# Homework Help: Inclined plane question?

1. Mar 3, 2012

### nesan

1. The problem statement, all variables and given/known data

Boom-Boom Slapshot, Canadian hockey star, slides down a 50m long ice covered hill on his skates. The frictionless hill is inclined at 35° to the horizontal. Once he reaches the bottom of the hill, the ice is covered with deep snow that has a coefficient of kinetic friction of .50
How far into the snow will BoomBoom go before coming to rest?

3. The attempt at a solution

I've figured out acceleration on the inclined plane which is 5.62m/s^2. But, I was having trouble with including friction into the other part as well as normal force.

a = sin35 * 9.8 = 5.62m/s^2

Using the formula Vf^2 = Vi^2 + 2(a)(d)

I found the velocity which is 23.71m/s

I got an answer of 175m but the text book says the answer is 57m. The textbook has been wrong few times, I just want to confirm this, thank you. :)

2. Mar 3, 2012

### Browne

How did you use that velocity to get 175m?
When I calculated it, I got 57m.

3. Mar 3, 2012

### nesan

a1 = g*sin35° = 5.62 m/s²

Vmax = √[2*a1*s] = √[2*5.62*50] = 23.7 m/s

a2 = -g[sin35° - µ*cos35°] = -1.607 m/s²

d2 = -Vmax²/(2*a2) = 174.76 m

Did I do something wrong?

4. Mar 3, 2012

### Browne

Ah, the problem is the interpretation of the question.
I think the problem was intended to mean that he starts travelling horizontally (at the bottom of the hill) when he enters the snow.

5. Mar 3, 2012

### nesan

Can you expand on that please? I'm really confused after trying 4 times.

Thank you. Just pointing me in the right direction will do. :)

6. Mar 3, 2012

### Browne

After he reaches Vmax, the incline becomes 0°

7. Mar 4, 2012

### nesan

I need more help please, I'm kind of lost. =\

8. Mar 4, 2012

### SammyS

Staff Emeritus
What is it that you don't understand regarding what Browne said in his last two posts?

It would be difficult to restate what he said in a manner that's any clearer !