Question

One of the new events in the 2002 Winter Olympics was the sport of skeleton (see photo). Starting at the top of a steep, icy track, a rider jumps onto a sled (known as a skeleton) and proceeds-belly down and head first-to slide down the track. The track has fifteen turns and drops 118 m in elevation from top to bottom.

(a) In the absence of nonconservative forces, such as friction and air resistance, what would be the speed of a rider at the bottom of the track? Assume that the speed of the rider at the beginning of the run is relatively small and can be ignored.

(b) In reality, the best riders reach the bottom with a speed of 35.8 m/s (about 80 mi/h). How much work is done on an 80.2-kg rider and skeleton by nonconservative forces?

I got a) 48.1 m/s for b) 41382 J.

It said A is correct. B incorrect

Answer 1

When there is no work done by non-conservative forces, we can say Ei = Ef
Ei = Potential energy = mgh = mg(104)
Ef = Kinetic energy = 1/2 mv^2

Equate the two and solve for v, note m's cancel

In part b, since non-conservative work is done, we get

Wnc = Ef - Ei
Again Ei and Ef are PE and KE, but now use 35.8m/s for speed and m's do not cancel, m=85.8kg
Note since the non conservative forces remove energy from the rider, we expect final energy to be less than initial, so Wnc is less than zero.