In the sport of skeleton a participant jumps onto a sled (known as a skeleton) a
ID: 1614853 • Letter: I
Question
In the sport of skeleton a participant jumps onto a sled (known as a skeleton) and proceeds to slide down an icy track, belly down and head first. The track has sixteen turns and drops 122 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 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 91 mi/h). How much work is done on a rider and his sled (assuming a total mass of 83.7-kg) by nonconservative forces?
Explanation / Answer
a)
Ignoring friction .. GPE lost = KE gained
mgh = 1/2mv^2
v = sqrt(2gh)
v = sqrt(2 x 9.80 x 122m)
v = 48.9 m/s
b)
KE gained without friction = 1/2m(48.9 m/s)^2
KE gained with friction = 1/2m(35.8m/s)^2
Difference in KE gains = WD against friction
KE = (1/2*83.7kg) {35.8^2 - 48.9^2}
WD = -4.6435 x 10^4 J