Skier skies down slope, including friction, calculate final speed

[wiggle]

Homework Statement

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.

Homework Equations

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

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.

 

[Doc Al]

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]

Looks good to me.