Horizontal Work

[JFMichael]

From http://science.howstuffworks.com/fpte4.htm :

Work is simply the application of a force over a distance, with one catch – the distance only counts if it is in the direction of the force you apply. Lifting a weight from the ground and putting it on a shelf is a good example of work. The force is equal to the weight of the object, and the distance is equal to the height of the shelf. If the weight were in another room, and you had to pick it up and walk across the room before you put it on the shelf, you didn’t do any more work than if the weight were sitting on the ground directly beneath the shelf. It may have felt like you did more work, but while you were walking with the weight you moved horizontally, while the force from the weight was vertical.

My physics textbook also mentions this aspect of work. But what about the force it requires to accelerate the weight horizontally? Why is that force different than the force of gravity (weight)?

[Shade]

Firstly, work done=energy converted, so their answer must be right, as assuming no friction, no potential, kinetic, heat or any other energy is made by the horizontal movement, so work done must be zero.

I can’t recall quite how it works, but I think you need to take into acount the direction of the force. So when you start the block moving north, you do do work (which shows up as kinetic energy) but when you stop it, you exert a force south, and the two works cancel out. I can’t quite remember how this fits with energy units, but I think it must be almost right.

[sturmhauke]

That really doesn’t sound right. Even disregarding friction, you start out at rest, move to the shelf, and come to rest again, so clearly energy is used to accelerate and decelerate. Perhaps the work is along the vector from start to finish - that is, in a diagonal line.

[Trinopus]

No, you guys are right; the textbook oversimplified. It takes some work to accelerate the stationary mass against its own inertia, and some more work to stop it again.

Trinopus

[Rabid_Squirrel]

It’s a pretty crappy article - of course moving a mass sideways woudl result in work being done. Carting a mass horizontally would require a force; ie, an acceleration sufficient to move the object from resting to your movement velocity. The classic example sould be an object being dragged at an angle, but that goes into vectors, which may be out of your league.

It’s just that the force required would be difficult fo find. In senior high-school , you get to find you that acceleration equals half of the final velocity times the acceleration time (0.5vt) assuming constant acceleration from rest.

The only notable difference between bog standard work and the work done by gravity, is that gravity always acts in a constant direction.

Incidentially, if the force is acting at an angle, you can break it up into a force vector co-current with the direction of the movement and another vector at 90° to the movement using basic trig.

[Shade]

Classical mechanics

The wikipedia article doesn’t go into much detail but says:

If a force F is applied to a particle that achieves a displacement (delta_r), the work done by the force is the scalar quantity

delta_W=F.delta_r

Suppose the mass of the particle is constant, and delta_W total is the total work done on the particle, which we obtain by summing the work done by each applied force. From Newton’s second law, we can show that

delta_W[sub]total[/sub] = delta_T, 

where T is called the kinetic energy.

(I’ve used “delta_” as I can’t remember what way of making symbols works on all browsers.)

This supports my previous guess in that:

(1) delta_W can be negative, if the force opposes the motion
(2) the total work is the total change in kinetic energy

So, I think I can say There is work done at both the start and the end of the horizontal motion, but the total work is zero.

I can’t really justify this, but I’m pretty sure that’s how it does work.

[Not_In_Anger]

I am so pleased to be the first to say: I thought you meant something different by “horizontal work.”

[sturmhauke]

*Originally posted by Shade *
**http://en.wikipedia.org/wiki/Classical_mechanics

The wikipedia article doesn’t go into much detail but says:(I’ve used “delta_” as I can’t remember what way of making symbols works on all browsers.)

This supports my previous guess in that:

(1) delta_W can be negative, if the force opposes the motion
(2) the total work is the total change in kinetic energy

So, I think I can say There is work done at both the start and the end of the horizontal motion, but the total work is zero.

I can’t really justify this, but I’m pretty sure that’s how it does work. **

So if you’re on the International Space Station and you move something from one end to the other, have you performed no work? If you walk along a completely flat stretch of road, have you performed no work?

[Shade]

Ah, maybe it’s that you do do work, but none of it ends up being done on the object. For instance, as you walk, you lift yourself up - doing work - and then come down, but the gravitational potential energy from the top doesn’t go back into your muscles, but into heat and sound as you go ‘thud’; so the net effect vertically is that no work is done on you as an object (ie. no more kinetic or potential energy) but energy is transfered from your ATP via muscles and hight, to heat.

BTW, I think I’m right, but I’m not completely certain - I’m still waiting for an applied mathematician, or physicist here.

[Desmostylus]

*Originally posted by Shade *
Ah, maybe it’s that you do do work, but none of it ends up being done on the object.

That’s the key. You’re doing very well here.

You’re discovering the the concept of “useful work”, i.e., work done on the object in question.

[Shade]

Great! Thanks!

So, you going to actually help and explain the question, or just stand there and watch me flounder, trying to derive the basic laws of the universe from scratch?

[Chronos]

If you move an object from one end of the room (or Space Station, for that matter) to the other, you may well be doing work against friction (which is why you get tired moving big things around), but you’re not doing any work on the object. You do work at both ends to start and stop it, but the work you do at the beginning is positive (force and movement in same direction), and the work you do at the end is negative (force and work in opposite directions), so they cancel out. Even at that, you can make both of those works as small as you like, by moving the object slowly.

And I thought you meant something else as well, but physics is so much more interesting than the other sort of “horizontal work” ;).

[Shade]

And I thought you meant something else as well, but physics is so much more interesting than the other sort of “horizontal work”

Funnily enough, if you google, you get several other kinds of horizontal work on the front page, but neither of the two you’re thinking of

[JFMichael]

Thank you for the responses.

• Originally posted by Chronos*
  If you move an object from one end of the room (or Space Station, for that matter) to the other, you may well be doing work against friction (which is why you get tired moving big things around), but you’re not doing any work on the object. You do work at both ends to start and stop it, but the work you do at the beginning is positive (force and movement in same direction), and the work you do at the end is negative (force and work in opposite directions), so they cancel out. Even at that, you can make both of those works as small as you like, by moving the object slowly.

If I were sliding objects across the room, wouldn’t friction stop them for me? Would the force of friction (in the opposite direction to movement) count as the negative force to cancel out the work I do to accelerate it?