In the sport of skeleton a participant slides down an icy track, belly down and head first. In the 2010 Winter Olympics, the track had sixteen turns and dropped 132 m in elevation from top to bottom. The gold-medal winner (Canadian Jon Montgomery) reached the bottom in one run with a speed of 40.5 m/s (about 91 mi/h). How much work was done on him and his sled (assuming a total mass of 118 kg) by non-conservative forces during this run?
In the absence of friction, loss in potential energy = gain in the kinetic energy
So, m*g*h = (1/2)*m*v^2
=> v^2 = 2*g*h
=> v = sqrt(2*g*h) = (2*9.8*132) = 50.9 m/s
Now -
Kinetic Energy gained without friction = ½m(50.9 m/s)²
Kinetic Energy gained with friction = ½m(40.50 m/s)²
Difference in KE gains = WD against friction
?KE = (½ 118kg) {50.9² - 40.50²} = 59*(950.56) = 56083.04 = 56.08
kJ
Therefore, work done on him and his sled by non-conservative forces = 56.08 kJ.
In the sport of skeleton a participant slides down an icy track, belly down and head...
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. In the 2010 Winter Olympics, the track had sixteen turns and dropped 126 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...