Coefficient of Friction between slide/girl

AI Thread Summary
The discussion revolves around calculating the coefficient of friction for a girl sliding down a 4.5m slide inclined at 35 degrees. Key points include the calculation of potential energy (PE) at the top, which is 911 J, and kinetic energy (KE) at the bottom, which is 162 J. Participants highlight the discrepancy between initial PE and final KE, questioning how friction affects energy loss. There is confusion about whether the force calculated using the cosine formula represents the normal force or frictional force. The conversation emphasizes the need to account for work done by friction to accurately determine the coefficient of friction.
cdsnig
Messages
2
Reaction score
0

Homework Statement


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.



Homework Equations



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


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?
 
Physics news on Phys.org
cdsnig said:

Homework Statement


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.

Homework Equations



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

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.
 
cdsnig said:

Homework Statement


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.

Homework Equations



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

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?
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Back
Top