How to Calculate Work at a 30 Degree Angle on a Ski Slope?

[bearhug]

A skier of mass 71 kg is pulled up a slope by a motor-driven cable.
(a) How much work is required to pull him a distance of 55 m up a 30° slope at a constant speed of 1.1 m/s

My whole problem with this question is that I'm not sure if I need to include ∑Fx and ∑Fy (where ∑Fx = mgsinӨ=ma and ∑Fy= n-mgcosӨ) in order to find F. This is where I'm getting stuck to solve for W
W=FcosӨ Δr
I just need help in getting started with this problem because I keep getting stuck. Any help would be greatly appreciated.

 

[Rozenwyn]

I think you can just use the simple W = F \times d equation for this problem. The total distance can be calculated using trigonometry.
You know that F = m \times a \ \Longrightarrow \ F = m \times \frac{d}{t^2}.
And you can also calculate the total time it takes to move the object to the final destination since you know the distance and the velocity of it. I think this would work.

 

[Astronuc]

W = mg = 71 kg * 9.8 m/s².

However, the Weight points directly down with gravity. Weight (force) is a vector, and it can be resolved into two components, one normal to the slope surface, and one parallel, pointing down hill. The cable would be pulling up hill at constant speed. Constant speed means not acceleration so no net force.

 

[bearhug]

I'm still confused as to what you are describing. Plus I figured vectors were involved however my book doesn't really explain how to use vectors in relation to this sort of problem. Could you explain a little more any help is appreciated.