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 the beginning of the run is relatively small and can be ignored. (b) In reality, the gold-medal winner (Canadian Jon Montgomery) reached the bottom in one heat 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 nonconservative forces during this heat?
To answer part (a) of the question, we can use the principle of conservation of mechanical energy. In the absence of nonconservative forces, the total mechanical energy of the rider-sled system would be conserved throughout the descent.
The mechanical energy of an object is the sum of its kinetic energy (KE) and potential energy (PE). At the beginning of the run, the total mechanical energy is given by:
Ei = KEi + PEi
Since the speed at the beginning of the run is considered to be relatively small and can be ignored, we can assume the initial kinetic energy is zero:
KEi = 0
The potential energy at the beginning of the run is given by:
PEi = m * g * hi
Where m is the mass of the rider and sled, g is the acceleration due to gravity (approximately 9.8 m/s^2), and hi is the initial height of the track.
As the rider slides down the track, the potential energy is converted into kinetic energy. At the bottom of the track, when the rider and sled have reached their maximum speed, the potential energy is zero. Therefore, the total mechanical energy at the bottom is given by:
Ef = KEf + PEf
= KEf + 0
= KEf
So, to find the speed of the rider at the bottom, we need to equate the initial total mechanical energy to the final kinetic energy:
Ei = Ef
KEi + PEi = KEf
0 + m * g * hi = (1/2) * m * vf^2
Simplifying the equation:
g * hi = (1/2) * vf^2
Solving for vf:
vf = sqrt(2 * g * hi)
Since the question specifies the height of the track drop as 126 m, we can substitute these values into the equation:
vf = sqrt(2 * 9.8 m/s^2 * 126 m)
vf ≈ 39.70 m/s
Therefore, in the absence of nonconservative forces, the speed of the rider at the bottom of the track would be approximately 39.70 m/s.
To find part (b), we need to determine the work done by nonconservative forces (such as friction and air resistance) during this heat. The work done by nonconservative forces is equal to the change in mechanical energy.
Work = ΔE = Ef - Ei
Work = KEf - KEi + PEf - PEi
Given that the speed of the rider at the bottom is 40.5 m/s, we can determine the final kinetic energy:
KEf = (1/2) * m * vf^2
KEf = (1/2) * 118 kg * (40.5 m/s)^2
Therefore, the work done by nonconservative forces during this heat can be calculated as:
Work = KEf - KEi + PEf - PEi
Since the question doesn't provide any information about the initial height of the track, we assume it to be zero. This means the initial potential energy is also zero.
Work = KEf - 0 + PEf - 0
Work = KEf + PEf
To find the work done on the rider and sled, we need to calculate the final potential energy:
PEf = m * g * hf
Substituting the given values, where hf is the height of the track at the bottom:
PEf = 118 kg * 9.8 m/s^2 * (-126 m)
Now, we can calculate the work done by nonconservative forces:
Work = (1/2) * 118 kg * (40.5 m/s)^2 + 118 kg * 9.8 m/s^2 * (-126 m)
Simplifying the equation:
Work ≈ -355088 J
Therefore, the work done on the rider and his sled by nonconservative forces during this heat is approximately -355,088 J.