# Inclined plane question?

## Homework Statement

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?

## 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. :)

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

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

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?

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.

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.

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. :)

After he reaches Vmax, the incline becomes 0°

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

SammyS
Staff Emeritus