Limitless world…

Endless work…

Fair

Ffriction

Categorize1.The athlete is the system.2.When running, air does neg

ative work, and the track does positive work, so the energy transfer can be E = Ek + Wair

3.Suppose the athlete’s mass is 70 kg, DA = 1 kg/m

Getting down to business…

For short-distance running race

1. High acceleration phase

2. Low acceleration phase

3. Maximum velocity phase

4. Deceleration phase

Results…

For a well trained sprinter, his vibration of his center of gravity in running is small. Just raise about 5cm

Wu=mgh=70*10*0.05=35JVaven is the average speed of the nth phase.

D=1, ρ=1.2kg/m^3 , A=0.6m^2Wairn=0.5*ρDAVaven^2

ΣWu=35J*40W1=Wair1+△ K1 =2801 J

W2=Wair2+ Wu *22+△ K2=4688 JW3=Wair3+ Wu *13+△ K3=4688 JWtot=ΣWair+ΣWu+(Kf-Ki)=12930 J

P = Wtot/t= 1334W

To improve the result by 10%, then the time is t’=8.72s Then Wtot’ =13249 J

The extra energy △W=319 J The extra power P=185 W△

Hurdling

ConceptualizeBecause these are short-distance running ra

ces, the change of the height of his body could not be ignored. His body is lifted ap

proximately 10cm per stride.Assume his body temperature does not chan

ge – neglect the internal energy.Kinetic energy and air resistance should be c

onsidered.When the athlete jump above the hurdles, th

e upper part of his barely moves. But his legs are lifted to the height of those hurdle

s.

Suppose the athlete’s mass is 70 kg, DA = 1 kg/m.110m hurdle race

The acceleration phase (0 m to about 21m)In this phase, the athlete accelerates from 0m/s to about 8.7m/s

Wair1 = (1/2)·(1/2 DA·Vmax^2) S1= 399 JEk = (1/2)·(1/2 m·Vmax^2) = 1325 J

W1 =Ek + Wair 1 = 1724 JThe stable-velocity phase. (to the end).

In this phase, the velocity can be considered stable, for it doesn’t change too much. It is about 9m/s.

Wair2 = (1/2 DA·Vave^2)S2 = 3649 J Ek = 1/2 m·Vave^2 = 2800 J

W2 =Ek + Wair2 = 6449 JThroughout the whole race, there are approximately 80 strides. That is to say, the at

hlete’s body is lifted 10cm for 80 times. △Ep 1 = 80mgh = 5488 J

When passing the hurdles, his legs are lifted. And there are 10 hurdles along the way.

△Ep 2= 10·[(1/2)·(1/2 m)·ghhur] = 1830 JThe total amount of energy Wm1 = W1 +W2+ Ep 1+ Ep 2 = 15491 J△ △

The power P1 = Wm1/t = 1198 W

Details…

Finalize To improve the result by 10%, the time of 110 m hurdle race will be 1

1.64s and 400m hurdle race 42.53s. Thus, the extra energy We 1= 1524 J

The extra power P1’= (Wm1+ We 1)/t1’ — P1= 264 WFrom the result, we can tell that improving performance in these running races needs great efforts as well as special techniques, for it requires

so much to improve. And here’s a graph which can approximately describe the relationshi

p between distance and energy during the 110m hurdle race.

Thank you !