Coefficient of friction-Puck travelling up an incline

[ThomasMagnus]

Coefficient of friction--Puck traveling up an incline

Homework Statement

A hockey puck is moving at 13m/s when it starts climbing an incline. It travels 7.1m from the base of the incline before sliding back down the slope, as shown

Find the coefficient of friction between the puck and the incline.

The Attempt at a Solution

First find acceleration of the puck
Vf=final velocity=0
Vo=initial velocity= 13m/s
d=distance=7.1
V_f²=V_o²+2ad
-169=14.2a, a=11.9m/s²
\SigmaF_x=ma_x
Since the puck is accelerating up the incline, the Force of Friction and the Force Parallel will be in the same direction.

F_F+F_||=ma_x

F_F=(\mu) F_N

F_N=mgcos(\theta)

F_||=mgsin(\theta)

(\mu)mgcos(\theta) + mgsin(\theta)= 11.9m

m(\mugcos(36)+gsin36)=11.9m

Divide both sides by m, m cancels.

\mu=11.9/9.8cos36+9.8sin36

\mu=.87

The answer in the book is .77. Can you help me with where I am going wrong?

Thanks! =)

 

[tiny-tim]

Hi Thomas!

(have a mu: µ and a theta: θ and a sigma: ∑ and a degree: ° )

m(\mugcos(36)+gsin36)=11.9m
…
\mu=11.9/9.8cos36+9.8sin36

Nooo

 

[ThomasMagnus]

I have honestly tried this over and over and keep ending up here. Where am I going wrong?

Thanks!

 

[ThomasMagnus]

I think I see the error.

mu=11.9-9.8sin36/(9.8cos36)

 

[ThomasMagnus]

11.90140845070423-5.760295472466237/(7.928366544874485)

6.141112978237993/7.928366544874485
=.77!

Thanks!

 

[tiny-tim]

mu=11.9-9.8sin36/(9.8cos36)

hmm …

i think you'll get less confused in future if you get into the habit of putting brackets in the right places!

 

[ThomasMagnus]

Sorry I was typing it On my iPod :)