Work and Energy: Applications to Sports



Work and Energy of a Sprinter

The work done to accelerate a sprinter and increase her/his kinetic energy comes mainly from the compression and extension of muscle fibers and tendons, which act like springs. The work done by a muscle in a contraction is the product of the force it exerts and the distance it shortens. The amount of force a muscle can apply for a given contraction depends on the speed of contraction; the slower the contraction the greater the force.

Film of a male sprinter (mass = 70 kg) showed that he accelerated from rest to 3.0 m/s in his first step, to 4.2 m/s in his second step (while his other foot was on the ground) and to 5.0 m/s in the third step.

The kinetic energy of the sprinter can be calculated for each step.
 

Step u (m/s) KE (J) DKE (J)
0 0 0 -
1 3.0 315 315
2 4.2 617 302
3 5.0 875 258
Each change in kinetic energy came from work done by his muscles. The change decreases slightly for each step because there is less time between steps and this decreases the amount of work his muscles can do for the same contraction. Thus the increase of energy cannot go on forever! The work done by the muscles while the sprinter is moving with constant speed is serving to make up for various energy losses: friction in the joints, air resistance, and heat losses in the tendons as they stretch and recoil.

About 2.5% of the metabolic energy (measured by oxygen consumption) is used to overcome air drag at a speed of 6 m/s in still air, about 13% at a sprinting speed of 10 m/s. This is consistent with the fact that air resistance is a function of speed.
 

How High Can You Jump?
 
 
Suppose that you want to jump as high as possible from a standing start. You bend your legs and then extend them rapidly. See how your center of gravity changes while you are still on the ground: 
 
Position
Center of Gravity-to-Ground Distance
1. Standing erect 1.0 m
2. Knees bent 0.65 m
3. End of take-off 1.05 m
How far have you accelerated your center of gravity? (Position 3 - Position 2) _____________

Studies of athletes jumping from force plates indicate that the maximum force exerted against the ground is about 2.3 times body weight W and that the average force is about 2W. (The average reaction force of the ground against the feet is also 2W and is the accelerating force on the jumper; Newton’s third law tells us this.)

 Equate the work done by the jumper to the potential energy change of the jumper-earth system to arrive at the final position of the jumper’s center of gravity. What value do you get?

 
a
V  u
        l
e         t
l            i
o              n
P                  g
 
 

A good estimate of the height reached by a pole vaulter comes from a conservation of energy calculation. Here is an example. The final speed of a vaulter (just as the pole is planted in the ground) is 9.5 m/s. The jumper’s center of gravity is 0.9 m above the ground. At the top of the jump, the center of gravity is 0.1 m above the bar. Use conservation of energy to find the height of the bar above the ground. (The world record is a little over 6 meters.) Why don’t you need to know the athlete’s mass?

The energy transfer processes involved in a pole vault are somewhat complicated, but the basic calculation done above gives a good estimate of the final result. The athlete does some work, pulling down on the pole while rising. Much of the work on the athlete is done by the pole, through the release of elastic potential energy. There is some loss of energy — the pole warms up a little. And the vaulter has a small horizontal velocity at the top, and so retains some kinetic energy. But the basic energy conversion relationship holds:

Kinetic energy -> elastic strain energy -> gravitational potential energy.

Return