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 100 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.
m/s
(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 84.4-kg rider and skeleton by nonconservative forces?
a. V^2 = Vo^2 + 2g*h.
V2 = 0 + 19.6*100 = 1960
V = 44.3 m/s.
To solve part (a) of the problem, we can apply the principle of conservation of mechanical energy. In the absence of nonconservative forces like friction and air resistance, the total mechanical energy of the system would be conserved.
The mechanical energy of the system is given by the sum of the potential energy (PE) and the kinetic energy (KE):
E = PE + KE
At the top of the track, all of the potential energy is converted into kinetic energy. Since the potential energy is given by mgh (mass times acceleration due to gravity times height), the initial potential energy at the top is mgh, where m is the mass of the rider, g is the acceleration due to gravity, and h is the height of the track. The initial kinetic energy can be ignored if the rider's initial speed is relatively small.
Therefore, when the rider reaches the bottom of the track, the only energy present will be the kinetic energy. The kinetic energy can be calculated using the equation KE = (1/2)mv^2, where m is the mass of the rider and v is the velocity at the bottom.
Setting the initial potential energy equal to the final kinetic energy, we have:
mgh = (1/2)mv^2
Simplifying, we find:
gh = (1/2)v^2
Solving for v, we get:
v = sqrt(2gh)
Substituting the given values, we have:
v = sqrt(2 * 9.8 m/s^2 * 100 m)
v = sqrt(1960 m^2/s^2)
v ≈ 44.27 m/s
So, in the absence of nonconservative forces, the speed of the rider at the bottom of the track would be approximately 44.27 m/s.
To solve part (b) of the problem, we need to find the work done by nonconservative forces, which is given by the difference in mechanical energy before and after the ride on the track:
Work = ΔE = Efinal - Einitial
Since the rider reaches the bottom with a speed of 35.8 m/s, we can calculate the final kinetic energy as:
KEfinal = (1/2)mvfinal^2
Substituting the values, we have:
KEfinal = (1/2)(84.4 kg)(35.8 m/s)^2
Calculating this, we find:
KEfinal ≈ 53,562.36 J
Using the same equation, we can calculate the initial kinetic energy using the speed from part (a):
KEinitial = (1/2)mvinitial^2
Since the initial speed is relatively small and can be ignored, we can consider the initial kinetic energy to be zero.
Therefore, the work done by nonconservative forces is given by:
Work = KEfinal - KEinitial
Work = 53,562.36 J - 0 J
Work ≈ 53,562.36 J
So, the work done on the 84.4-kg rider and skeleton by the nonconservative forces is approximately 53,562.36 Joules.
(a) To find the speed of the rider at the bottom of the track in the absence of nonconservative forces, we can apply the principle of conservation of mechanical energy. At the top of the track, the rider has gravitational potential energy, which is converted into kinetic energy at the bottom of the track.
The gravitational potential energy at the top of the track is given by the formula:
PE = mgh
where m is the mass of the rider, g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the height of the track (which we know to be 100 m).
PE = (84.4 kg)(9.8 m/s^2)(100 m)
PE = 82,872 J
This gravitational potential energy is converted into kinetic energy at the bottom of the track, given by the formula:
KE = (1/2)mv^2
where m is the mass of the rider and v is the speed of the rider at the bottom of the track.
82,872 J = (1/2)(84.4 kg)v^2
v^2 = (2 * 82,872 J) / (84.4 kg)
v^2 = 1,959.12 m^2/s^2
Taking the square root of both sides, we find:
v = √(1,959.12 m^2/s^2)
v ≈ 44.24 m/s
Therefore, in the absence of nonconservative forces, the speed of the rider at the bottom of the track would be approximately 44.24 m/s.
(b) To find the work done by nonconservative forces, we can use the work-energy principle, which states that the net work done on an object is equal to its change in kinetic energy.
The change in kinetic energy can be calculated as:
ΔKE = (1/2)mv_f^2 - (1/2)mv_i^2
where m is the mass of the rider, v_f is the final velocity (35.8 m/s), and v_i is the initial velocity (which we can assume to be zero since the problem states the initial speed is relatively small and can be ignored).
ΔKE = (1/2)(84.4 kg)(35.8 m/s)^2 - (1/2)(84.4 kg)(0 m/s)^2
ΔKE = (1/2)(84.4 kg)(35.8 m/s)^2
ΔKE = 68,209.048 J
Therefore, the work done by nonconservative forces is equal to the change in kinetic energy, which is approximately 68,209.048 J.