Work Done

[budmball]

I am sure this topic has been discussed before. However, I would like to consult the members of the forum about it.

 

The whole class had a debate with the physics teacher and it left all of us even the physics department baffled.

 

Is running a marathon considered work done considering that the runner has exerted a force and due to such an exertion it made his body moved.

 

I would prefer a detailed answer to my question as I hope to take the replies to class so that we can debate on it further.

[Severian]

To be honest, I don't understand how a physics teacher could be confused about this. Work is done - the runner has exerted a force in order to overcome the frictional/disapative forces in his muscles and joints. Now if your marathon runner were on perfect, frictionless rollerskates, and is pulled by a friend then the total work done in moving the rollerskater from the start of the marathon to the end is just [math]mgh[/math] where [math]m[/math] is his mass, [math]g[/math] is acceleration due to gravity and [math]h=h_{\rm finish}-h_{\rm start}[/math] is the height difference between the start and finish. So even if he is frictionless, there will probably be work done. To have no work done, he would have to be frictionless, and finish at the same height.

 

To put it another way, work done = Force x distance. Ignoring friction, moving along a completely horizontal track requires no force (because there is no force opposing his motion). If the track slopes upwards, you have to apply a force to oppose gravity, and the work done will be [math]mgh[/math]. If the track slopes downwards, you are given acceleration by gravity and the work done is negative - he will end up with kinetic energy equal to [math]-mgh[/math]. When a frictional force is present you always need to overcome this force in order to move him - lets say the force is [math]F_f[/math] - so you do work [math]F_f s + mgh[/math] (where s is the distance he moves).

[swansont]

Permit me to play Devil's advocate. Or God's advocate. Who knows?

 

No net work has been done. Ignore, for a moment, that the subject is a runner. If one were told that an object moves at constant speed over flat terrain, one would say no work has been done on the object, since the net force is zero. It matters not whether the surface is frictionless - that only tells you if some sort of propulsion is required. If friction is involved, then propulsion is required, and some sort of energy must be transferred; since friction is doing work on the system, the propulsion must provide that amount.

 

You can talk about the work done by any particular force, however, and that's what Severian has done. The runner has done work on the road when he pushes off. But the road has also done work on the runner when the foot lands. At constant speed, the two cancel.

 

 

The runner expends energy, but does no work. And, the runner does work on the road. Depends on how you ask the question. To the extent that it baffles physicists, it is probably because the question is not well-defined.

[swansont]

The monkey wrench in all I just wrote is probably this - even if the average speed of the runner is zero, he is probably always undergoing some sort of acceleration, pushing forward and landing with each stride, so work is being done in that case. So a correlary to the ill-defined problem bafflement is how much approximation is done, and how much over-analysis is done (i.e. can you actually assume a spherical cow)

[LazerFazer]

Wait, I thought work was a change in energy also?

 

So would it be correct in saying:

 

[math]

W = \Delta E

[/math]

 

[math]

\Delta E = mg \Delta h + \frac 12m \Delta v^{2}

[/math]

 

[math]

W = mg \Delta h + \frac 12m \Delta v^{2}

[/math]

 

Also, isn't the runner applying a force onto the road, therefore HE/SHE is doing work ON the road. At the same time, the ROAD is doing work on HIM/HER? And since work is a scalar quantity, do they really cancel out, or do they just add to each other?

 

LF