when i solved i got a) 45.14m/s b)-31,977.51J but it was WRONG:(, if you could h
ID: 1912343 • Letter: W
Question
when i solved i got a) 45.14m/s b)-31,977.51J but it was WRONG:(, if you could help me get the RIGHT answer that woulbe great One of the new events in the 2002 Winter Olympics was the sport of skeleton. 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 104 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 a track with a 84 m drop in elevation? 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 of a 104 m drop with a speed of 35.8 m/s (about 80 mi/h). How much work is done on an 84.6 kg rider and skeleton by nonconservative forces?Explanation / Answer
by conservation of energy,
for 84m drop
mgh = 1/2mv^2
9.8*84 = 0.5*v^2
v = 40.58 m/s
****************************************
for 104m drop
mgh = 1/2mv^2
v = 45.15 m/s
work done by nonconservative force = change in kinetic energy
= -1/2*m*(45.15^2 - 35.8^2)
= -32010 J
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