-1
$\begingroup$

If I run and push a car, the force considered is the one felt in my arms not in my legs? The distance the one marked on the field? If the energy outputs is constant and linear with time, why isn't work also linear with the time and increase so exponentially? If I stop applying the force on the car, stop doing work on it, d is still increasing because of the velocity. then the formula F.d is integrating something superfluous that it shouldn't count.

Consider a rocket trajectory in space for 10 sec from distance d0 to d. From the earth I know it has traveled a greater distance in the last seconds because of the stored momentum. If I apply the work formula F.d to measure the energy transfer. It means the engine has done more work from second 9 to 10 than between second 0 to 1?

My physic book doesn't seem to care whether it correlates with the chemical energy output or whether I understand the formula. It says the force is the same all over the distance and we multiply it by the distance traveled. Imagine the flow rate of the propellant is constant, it outputs X watts (I guess this is the condition for the force be the same), does the ergol have more potential when it goes faster?

I've seen momentum, impulse so far, I can't find any calculus that lead us to F.d, that would justify we decided to scale all our energy quantification with the work and kinetic energy formula, I understand the relation between those two. But I'd like to see some calculus starting from the momentum theorem which I understand, and by changing the variable from t to d, or by integrating the impulse.

What bothers me is that if work is linear to the distance, it can't be linear to time (which is the distance traveled by my hand in my watch but which has been used as a reference to measure the acceleration of my rocket and the power of my propellant). (I see that using d instead of t may be convenient with torques but even there I see no proof that the time to torque represents the total time of my impulses on both sides, maybe they are just less or more frequent depending on the time the normal force to come back from the center)

I think I would have understood if the force had be defined with m.t/d² from the beginning.

| cite | improve this question | | | | |
$\endgroup$
  • $\begingroup$ work = force x distance = power x time, which is obvious because work has units of kg m²/s², force of kg m/s² and power of kg m²/s³ $\endgroup$ – Yukterez Mar 29 at 21:57
  • $\begingroup$ I think I'm confused with power and force, I see both as a potential of acceleration and both are constant over time but it represents energy right? Units may be tricky and I think they came after we made the formula not the other way. I really have difficulties to express my problem with the concept of work. $\endgroup$ – Sarah Mar 29 at 22:04
  • 2
    $\begingroup$ I'm afraid your presentation is, at least to me, very confusing and lacking focus. $\endgroup$ – Bob D Mar 29 at 22:06
  • $\begingroup$ I'm afraid about that too. If something is able to deliver the same amount of energy per second why is the work done(energy transferred) increasing so exponentially over time $\endgroup$ – Sarah Mar 29 at 22:13
  • 1
    $\begingroup$ @Dale yes thank you, it really helps. $\endgroup$ – Sarah Mar 29 at 23:39
1
$\begingroup$

If the energy outputs is constant and linear with time, why isn't work also linear with the time of the net force (impulse)?

You're making an assumption that the energy (power) output is linear with time, and it's not (or if it is, then the force isn't constant). As an example, a car engine can (for some regimes) be considered a constant power device. This means that as it goes faster, the force it can apply is reduced and the acceleration drops.

The same will happen to you as you push the car. Once you get the car up past walking speed, it's going to be much more difficult to maintain the constant force you mentioned. In fact, it will take you more power to do so.

If I apply the work formula F.d to measure the energy transfer. It means the engine has done more work from second 9 to 10 than between second 0 to 1?

Rockets are tricky. The engine has done more work on the rocket during the end of that period than the beginning. But it has done less work on the exhaust. Before it lifts off, the engine is accelerating the fuel from rest to a very high speed. As it moves forward the exhaust speed (relative to earth) drops a bit. So it's doing more work on the rocket and less on the exhaust.

More power because I have to keep behind the car but if I can't keep the speed it's obvious the power of my legs do no work and shouldn't be count.

Don't think of it as "keeping the speed". Think of it as your body (or whatever engine) trying to apply a force between two things: one is the car, one is the ground. When the objects are at the same speed, applying the force is easy. When they are moving relative to each other, applying the force is more difficult.

if the amount of power or force can't be measured if you don't know if you are accelerating?

You'll need to explain this more. You can certainly measure power and force. But to be consistent, you need to do all your measurements in the same inertial frame.

| cite | improve this answer | | | | |
$\endgroup$
  • $\begingroup$ Thank you for your answer. More power because I have to keep behind the car but if I can't keep the speed it's obvious the power of my legs do no work and shouldn't be count. I can't believe that if power is constant then force isn't anymore. It seems so counter intuitive. Because I thought we could apply newton's mechanics whatever the inertial reference frame except rotating objects. So does it mean the amount of power or force can't be measured if you don't know you are accelerating? Then how do you measure the power of a car on the earth properly if it has already a velocity? $\endgroup$ – Sarah Mar 29 at 22:38
  • $\begingroup$ @Sarah Note that no physics concept is made more clear by including human bodies in the analysis, (with the possible exception of Feynman's Rope Trick). $\endgroup$ – JEB Mar 30 at 0:53

Not the answer you're looking for? Browse other questions tagged or ask your own question.