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Physics Question - ma = mg
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TimFox:
You have merely re-stated Newton's Third Law.
Weight is defined as the force of gravity on a mass, which in a uniform gravitational field (i.e., over a region much smaller than the distance to the heavy source of local gravitation, such as within a meter of the Earth's surface) is equal to the product W = m x g, where g is the acceleration of gravity that is essentially uniform over this small region.  On the moon, a similar region will have a different value for g, and therefore the mass will have a different weight.
In freshman physics labs, in order to demonstrate the acceleration of gravity, a useful teaching method is the "Atwood Machine"  https://en.wikipedia.org/wiki/Atwood_machine , where two masses with a known relatively small difference are connected by strings through pulleys, so that the net force is the difference in their weights. 
Yes, if the mass is sitting on a table, the table resists the force of gravity and the mass does not accelerate or move (if the surface is spongy, the mass will sink into the surface until the spring force of the foam rubber balances the weight).
However, if you drop the mass, and it falls freely with nothing resisting it until it hits your foot, it accelerates according to Newton's Second Law, where the accelerating force equals the weight (defined above) and the mass is the mass.  The velocity when it hits your foot is left as an exercise for the reader.
bostonman:
Many anomalies exist while picking up a box fast versus slow. I'm trying to leave out the human factor in all this, air resistance (which I think has been ignored in the last replies), etc...

Also, my physics background is only two or three courses (physics 1, 2, and 3), so I'm limited in knowledge, but obviously these discussions (and reading) have helped over the years.

Leaving out recouping energy or methods of picking up a box as in holding the box out in front of me. If say a 100kg box is on the ground, I hover over it, pick it up exactly vertical 0.5m, and drop it. The Force on the box is 980N (100kg * 9.8m/s^2), and the Work is 490J (0.5m * 980N).

Now if I picked up the box the same height in 1s versus 10s, the acceleration is still only 9.8m/s^2 regardless? So the fact I've accelerated the box faster or slower has no bearing on how many Joules my body used (using a conversion from Google: 117 calories).

It makes sense to a degree from the chasing the horse analogy. The horse may outrun a human, but it burned the candle at both ends and runs out of energy 100 miles down the road after only 1hr (let's assume the animal and human is unable to take a "breather" and start running again). Whereas the human took 50hrs to run 100miles. In the long run, both have burned the same calories based on whatever factors would go into calculating Joules of a human/animal running - something I'm not asking to be calculated.

I know from years of going to the gym that some people think running sprints periodically for short duration burns more calories than those who run a constant regular speed. But I imagine it would be the same thing. The person next to me on the treadmill will have run 10miles in a short time, but they didn't (in theoretical physics definitions) haven't burned anymore calories than me who took five times longer to run the same 10miles.

Getting back to the box analogy, as someone said, if I now hold the box (maybe say hugging it against my chest), no Work is done. How could this not be true? My arms would get tired, and I'm assuming (again, leaving out the human factor) that my body is burning calories in order to overcome gravity. By the formula Work = Fd, the Work is 0J, and, if it were a motor holding something, the heat lost would be due to wire resistance but no Work done by the motor, but it seems my body would need energy to hold that box still.
Nominal Animal:

--- Quote from: bostonman on June 24, 2021, 01:13:39 am ---Now if I picked up the box the same height in 1s versus 10s, the acceleration is still only 9.8m/s^2 regardless? So the fact I've accelerated the box faster or slower has no bearing on how many Joules my body used (using a conversion from Google: 117 calories).

--- End quote ---
You are confusing work and the energy spent by your body; the same mistake IanB made above.

To quote from Wikipedia:
    Work is the energy transferred to or from an object via the application of force along a displacement.

Perhaps the error is my fault, though.  I should have been careful to always use the word only, as in "the work in only lifting a box upwards one meter in constant gravity"; but I just didn't think of properly excluding adding (unrelated/unnecessary) kinetic energy to the object.  (Kinetic energy is not just linear motion, it includes rotation also.)

If we have a box massing ten kilograms, standing still on the floor, with a table top nearby with its top one meter off the floor, and you move the box from the floor to the table, from standstill to standstill, the work done – the energy transferred to the object – is always 10 joules (10 J = 10 kg m² s⁻²), regardless of whether you do so in a fraction of a second, or take a year to do it.  Or even if you do it twice, moving the box back to the floor in between.

This amount, 10 joules, is also the minimum amount of energy you will need to spend to achieve this result.  There is no upper limit, but the difference in energy will not be transferred to the object; it goes somewhere else.  Often waste heat.

If the box starts from standstill, but does not end at standstill, we can increase the work without limit by transferring extra energy as kinetic energy (classically, 0.5 m v², where m is the mass, and v is the change in velocity, for a non-rotating object).

If the box does not start at standstill, but does end at standstill, then the work depends on the direction and speed (and rotation, if rotating) of the box.  We could even end up with "negative" work, ending up gaining more energy from the box by stopping it than we need to transfer to it to move it a meter higher.  A perfect example is using a trebuchet or similar device, and consider the situation where the box has just been "lobbed" as the starting point.  Given a suitable trajectory, the box will end up on the table without any energy transfer, any work at all; the box just converts some of its kinetic energy to potential energy (by moving upwards in the local gravity potential) without any further external assistance.

