What is the average coefficient of friction for a skier on a 15.2° incline?

AI Thread Summary
The discussion focuses on calculating the average coefficient of friction for a skier on a 15.2° incline after gliding 12.0 m. Initial calculations yielded a coefficient of friction of 0.524, but upon further review and corrections, it was recalculated to 0.549. Participants debated the correct approach to the equations, particularly regarding the signs and the net acceleration. After additional adjustments, a final value of 0.4275 was proposed, which received validation from another participant. The thread emphasizes the importance of careful equation setup and sign conventions in physics problems.
mparsons06
Messages
60
Reaction score
0

Homework Statement



A skier traveling 12.6 m/s reaches the foot of a steady upward 15.2° incline and glides 12.0 m up along this slope before coming to rest. What was the average coefficient of friction?

Homework Equations





The Attempt at a Solution



D = 12m
Velocity initial = 12.3 m/s
V final = 0 m/s
Angle = 15.2° incline

Use the equation Vf^2 = Vi^2 + 2*a*d
Find a

0 = 12.3^2 + 2*a*12
2*a*12 = -12.3^2

Divide both sides by 2*12
a = 6.30375 m/s^2


Force parallel = mass * g * sin θ
Force normal = mass*g* cos θ
Friction Force = µ * mass*g* cos θ

Going up an incline 2 forces decrease your velocity
Force parallel and Friction, so we add these 2 forces to find the total force decreasing your velocity. Since these forces decrease your velocity they are negative.
a = 6.30375 m/s^2


∑ Forces = mass * acceleration
(-mass * g * sin θ) + -(µ * mass*g* cos θ) = mass * 6.30375
Notice mass cancels


(- g * sin 15.2°) + (-µ * g* cos 15.2° ) = 6.30375
6.30375 = -2.57 + -9.457 µ
6.30375 = -12.027 µ
µ = 0.524

But the answer is incorrect. Can someone please tell me where I messed up?
 
Physics news on Phys.org
By how much? Your initial velocity suddenly went from 12.6 m/s to 12.3m/s.
 
I just recalculated my mistake, and I got \mu = 0.549.

Does that seem right to you?
 
(- g * sin 15.2°) + (-µ * g* cos 15.2° ) = 6.30375
6.30375 = -2.57 + -9.457 µ
6.30375 = -12.027 µ
µ = 0.524

But the answer is incorrect. Can someone please tell me where I messed up?[/QUOTE]

In your original calculation you have put the net acceleration +ve which isn't the case as forces are along the acceleration.
 
So, I'm slightly confused. Do you mean it should be:

(- g * sin 15.2°) + (-µ * g* cos 15.2° ) = 6.615
6.615 = -2.57 + 9.457 µ
6.615 = 6.887 µ
µ = 0.961

I'm drawing a blank. =/
 
No I meant the opposite. It should be
-g*sin15.2 - µ*g*cos15.2 = -6.615.
I had only quoted you.
One more thing, why are you always adding the the terms on left when there is a unknown
µ with one of the terms. Like here(this is yours):
6.615 = -2.57 + 9.457 µ
6.615 = 6.887 µ​
[/CENTER]

I noticed it in the first post too.
 
aim1732 said:
No I meant the opposite. It should be
-g*sin15.2 - µ*g*cos15.2 = -6.615.
I had only quoted you.
One more thing, why are you always adding the the terms on left when there is a unknown
µ with one of the terms. Like here(this is yours):
6.615 = -2.57 + 9.457 µ
6.615 = 6.887 µ​
[/CENTER]

So I redid it:

-g*sin15.2 - µ*g*cos15.2 = -6.615.
-2.572 - 9.457 µ = -6.615
-9.457 µ = -4.043
µ = 0.4275

Is that correct?
 
Yes looks good to me.
 
Back
Top