Work in the middle of space

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Main Question or Discussion Point

I have trouble grasping the concept of work as a concept that applies anywhere in the universe. It seems to me it only applies in places with constant friction or other means of constant energy transfer.

If I am in the middle of space and I apply a force to an object, how can I know how much work I did on it if i goes on out of sight with nothing to slow it down?

Does this mean that work is a physics concept that can only be used in systems where there are constant "slowdown" forces?
 

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  • #2
malty
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I have trouble grasping the concept of work as a concept that applies anywhere in the universe. It seems to me it only applies in places with constant friction or other means of constant energy transfer.

If I am in the middle of space and I apply a force to an object, how can I know how much work I did on it if i goes on out of sight with nothing to slow it down?

Does this mean that work is a physics concept that can only be used in systems where there are constant "slowdown" forces?
Um, Not quite sure what exactly you mean, but work is merely the change in kinetic energy of a body or partice, so I'd imagine (or at least think) that would imply that work applies everywhere, because everyparticle has kinetic energy.

Edit: Should have added as long as the frames of reference are inertial.
 
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  • #3
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Um, Not quite sure what exactly you mean, but work is merely the change in kinetic energy of a body or partice, so I'd imagine (or at least think) that would imply that work applies everywhere, because everyparticle has kinetic energy.

Edit: Should have added as long as the frames of reference are inertial.
Work is not necessarily the change in kinetic energy. Work is one of two ways of transferring energy, the other being heat. In classical terms, a particle or object can possess zero kinetic energy but possess more than one kind of potential energy.
 
  • #4
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I have trouble grasping the concept of work as a concept that applies anywhere in the universe. It seems to me it only applies in places with constant friction or other means of constant energy transfer.

If I am in the middle of space and I apply a force to an object, how can I know how much work I did on it if i goes on out of sight with nothing to slow it down?

Does this mean that work is a physics concept that can only be used in systems where there are constant "slowdown" forces?
To answer your question, you observe the change in the object's velocity (and thus kinetic energy) relative to you or some other non-accelerating (inertial) reference frame.
 
  • #5
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What I meant is since work is the amount of force times the distance traveled by the object in the direction of the force,

W=Fd

in a setting where that distance cannot be measured since the object would go on forever or for ridiculously large distance (say in the middle of space), then work seems to not be a very useful concept. But here in a gravitational field where friction slows it down, work is more useful.
 
  • #6
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What I meant is since work is the amount of force times the distance traveled by the object in the direction of the force,

W=Fd

in a setting where that distance cannot be measured since the object would go on forever (say in the middle of space), then work seems to not be a very useful concept. But here in a gravitational field where friction slows it down, work is more useful.
You wouldn't need to know the distance in that case since it would be an isolated system - just you and the other object - so conservation of energy would apply.

You could easily apply W + KE + PE = KE(2) + PE(2)

The PEs would be zero, and if you know the velocity of the object before and after the work is applied, you can determine the work done on the object.
 
  • #7
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You wouldn't need to know the distance in that case since it would be an isolated system - just you and the other object - so conservation of energy would apply.

You could easily apply W + KE + PE = KE(2) + PE(2)

The PEs would be zero, and if you know the velocity of the object before and after the work is applied, you can determine the work done on the object.
I see, so maybe it is the W=Fd method of determining work done that is not useful in that case.
 
  • #8
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I see, so maybe it is the W=Fd method of determining work done that is not useful in that case.
That and also because the work applied by the person was done at one time; it wasn't done over a physical distance or anything. Since the system was out in the middle of space, there were no other bodies doing work on the object.
 
  • #9
malty
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Work is not necessarily the change in kinetic energy. Work is one of two ways of transferring energy, the other being heat. In classical terms, a particle or object can possess zero kinetic energy but possess more than one kind of potential energy.
Oh right my misunderstanding (and apologies if it was misleading), thanks, you see that what I love about these forums, they constantly add to your understanding of concepts and help clear up/find out any misconceptions that you may have. Just hope I can return the favour someday, thanks Shackleford :)
 
  • #10
russ_watters
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I see, so maybe it is the W=Fd method of determining work done that is not useful in that case.
No. The distance is only the distance it travels while the force is being applied. So there is no difference in measuring/calculating work applied to an object in space/on earth. If the force isn't being applied, it isn't doing any work.
 
