What is the Work Done During a 100m Sprint?

[dboice95]

Homework Statement

How much work per step per leg in the race? How much work must he (single leg) squat to equate 1 step of work in the race?

The given variables are:
Displacement: 100 meters
Time: 9.65 seconds
runners weight in kg= ?

100 steps to complete 100m race, alternating legs.

Homework Equations

Force= Mass x Acceleration

Work = Force x Displacement

Acceleration = change in velocity/ change in time

velocity = displacement/ time

acceleration due to gravity = 9.81 m/s[/B]

The Attempt at a Solution

The problem was for usian bolt. so I google usain bolts mass in kg to be 94.

I calculated force per leg to be 94 x 9.81 = 922.14N

I then calculated work to be 922.14N x 1 (one step for one meter) = 922.14 JI'm not sure if any of this is correct or where to go from here. Can anyone help out?[/B]

 

[phinds]

Displacement: 100 meters

? I get zero displacement, assuming he starts off standing up and ends up standing up and the track is level (which they all are)

 

[haruspex]

Work = Force x Displacement

Only if force and displacement vectors are parallel. Otherwise you need to multiply by the cosine of the angle between them.
You have used a vertical force and a horizontal displacement.

Where does the question come from? Is this the entire question, word for word? There is so far insufficient information to provide an answer.

 

[dboice95]

This is supposed to be an extra credit problem from my professor's powerpoint slides. This is everything word for word. I thought there was insufficient information as well.

In the rest of the powerpoint slide there iss another equation for work, maybe this relates?
F= force, d= displacement, I'm not sure what variable r stands for.

Fd = 1/2m(vf-vi)^2 + mg(r final - r initial)Phinds I think the 100m displacement refers to the 100m length of the track.

 

[lekh2003]

Homework Statement

How much work per step per leg in the race? How much work must he (single leg) squat to equate 1 step of work in the race?

The given variables are:
Displacement: 100 meters
Time: 9.65 seconds
runners weight in kg= ?

100 steps to complete 100m race, alternating legs.

I don't think there is sufficient information. You need to know at what angle Usain Bolt pushes of from the floor. If the force he put was simply his weight, than he would just keep jumping up every step. You need to know how he kicked off. Then you can find the correct force vector. Ask your teacher for more information or whether they want some other kind of solution.

 

[haruspex]

Fd = 1/2m(vf-vi)^2 + mg(r final - r initial)

Good grief! I do hope the slide does not say that. It is quite wrong. It should say
Fd = 1/2m(vf²-vi²)+ mg(r final - r initial)
Also, note that the r values are heights.
Anyway, this equation is more useful for your question. You are given a time. Can you deduce anything about velocities?

 

[phinds]

Phinds I think the 100m displacement refers to the 100m length of the track.

Yes, and had it been stated as a length I would have no argument, but it was stated as a displacement, and I think of that as the distance through which an object is moved AGAINST the force acting on it. For example, if you take a 100lb box off of the back of one truck and carry it 100 yards at the same height to the back of another truck, then you have done no work because there is no displacement.

 

[CWatters]

Poor question in my opinion. All I can suggest is to note that his kinetic energy isn't the same at the start and finish.

 

[haruspex]

Poor question in my opinion. All I can suggest is to note that his kinetic energy isn't the same at the start and finish.

I regard it as quite a good question. As you say, and as I hinted in post #6, the right way to attempt the question is via KE. The temptation to (mis)use force x distance is a trap for the unthinking.

The worrying part is the incorrect equation for change in KE in post #5. Maybe it was misquoted.

 

[haruspex]

it was stated as a displacement

Well, it is a displacement, just not in the same direction as the weight.

 

[I like Serena]

I regard it as quite a good question. As you say, and as I hinted in post #6, the right way to attempt the question is via KE. The temptation to (mis)use force x distance is a trap for the unthinking.

The worrying part is the incorrect equation for change in KE in post #5. Maybe it was misquoted.

What do we need KE for? And what's wrong with force x distance?
If we assume constant acceleration a, we have W=F_net d, which happens to be equal to the final kinetic energy (neglecting friction and such). It seems to me that the kinetic energy is irrelevant though.
Given the distance and the time, we can calculate the acceleration a, and consequently the total work.

 

[haruspex]

what's wrong with force x distance?

Let me clarify..
If assuming uniform acceleration, there's nothing wrong with force x distance if that force is to be calculated from the necessary acceleration given the distance and time, as you suggest. Indeed, that is completely equivalent to using KE under that same assumption.
I wrote "(mis)use", i.e., in this case, using the weight of the runner as the force.

That said, it would not be correct to assume uniform acceleration. It would be more accurate to assume uniform power, and even then would overstate the final acceleration. The acceleration is much greater at the start, and almost nothing near the end. Even assuming uniform speed would probably give a more accurate result than using uniform acceleration.

 

[I like Serena]

Let me clarify..
If assuming uniform acceleration, there's nothing wrong with force x distance if that force is to be calculated from the necessary acceleration given the distance and time, as you suggest. Indeed, that is completely equivalent to using KE under that same assumption.
I wrote "(mis)use", i.e., in this case, using the weight of the runner as the force.

That said, it would not be correct to assume uniform acceleration. It would be more accurate to assume uniform power, and even then would overstate the final acceleration. The acceleration is much greater at the start, and almost nothing near the end. Even assuming uniform speed would probably give a more accurate result than using uniform acceleration.

I see what you mean now - assuming constant power instead of constant acceleration.
Either way, more realistic or not, doesn't it make the problem more difficult to solve than would be intended?
For the OP I would suggest to go with the constant acceleration approach.
I think that at this stage the important part is that we should not take gravity into account - even though it is given, and more generally to get the fundamentals of the work-force relation understood.

 

[haruspex]

Either way, more realistic or not, doesn't it make the problem more difficult to solve than would be intended?

Constant power does, yes, but constant speed is likely better as well as simpler than constant acceleration.

 

[CWatters]

What do we need KE for? And what's wrong with force x distance?

The displacement is horizontal. The only horizontal forces acting on him are inertia (when he's accelerating) and air resistance (which is unknown).

 

[dboice95]

So far this is where I am at:

I googled his mass because I'm not sure how else we are supposed to find usain bolts mass given the information.

one way I did this was with the equation W = Fd where F=ma and velocity = 10.36

acceleration = 10.36/9.65 and the 1 meter represents the distance he traveled for 1 step.

so I got work per step= (94kg) (1.07)(1m) = 100.58 J (work per step)The other way I did this is with the equation W = change in KE + change in PE + change in SE

PE= 0
SE= 0

Change in KE = .5(94)(10.36)^2 - .5(94)(0)^2

W = 5044.49J

I then divided 5044.49/50 because each leg steps 50 times.. and that equals 100.89J per step

The last part of the question asks how much "work" does he have to squat to equate to one step of the race. I don't understand what's being asked. If someone PM's me I'll send a screenshot of the problem.

 

[haruspex]

Change in KE

Out of interest, what answer do you get if you assume instead that all the acceleration takes place in the first step or so?

how much "work" does he have to squat

This is indeed a strange question. The obvious answer is the same as in the first part. It would have made more sense to have asked for the height of such a squat, i.e. how much raising of the mass centre takes the same work as one step of the run.