A skier starts at rest at the top of a slope 65m high. If frictional force is -10.5kj how fast is she going?

Question

ok if she 60 kg how fast is she going

Answer 1

You need to know the mass of the skier in order to compute. By the way, kj are units of energy, not force (N)
I will give the method and express the answer in terms of the mass, m, of the skier using conservation of energy including the frictional energy loss

m*g*h=.5*m*v^2-10500
v=sqrt(2*g*h+10500*2/m)
v=sqrt(1275.3+21000/m)

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Answer 2

I don't know about you, but I really like to know the physics behind equations. In your case:

potential energy - lost energy due to friction = mgh - Fd = 1/2 mv^2 = kinetic energy = KE. m = mass of the skier, h = height of the hill = 65 m, v = velocity (speed) at the bottom of the hill (the answer you are seeking), and Fd = the friction force (Fd = kNd = kWcos(theta)d = kmg(cos(theta))d.) times the distance (d) the skier travels down the slope. d = h/sin(theta) where theta is the slope of the hill relative to the horizontal.

Fd is the so-called work function. Work is equivalent to energy expended. Thus, as the skier fights the friction force, Fd Newton-meters in energy are sapped from the PE the skier started with at the top of the hill. This makes v, at the bottom, somewhat less than it would have been without friction.

Notice you need to know the slope of the hill. Think about it. In your experience, did you slide faster down a steep hill or down a shallow hill? Uh huh...that's why you need to know the slope.

And thanks for the mass, but that's not really needed...contrary to what one person said. Check this out: PE - Fd = mgh - kmg(cos(theta))d = 1/2 mv^2 = KE . Notice this reduces to gh - kg(cos(theta))h/(sin(theta)) = 1/2 v^2 = gh(1 - k cot(theta)); so that v^2 = 2gh (1 - k cot(theta)) and you can solve for v without ever knowing the mass of the skier.

So now you know the rest of the story...the physics behind the equations. That is, potential energy is converted into kinetic energy. But some of the PE is lost to friction energy (work); so the velocity at the bottom of the hill would be less than it would be otherwise without friction.