One of the new events in the 2002 Winter Olympics was the sport of skeleton (see the 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 104 m in elevation from top to bottom. (a) In the absense of non-conservative 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 82.0-kg rider and skeleton by non-conservative forces?
To solve this problem, we can make use of the principle of conservation of mechanical energy. According to this principle, the total mechanical energy of a system remains constant in the absence of non-conservative forces.
(a) In the absence of non-conservative forces, the total mechanical energy of the rider-sled system is conserved. The mechanical energy of an object includes its potential energy (PE) and kinetic energy (KE).
At the top of the track, the rider-sled system has only potential energy. The potential energy at the top can be calculated using the formula:
PE = mgh
Where:
m = mass of the rider + sled system
g = acceleration due to gravity (approximately 9.8 m/s^2)
h = height (change in elevation) of the track (104 m)
Substituting the given values:
PE = (82.0 kg)(9.8 m/s^2)(104 m)
PE = 83,578.4 J
At the bottom of the track, all of the potential energy is converted into kinetic energy. The kinetic energy of an object can be calculated using the formula:
KE = (1/2)mv^2
Where:
m = mass of the rider + sled system (82.0 kg)
v = velocity of the rider-sled system at the bottom of the track (unknown)
Since the total mechanical energy is conserved, we can equate the potential energy at the top to the kinetic energy at the bottom:
PE = KE
83,578.4 J = (1/2)(82.0 kg)(v^2)
Solving for v:
166,156.8 J = 41.0 kg (v^2)
v^2 = 166,156.8 J / 41.0 kg
v^2 = 4,048.9659 m^2/s^2
Taking the square root of both sides:
v ≈ 63.63 m/s
Therefore, in the absence of non-conservative forces, the speed of the rider-sled system at the bottom of the track would be approximately 63.63 m/s.
(b) In reality, due to non-conservative forces like friction and air resistance, the best riders reach the bottom of the track with a speed of 35.8 m/s.
To calculate the work done by non-conservative forces, we need to find the change in mechanical energy of the rider-sled system:
ΔE = KE_final - KE_initial
ΔE = (1/2)(82.0 kg)(35.8 m/s)^2 - (1/2)(82.0 kg)(63.63 m/s)^2
Calculating this expression gives us:
ΔE ≈ -53,253.609 J
The negative sign indicates that work is done on the rider-sled system by non-conservative forces. The magnitude of the work done is approximately 53,253.609 Joules.
Therefore, the work done on the rider-sled system by non-conservative forces is approximately 53,253.609 J.