Easy Conservation of Energy Problem

[BioBabe91]

Homework Statement

A skier of mass 55.0 kg slides down a slope 11.7 m long, inclined at an angle θ to the horizontal. The magnitude of kinetic friction is 41.5 N. The skier's initial speed is 65.7 cm/s (0.657 m/s) and the speed at the bottom of the slope is 7.19 m/s. Determine the angle θ from the law of conservation of energy. Air resistance does not matter.

Homework Equations

Ek = 1/2*m*v^2
Eg = mgh
Ethermal = W = Fk*cosθ*Δd

The Attempt at a Solution

Eth= Fk*cosθ*Δd
Substituted in the values, then used conserv. of energy:
mgh + 1/2*mv^2 = 1/2*mv^2 + Eth
where h = 11.7/sinθ.
But I got stuck solving for the angle. Where did I go wrong?

 

[LowlyPion]

Welcome to PF.

I think the work from the kinetic friction is taken over the length of the incline because the cosφ relative to the motion of the skier is 0, along the plane of the slope, not the angle θ of the incline.

Otherwise, I think you have the right idea.

 

[BioBabe91]

That's true. But in what cases would cosφ not be equal to 0, for future questions like this?

Thanks.

 

[LowlyPion]

That's true. But in what cases would cosφ not be equal to 0, for future questions like this?

Thanks.

For friction I'd guess that the angle φ is usually always 0, if it acts against the direction of motion. Now on circular motion problems, friction that would be say keeping a car from sliding down an incline would contribute no energy loss because there the friction would be at 90°

Work = F ⋅ D

(which is the dot product if you are familiar with vectors)