QUESTION: When we apply, say, chemical energy to accelerate an object from rest to some final velocity v, there seems to be an inconsistency. To illustrate my question, let's say the acceleration process takes 10 seconds and the object accelerates uniformly throughout. Obviously, at the 5 second mark, the object achieves a velocity equal to v/2. In terms of work, the work done during the last 5 seconds is three times as great as was done during the first 5. This is very standard and in keeping with the usual lessons. My problem is that the application of chemical energy does not match the values of work done. I say this because both acceleration and force are a constant for the entire 10 seconds. From this it naturally follows that the amount of chemical energy converted into mechanical occurs at a linear rate. So, this then means that we used half the chemical energy to get to v/2 and thus we will use an equal amount of that chemical energy to double the object's velocity. Is it possible that physics courses have been teaching something that is not true (work and ke) for a very long time? I can't reconcile this inconsistency and it is driving me nuts; it is not the only one I have noticed.
ANSWER: Hello Bob,
It's clear that you expect me to tell you that the physics teaching has in fact been correct. But you want it to make sense to you so that the appearance to you of inconsistency is gone. I applaud that.
There is a flaw in your analysis, but finding it was difficult. The flaw is in the statement "the amount of chemical energy converted into mechanical occurs at a linear rate." I bought into that idea for a while. There is a formula that relates power and velocity. If the force and the velocity are in the same direction,
Power = force*velocity
Go to the site
Notice the first line of text and the 2nd formula. So as the object speeds up, the rate of consuming energy must increase.
Consider that you are pushing a wagon with constant acceleration for 10 s. As the wagon speeds up, you have to speed up and continue pushing with the same force. Eventually you are running just to keep up and you have to push, still with the same force. Your chemical energy is consumed at a higher rate in the later stages of this exercise.
Since energy = power*time = force*velocity*time, if you integrate over the first 5 seconds and again over the last 5 seconds, you should find that the second result is 3X as great as the first.
I hope this helps -- thank you for the good puzzle,
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QUESTION: Steve, the puzzle still remains. By using "energy = power*time = force*velocity*time" you have only obscured the situation. That definition for energy reduces, unit-wise, back to work and will naturally produce the 3 to 1 ratio we already know about.
I've been thinking about it and I came up with a real world situation instead of the theoretical idea I mentioned first. Think about a linear motor, or a rail gun. The electrical resistance is very small and a heavy current only flows when the rail gun operates. More importantly, the amount of current is uniform as the projectile accelerates taking us back to my original conflict/anomaly. Using the usual lessons gives the 3 to 1 ratio but when we look at electrical current, half the total current will flow during the period when the projectile reaches v/2 leaving an equal amount to double it.
Would you say that it is impossible for there to be an mistake / an error of any kind with the concept of work and kinetic energy? I know the mathematics of the work energy theorem work out beautifully but does that necessarily mean that force*distance and 1/2 mv^2 should represent energy?
I would choose to say "That definition for energy reduces, unit-wise, back to Joules which are also the proper units for chemical energy."
But, regarding your rail gun real world situation:
I don't accept that constant current assures constant force on the armature. Go to this web site:
Note what it says in the paragraph below Figure 1. I think therefore this is not a convincing situation to apply to your suspicions.