The Relationship Between Newtons and Joules

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Discussion Overview

The discussion revolves around the relationship between Newtons and Joules, specifically exploring how work is calculated when a force is applied to an object at different velocities. Participants examine the implications of applying a constant force over varying distances and speeds, and how these factors influence the work done, as well as the frame-dependence of work and energy concepts.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant claims that applying 1 Newton to 1 kilogram at rest for one second results in 0.5 Joules of work, while applying the same force to a moving object at 1,000 m/s may yield about 1,000 Joules of work.
  • Another participant agrees that more work is done when pushing an object over a greater distance, emphasizing that it is the work done with the force that matters, not just the force itself.
  • A question is raised about whether the work performed by a Newton is proportional to the square of the velocity during the time the force is applied.
  • One participant argues that work is proportional to the distance traveled during the force application, which depends on average velocity rather than the square of the velocity.
  • There is a discussion about the definitions of acceleration and how it relates to velocity, with clarifications on instantaneous versus average acceleration.
  • One participant notes that work is frame-dependent, providing examples of how the work done can differ based on the observer's frame of reference.
  • Another participant seeks clarification on whether the magnitude of work in Joules changes based on the frame of reference and if it is affected by relativistic effects.
  • Concerns are raised about the dimensional nature of terms like acceleration and how they relate to work and power, with a participant expressing confusion over the units involved.

Areas of Agreement / Disagreement

Participants express differing views on the relationship between force, velocity, and work, with some agreeing on the frame-dependence of work and energy concepts while others contest specific interpretations of acceleration and its implications for work calculations. The discussion remains unresolved on several points, particularly regarding the influence of frame of reference on work and energy.

Contextual Notes

Participants highlight limitations in their understanding of the concepts discussed, particularly regarding the definitions and relationships between force, work, and energy in different frames of reference. There is also mention of unresolved mathematical steps and the need for clarity on dimensional analysis.

BitWiz
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If I apply 1 Newton to 1 kilogram (at rest) for one second, it will have accelerated to 1 m/s^2 and traveled 0.5 meters. I have therefore done 0.5 J of work. Is this correct?

If I apply 1 Newton to 1kg traveling at 1,000 m/s for one second, have I now done about 1,000 joules of work with the same Newton?

Thanks,
Bit
 
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BitWiz said:
If I apply 1 Newton to 1 kilogram (at rest) for one second, it will have accelerated to 1 m/s^2 and traveled 0.5 meters. I have therefore done 0.5 J of work. Is this correct?

If I apply 1 Newton to 1kg traveling at 1,000 m/s for one second, have I now done about 1,000 joules of work with the same Newton?
Sure. You do a lot more work on the mass pushing it for 1,000 m compared to only pushing it for 0.5 m. It's not the force that counts, it's the work you do with that force.
 
Thanks, Doc,

So is the work "performed" by a Newton proportional to the square of the velocity during the time interval the Newton is applied?

Bit
 
BitWiz said:
So is the work "performed" by a Newton proportional to the square of the velocity during the time interval the Newton is applied?
The work is proportional to the distance traveled during that interval (assuming the push and the velocity are in the same direction). And the distance would depend on the average velocity, not the square of the velocity.
 
Thanks again, Doc,

OK, to make sure I've got this straight:

given distance(d), time(t), and mass(m)

Code:
velocity(v)                                             = d/t
acceleration(A)                               = v/t     = d/t^2
force (F)                           = Am      = vm/t    = dm/t^2
work(W)                   = Fd      = Amd     = vmd/t   = d^2m/t^2
power(P)        = W/t     = Fd/t    = Amd/t   = vmd/t^2 = d^2m/t^3
Other than that I have some things in unconventional order, am I OK so far?

Thanks,
Bit
 
In your table of equations, v/t isn't constant if there is acceleration. Acceleration = dv/dt.

Instantaneous power at any point in time can also be considered to be force time speed (if in the same direction).
 
Hmmm ...

I would think v/t is acceleration.
 
BitWiz said:
Hmmm ...