Basically, if we do not assume standstill at both starting and ending points, we can stuff as much kinetic energy in the non-standstill case as we want, and increase the work (energy transferred to or from the object) without limit.

I so wish I could explain this stuff better!


For simplicity, let's assume that we apply a constant lifting force for as long as possible, then drop the lifting force to zero at the precise moment, so that the box ends up with zero speed at the one meter elevation point.  We ignore friction and buoyancy due to air, and both initial and final positions have zero velocity.  No rotation either; the box has the same kinetic energy at the end that it had at the start.

During the lifting phase, the acceleration (positive upwards) of the box is a-g > 0, where g is the (downwards) acceleration due to gravity, 9.80 m/s², and a = F/m, F being the lifting force applied and m being the mass of the box (10 kg).  During the coasting part, the acceleration is -g.

When a constant acceleration a is applied for duration t, the change in velocity is at, and change in position is vt+0.5at², where v is the initial velocity when the acceleration started.

When a force acts on an object, the integral of the force (vector) over the interval, describes the linear momentum imparted to the object by that force during that interval.  This is called the impulse.  Its units are Newton seconds (1 N s = 1 kg m s⁻1).

(Remember that the box standing still on the floor has two major forces acting on it: one is gravity, and the other is the static force of the floor resisting deformation.  Their impulses cancel out, whenever the box stays in the same location.  So, if one does not consider all forces acting on an object, the impulses due to only some of the forces acting on it aren't really relevant in the real world: it is like measuring the velocity of a car with respect to an aeroplane flying overhead.  By picking a suitable subset of forces, you can make the impulses whatever you want, just like you can make said car velocity whatever you want by picking a suitable aeroplane as the reference.)

I suspect, but have no proof, that the chemical energy spent in human musculature is, as a first approximation, roughly linearly dependent on the impulse the forces generated by those muscles, for a relatively wide (but not full) range of human muscle power.  In the absense of a better model, I'll use this.

So, to minimise the human energy spent, we don't try to minimise the force, but the integral of the lifting force over time; i.e, we minimize the impulse due to the forces the muscles manage to generate.

If you work out the math, and define the upwards lifting force F as F=kmg, i.e. as k times the force due to gravity (and k>1), then the lifting impulse is sqrt(2gm²k/(k-1)); the force kmg is applied for duration sqrt(2/(gk(k-1)).  For k>1, this impulse is a monotonically decreasing function, so in this model, the harder the constant force you can apply, the less chemical energy your muscles will burn.

For k=10, and g=9.80m/s², the duration is just 50 milliseconds (0.048 seconds); for k=5, 100 milliseconds (0.101 seconds); for k=2, 320 milliseconds (0.319 seconds). (The k scale factor is nice, because then the mass m drops out from the duration.  The units do not seem to match, but that's just because the formulae has the fixed height to be lifted upwards, one meter, "hidden" in the constants.  If you write it out using proper dimensions, it does work out correctly.)

It does make sort-of intuitive sense – at least, if you've ever compared the effort needed to do say pull-ups slowly versus fast.  I can do a dozen fast pullups (not the swingy ones – those are cheating via angular momentum –, nor using your legs to kick in the air to get some extra help; just fast but proper ones), but ask me to stay stationary midway, and I'll drop in a few seconds, much faster than it would take to do those pullups.  I think.  Need to verify at the gym (on separate days).

However, this result is not because of work done to the box, i.e. energy transferred to the box, but because we assume the energy spent in the muscles is directly proportional to the impulse applied (by the forces generated by those muscles, as measured where those limbs grasp something).
CatalinaWOW:
This longish thread has shown how much meat there is in even simple problems.  And how important it is to frame the question properly.  Things which seem obvious to those with long exposure can be either quite confusing or devilishly simple depending on presentation and/or point of view. 

Pertinent topics which have been touched on are the relation of weight and mass, the inability to distinguish (at non relativistic speeds) the difference between a fixed acceleration and being in a static gravitational field and the differences between work and power and the difference between power stored in a system through changes in it's gravitational potential and work elsewhere (muscles and the like).

Just think of all the complexity a hard problem would bring to the table.
Nominal Animal:
Classical physics is easy and straightforward.   Until you try to truly understand it, or to apply your understanding to solve a problem.  It's only easy if you've already worked it all out before.

Quantum physics is where you need to discard whatever vestiges of intuition you have left, because intuition and its odd evolved rules it operates by becomes a severe hindrance for proper understanding here.

String theory is just mathematicians having a very long running joke on physicists.

You can construct similar sequences for any topic you choose.  If you disagree with my three choices above, just replace them with something more appropriate in your experience; the difference is just due to the differences in our personal experiences, and sequences that apply to me probably do not apply to you and vice versa.

Learning is so darned subjective.
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