Master Answers.
| question | possible points | points |
| 1 | 10 | |
| 2 | 10 | |
| 3 | 40 | |
| 4 | 20 | |
| 5 | 30 | |
| 6 | 30 | |
| 7 | 20 | |
| 8 | 20 | |
| total | 180 |
You may use one sheet of helpful information for this test. This information should be made by the student and could contain useful information like conversion factors and formulae as well as example problems. You will be permitted one sheet for each of the tests. No books or notebooks are permitted.
You may leave your answers in unrationalized form. eg Kinetic E = 1/2 * 4 *30^2 Joules. There is no need to use a calculator to multiply out the numbers. The quantities must be in a consistent set of units. The preferred units are the International units….. m, sec, joules and watts. Answers with mixed units will be marked-off.
To set the locomotive of very large mass into motion, the mass must be accelerated. If the wheels/track combination are frictionless, the acceleration of the mass can be obtained with any force above zero. a = F/m. The smaller the force, the smaller the acceleration so that it would take a long time for very small forces to move 10m.
2. A net force of 25 N in the +x direction is applied to a 70kg mass for a time of 20 seconds. What is the velocity of the mass at the end of this time? (magnitude and direction)
a=F/m and v = at = F/m*t = 25/70*20 = 7.1m/s. A rather modest force applied for a longish time obtains a velocity of a marathon runner.
3. A 70 kg athlete thrusts with his legs to achieve a jump of 0.5 m.
a)What is the takeoff speed?
To rise the 0.5m, the take off speed can be calculated from the landing speed v^2=2gy = 2*10*0.5 = 10, v = 3.3m/s
b)If the jump is accomplished with legs only, by thrusting the CM a distance of 0.5m, what acceleration is necessary?
This is the same calculation as above, but we solve for a in v^2 = 2ay. a = 10m/s/s
c)What is the net force during the thrust? ( Magnitude and direction.)
F = ma = 70*10 = 700N in the upward direction.
d) What is the thrust force with the legs?
The net force results from the athletes weight plus the reaction force from the floor. 700 = Reaction – wt = Reaction – 700.
Reaction force is 1400 N. The thrust from the legs downward must be 1400 N.
4. A 70 Kg athlete can jump 0.6m and wants to know how high he can jump if he weighed 60 kg where the weight loss is all in fat. He retains all his muscle mass.
a) How high can he jump as a 60 Kg athlete?
Assuming that he can produce the same energy at both weights, the smaller weight can rise higher.
Mgh = m’g h’. h’ = m/m’ * h = 70/60 * 0.6 = 0.7 meters.
b) How much higher can he jump if he were to increase the force of his legs by 20% through weight training? [without weight gain].
Assuming his mass remains the same and he increases the force obtained by 20%, the energy of the jump increases by 20% . The jump height is increased by 20% from 0.6m to 0.72m. If the athlete had the weight loss plus increase the force of the thrust, the jump would increase from 0.6 m to 0.84m!! [Both a) and b) analysis are simplified. For the athlete to take full advantage of the increase force and decrease in mass, he must also increase his muscle speed.] h’ = 1.2 m/m’* h for the general case.
5. Using the scaling laws for animals and for the muscles, indicate the height dependence on the following sports.
a) jump reach (raise CM a maximum height).
The scaling law for launch speed is independent of height so any person is competitive for the jump reach.
b) Gymnastics
There is a premium for acceleration in gymnastics. a varies as 1/L The athletes with the most acceleration are the smallest so the sport favours short athletes.
c) High Jump
Since the jump reach has no height dependence, the raising of the CM does not favour taller athletes. However, the taller athletes start with their CM higher, the higher bars can be cleared by the same rise in CM so the sport favours taller athletes.
6. A 1000 Kg car can reach 30m/sec in a time of 10 sec.
a) What is the average power required for this task?
The average power = energy/time = ½*mv^2/time = ½ * 1000 * 30^2/10 = 45,000 watts
b) What is the average force required?
F = m*a = m*v/t = 1000*30/10 = 3000N in the direction of the car’s motion.
c) How much average power is required to accelerate up to 30m/s in 5 seconds?
Same calculation as a) except time is decreased by ½ . average power = 90,000 watts
7. Two pole vaulters are competing. Vaulter A is 2 m tall and has a mass of 90Kg. Vaulter B is 1.5m tall and has a mass of 60 Kg. They both have top speeds of 10m/s and have perfect technique for the pole vault so that they can convert all of their running kinetic energy into height for the vault.
a) How high can each vaulter raise the CM?
For each vaulter ½*mv^2 = mgh. So h = ½ v^2/g = ½*10^2/10 = 5m. Independent of the mass of the vaulter.
b) Which vaulter will win the event?
Since the contest is to see who clears the bar , the vaulter who starts out with the higher CM will win. This means that the taller of the two vaulters will win the contest.
8. An athlete can be seen as a high performance machine that can convert chemical energy [food] into kinetic energy with a high rate and with a high efficiency and with a machine that doesn’t suffer a breakdown. The Tour de France, endurance bicycle race has each rider on each of about 30 successive days carry out the equivalent energy conversion of 2 marathon running races. We know that the marathon runners cannot run 2 marathons a day for 30 successive days. The machine breaks down. What is the main factor that enables the bike riders to carry out this incredible task of energy conversion from food to kinetic energy? What must the bike rider take in large quantities along with his food to make it possible. [In this question, you should ignore the shock of landing each step for the runner and its effect on the joints.] Hint. Remember that in the course of the Tour de France, the riders ride through the hot plains of France in midsummer. Think of what this would do to a marathon runner.
The continued production of mechanical energy by the athlete must be accompanied by heating of the body. If this heat is not completely dissipated, the body will heat up which results in the failure of the athlete when his temperature goes into the fever state. Even death can result from this. The marathon runners are usually flirting with disaster in the heating department because the radiation of heat is insufficient and the evaporation which is the most efficient heat dissipation element is not effective at the low speeds of runners. The cyclists on the other hand have quite high speeds so that the cooling by evaporation works very well… at least when they are on the flat. Their body is kept in good shape by maintaining a proper temperature. In order for cooling by evaporation to work, they must drink a lot of water [ 4 gallons/day on the tour 30 lb of water which is effectively dissipated each day. Those puny cups of water taken by marathoners at the watering stations show how little the marathoners dissipate in cooling water during the race] .