The problem states that a skier is on a 20m hill where the bottom makes a 20 degree angle with the horizontal. if the hill has a coefficient of friction of .210, and then the ground levels and the flatland has the same coefficient (.210) what is the horizontal distance the skier travels before coming to a stop. (horizontal after the hill)
The work of the non-constant force on the hill is just the negative force of friction times the
change of x. This equals (KE(f) - KE(i)) + (PE(f) -PE(i)). Since KE(i) and PE(f) are zero the whole equation becomes 0.210 * the normal force * the change of x = .5v^2 - (20 * 9.8).
(all the masses cancel out)
The Attempt at a Solution
Now to solve for the normal I used 9.8cos(20). The normal being mg on a 20 degree hill.
After the hill the normal became 9.8 (all mass cancels out). After solving for v then plugging into v^2 = Vo^2 +2ax I got 40.6. (a = 2.058) The answer in the book is 40.3m.
I redid the problem using the normal as just 9.8 for the friction force on the hill. This I got 36.8m.
The first two examples I was using the change of x to be the hypotenuse of the triangle.
(20/sin(20)) When I used the change of x to be just the x direction of the hill and the normal to be just 9.8, I got 40.3m which is the answer in the book. I was wondering why the non-constant force on the slope would just be .210m(9.8) times 54.9m instead of .210m(9.8)cos(20) times 58.5m.
I'm sorry all this reads confusing but if you write it out it makes it a lot easier.