I would think v/t is acceleration.
Careful. Acceleration is the change in velocity per time. More accurately, the instantaneous acceleration is dv/dt; the average acceleration is Δv/Δt.

Only if you start from rest with uniform acceleration will a = v/t, where v is the final velocity.
 
BitWiz said:
If I apply 1 Newton to 1 kilogram (at rest) for one second, it will have accelerated to 1 m/s^2 and traveled 0.5 meters. I have therefore done 0.5 J of work. Is this correct?

If I apply 1 Newton to 1kg traveling at 1,000 m/s for one second, have I now done about 1,000 joules of work with the same Newton?
From the perspective of an inertial frame in which that one kilogram object was (a) initially at rest versus (b) initially traveling at 1,000 m/s.

Consider the latter object that is initially moving at 1,000 m/s. (Side note: 1,000 m/s with respect to what? Even in Newtonian mechanics all velocities are relative.) From the perspective of an inertial frame in which the object is initially moving at 1,000 m/s, you have done about 1,000 joules of work. Now consider an inertial frame initially co-moving with that 1 kg object. From the perspective of this frame, you have done 0.5 joules of work.

What this means is that work is a frame-dependent concept. That shouldn't be all that surprising since (a) kinetic energy is obviously a frame dependent concept, and (b) work is related to energy via the work-energy theorem.
 
  • #10
Doc Al said:
Careful. Acceleration is the change in velocity per time. More accurately, the instantaneous acceleration is dv/dt; the average acceleration is Δv/Δt.

Only if you start from rest with uniform acceleration will a = v/t, where v is the final velocity.

Thanks, Doc. I was trying to understand it as a ratio without fixed values. Thanks for pointing out the concepts of instantaneous and average values. I understand those, but never really appreciated the notation. If I'd seen "delta" used this way in the past (top and bottom), I'm not sure I appreciated its significance at the time -- and I like it!

I'm still trying to get a handle on the "dimensional" nature of these terms, how they relate, and how they would extrapolate. For instance, Δ(A)ccleration is apparently d/t^3 if linear. Someone in a prior post pointed out that (P)ower/(v)elocity = (F)orce. That kind of thing.

I'm also trying to appreciate the frame of reference concept and how it applies to "Newtonian" situations. I think this is where I've gotten lost. Maybe you can tell me this: does a magnitude in joules change based on the frame? Is it affected by Lorentz?

Thanks,
Bit
 
  • #11
D H said:
From the perspective of an inertial frame in which that one kilogram object was (a) initially at rest versus (b) initially traveling at 1,000 m/s.

Consider the latter object that is initially moving at 1,000 m/s. (Side note: 1,000 m/s with respect to what? Even in Newtonian mechanics all velocities are relative.) From the perspective of an inertial frame in which the object is initially moving at 1,000 m/s, you have done about 1,000 joules of work. Now consider an inertial frame initially co-moving with that 1 kg object. From the perspective of this frame, you have done 0.5 joules of work.

What this means is that work is a frame-dependent concept. That shouldn't be all that surprising since (a) kinetic energy is obviously a frame dependent concept, and (b) work is related to energy via the work-energy theorem.

Thanks, D H. I think you just answered one of my questions above (on frames) to Doc Al. Should have read you first!

Can I ask you this: if I'm moving at a relativistic speed and accelerate, the space in front of me apparently contracts in inverse proportion to my momentum. Is this correct? If so, how does this affect "distance" units related to work and power from "my" frame and that of a fixed observer.

Thanks!
Bit
 
  • #12
BitWiz said:
I'm still trying to get a handle on the "dimensional" nature of these terms, how they relate, and how they would extrapolate. For instance, Δ(A)ccleration is apparently d/t^3 if linear.
Not sure what you mean here. Δa has units of a, which are d/t^2, not d/t^3.

I'm also trying to appreciate the frame of reference concept and how it applies to "Newtonian" situations. I think this is where I've gotten lost. Maybe you can tell me this: does a magnitude in joules change based on the frame? Is it affected by Lorentz?
As you noticed, D H has answered that. (Yes, kinetic energy and work are frame-dependent.)
 

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