  • #11
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No. The distance is only the distance it travels while the force is being applied. So there is no difference in measuring/calculating work applied to an object in space/on earth. If the force isn't being applied, it isn't doing any work.
So the more elastic the collision, the less work is being done since there is less time for the change in distance to be measured?

A purely elastic collision would mean 0 work is done?
 
  • #12
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So the more elastic the collision, the less work is being done since there is less time for the change in distance to be measured?

A purely elastic collision would mean 0 work is done?
It's not the time for the change in distance to be measured, but the distance over wich the force acts. I can see no reason why that has to be shorter with more elastic collisions.

if you have two equal masses, mass_1 has speed 0, and you throw mass_2 against it with speed v, such that mass_1 continues with speed v and mass_2 is now at rest, then mass_2 has done an amount of work of (1/2)mv^2 on mass_1, and mass_1 has done an amount of work of -(1/2)mv^2 on mass_2

Because a positive amount of work was done on it, the kinetic energy of mass_1 has gone up from 0 to (1/2)mv^2.
Because a negative amount of work was done on it, the kinetic energy of mass_1 has gone down from (1/2)mv^2 to 0.
With an elastic collision, the net work done on both of the masses is 0.
 
  • #13
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It's not the time for the change in distance to be measured, but the distance over wich the force acts. I can see no reason why that has to be shorter with more elastic collisions.

if you have two equal masses, mass_1 has speed 0, and you throw mass_2 against it with speed v, such that mass_1 continues with speed v and mass_2 is now at rest, then mass_2 has done an amount of work of (1/2)mv^2 on mass_1, and mass_1 has done an amount of work of -(1/2)mv^2 on mass_2

Because a positive amount of work was done on it, the kinetic energy of mass_1 has gone up from 0 to (1/2)mv^2.
Because a negative amount of work was done on it, the kinetic energy of mass_1 has gone down from (1/2)mv^2 to 0.
With an elastic collision, the net work done on both of the masses is 0.
I am in the same reference frame as ball 2 and I see ball 1 moving at 1 m/s weighing 1 kg striking ball 2 (which weighs 1 kg) perfectly elastically. If my task is to describe the work being done by ball 1 onto ball 2 in this perfectly elastic collision using only the F=ma and W=Fd equations (nothing with "(1/2)mv^2" which you showed can describe it quite well.), can I complete that task?

By the way, thanks for your help so far.
 
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  • #14
russ_watters
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So the more elastic the collision, the less work is being done since there is less time for the change in distance to be measured?

A purely elastic collision would mean 0 work is done?
The elasticity of a collision doesn't imply anything about the length of time for contact. That said, a purely elastic collision completely conserves momentum and causes the maximum transfer of energy. If the object is both hard and highly elastic (such as a steel ball bearing), the length of time for contact is extremely short, but the force is extremely large.
 
  • #15
russ_watters
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I am in the same reference frame as ball 2 and I see ball 1 moving at 1 m/s weighing 1 kg striking ball 2 (which weighs 1 kg) perfectly elastically. If my task is to describe the work being done by ball 1 onto ball 2 in this perfectly elastic collision using only the F=ma and W=Fd equations (nothing with "(1/2)mv^2" which you showed can describe it quite well.), can I complete that task?
I'm not sure why you would put such a constraint on the task, since those equations are derived from each other.

However, in order to use only f=ma and w=fd, you would need a way to model the dynamics of the collision. Due to the shape of the objects, the force isn't constant -- it doesn't even vary linearly. You could, however, measure the force and time (with a load cell) or acceleration and time (with an accelerometer) and numerically integrate to find the work done. I don't know of a good way to measure the distance traveled directly (except visually), so you'd probably use the newtonian motion equations anyway.

Modelling a collision of an object with a spring would be an easier task, as the force is proportional to the deflection. But then, you'd just find yourself deriving the spring energy equation.
 
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