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Physics Question - ma = mg

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bob91343:
Yes, mass changes with velocity.  So the proper equation is f = d(mv).  Or f = m(dv) x v(dm).  Or something like that.

RJSV:
 What I mean is, you have to start with integration, of the conventional little bits of matter, within a sphere (Earth). That's calculus, and I believe you can do this with a 'point mass', like a basketball for example.
   So you integrate that: The 3-D sphere with the point mass, to get the total force. The actual force equation has m1 x m2 and divided by r cubed (to 3rd pwr). And some sort of constant.
   After integrating that mess, you've got a force but no motion. Of course, any free mass will begin to 'fall' or accelerate downward. AND, any mass supported by the springs of a scale is going to exert that force as static.
At that point you have, simply, Velocity = Accel X Time
To me, it always looks like either stationary, with a static force, or it's going to be moving, in which case you've got Position / Velocity / and Acceleration.
  Non-relativistic masses, or slightly relativistic situations are, well, EXACTLY THAT. Although likely sensible to view mass as warping space - time, you need a lot of effect, to get, well, a lot of effect.
No swirls around a 'hyper-big' mass as we blast thru hyperspace.....

bostonman:
I've been away and haven't had a chance to reply.


--- Quote ---To lift the box, you need to do work.  If the force against which we do the work is constant, then the work W = F h.
--- End quote ---

To divert a bit, I know (or believe to be true) that if I lift a box fast or slow, the same work is still done. If W = F * h, and F = ma, why would picking up the box faster not use more Work?

Assuming no other forces except Earth gravity and the Force of the box, if I pick up the box 1m in 1s, my acceleration is faster than if I pick up the box 1m in 10s; thus making the Work larger.

T3sl4co1l:
The power is proportionally larger, the work is the same.

Mind that biological effort is irrelevant, for example you must spend effort just holding the box out in front of you (while doing no work on it).  This doesn't need to be the case, as you could rest it on your knee or shoulders or whatever and not expend any effort beyond keeping it balanced.  Or, y'know, rest it on the floor where it's obviously not doing any work.

Electrical analogs include a lot of actuators, which incur idle power to hold position.  Stepper motors, servos, solenoids, etc.  The only reason these dissipate power is wire resistance.  If they were wound with lossless wire (superconductor), the voltage drop would be zero while the holding force is proportional to applied current (which can circulate without decay).

Tim

CatalinaWOW:

--- Quote from: bostonman on June 23, 2021, 02:03:13 pm ---I've been away and haven't had a chance to reply.


--- Quote ---To lift the box, you need to do work.  If the force against which we do the work is constant, then the work W = F h.
--- End quote ---

To divert a bit, I know (or believe to be true) that if I lift a box fast or slow, the same work is still done. If W = F * h, and F = ma, why would picking up the box faster not use more Work?

Assuming no other forces except Earth gravity and the Force of the box, if I pick up the box 1m in 1s, my acceleration is faster than if I pick up the box 1m in 10s; thus making the Work larger.

--- End quote ---

The key to understanding the conundrum is recognizing that you accelerate and decelerate.  Doing both faster requires more power but the opposite signs mean that extra power doesn't translate into work, Only the net change in potential energy of the box is work. 

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