# Skier skies down slope, including friction, calculate final speed

#### wiggle

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

A 50Kg skier skies down a 25 degree slope. at the top of the slope her speed is 4m/s and accelerates down the hill. the coefficient of friction is 0.12 between skies and snow. ignoring air resistance calculate her speed at point that is displaced 60m downhill.

2. Relevant equations

E=KE+PE
E=.5*m*V^2+m*g*h

3. The attempt at a solution

I believe I solved it correctly using E0=.5(50)4^2+50(9.8)(60sin(25)) and getting 12650

then setting 12650 = .5*m*Vf^2 and I'm getting Vf=22.5 m/s

However I'm unsure how to go about solving while taking into consideration the friction, since there is a non conservative force at work using KE0+PE0=KEf+PEf would get me close but not all the way there.

Related Introductory Physics Homework News on Phys.org

#### Doc Al

Mentor
The work done by friction will reduce the total mechanical energy.

#### wiggle

Right.

So I've just broken it up into x and y components

Fx=Wsin25-.12(Wcos(25))=m*a

I calculated for acceleration in the x plane, and found a=3.08 m/s^2

Then I just did

Vf=sqr(2ax+V0^2)
sqr(2(3.08)(60)+4^2)= Vf= 19.6 m/s

Does this seem correct?

#### Doc Al

Mentor
Looks good to me.

### Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving