The Relationship Between Newtons and Joules

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Applying 1 Newton to 1 kilogram at rest for one second results in 0.5 joules of work, as it accelerates to 1 m/s and travels 0.5 meters. If the same force is applied to a 1 kg mass initially traveling at 1,000 m/s, approximately 1,000 joules of work is done from one inertial frame, while only 0.5 joules from another co-moving frame. This illustrates that work is frame-dependent, similar to kinetic energy, and is related to the work-energy theorem. The discussion also touches on the importance of understanding instantaneous versus average acceleration and how these concepts relate to force and work. Overall, the relationship between Newtons and Joules is complex and influenced by the observer's frame of reference.
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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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