PM

Coefficient of Friction between slide/girl

1. The problem statement, all variables and given/known data
A 36kg girl slides down a slide that is 4.5m long. At the bottom of slide she is moving at 3 m/sec. If slide is inclined at 35 degrees find the coefficient of friction between slide and girl.

2. Relevant equations

Ff = μ*Force normal
Total energy = KE + PE
KE = 1/2 mv^2
PE = mgh
Total energy = F cosθ

3. The attempt at a solution

KE = 162 using above formula, PE = 911.15. Total energy = 1073.15
F = 1310.32 using the cosθ formula.

And that is where I am stuck - is the F the normal force or the Ff (from friction) how would I solve for the coefficient from this point? And have I even approached it in the right way so far?

 PeterO Feb26-12 06:36 PM

Re: Coefficient of Friction between slide/girl

Quote:
 Quote by cdsnig (Post 3785735) 1. The problem statement, all variables and given/known data A 36kg girl slides down a slide that is 4.5m long. At the bottom of slide she is moving at 3 m/sec. If slide is inclined at 35 degrees find the coefficient of friction between slide and girl. 2. Relevant equations Ff = μ*Force normal Total energy = KE + PE KE = 1/2 mv^2 PE = mgh Total energy = F cosθ 3. The attempt at a solution KE = 162 using above formula, PE = 911.15. Total energy = 1073.15 F = 1310.32 using the cosθ formula. And that is where I am stuck - is the F the normal force or the Ff (from friction) how would I solve for the coefficient from this point? And have I even approached it in the right way so far?
Firstly; how did you get that PE figure - not enough to say using the formula above as I need to know what values you used for m, g and h.

Secondly; why did you add KE and PE together? Surely we had pE at the top and KE at the bottom, but never both at the same time [perhaps half and half on the way down, but you were considering the top and the bottom, I hope]
You said,
Total energy = KE + PE

At the top that would be Total energy = 0 + PE
At the bottom that would be Total energy = KE + 0

And those two totals are different, due to the friction.

Edit: Total energy = F cosθ makes you group of formulas look like a grab-bag of possibilities, rather than a reasoned list of what actually applies.

 gneill Feb26-12 06:41 PM

Re: Coefficient of Friction between slide/girl

Quote:
 Quote by cdsnig (Post 3785735) 1. The problem statement, all variables and given/known data A 36kg girl slides down a slide that is 4.5m long. At the bottom of slide she is moving at 3 m/sec. If slide is inclined at 35 degrees find the coefficient of friction between slide and girl. 2. Relevant equations Ff = μ*Force normal Total energy = KE + PE KE = 1/2 mv^2 PE = mgh Total energy = F cosθ 3. The attempt at a solution KE = 162 using above formula, PE = 911.15. Total energy = 1073.15 F = 1310.32 using the cosθ formula. And that is where I am stuck - is the F the normal force or the Ff (from friction) how would I solve for the coefficient from this point? And have I even approached it in the right way so far?
At the top of the slide the girl has PE = 911 J. At the bottom, her KE is only 162 J. What happened to the rest of the PE? How much work was done by friction?

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