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We derived two equations in class.

  1. The work done between two points $A$, $B$ is equal to the difference between the kinetic energy at the last point and the one at the first point.
  2. The work done between two points $A$, $B$ is equal to the difference between the potential energy at the first point and the one at the last point.

Now the thing is the following: if I have a car driving with constant velocity on the street and I use some work to accelerate it, then it is driving with a higher speed, so the kinetic energy will have been changed the way we said. But the potential energy is left unchanged, so I was wondering: when are these two equations true and when are they not applicable?

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Actually, the potential energy of the car has decreased, but in a less obvious way than you're probably used to. I assume you have only encountered work in the context of newtonian mechanics so far and haven't learned much about thermodynamics? The potential energy that has been decreased in the car is the bond energy of the fuel. Fuel is being 'burned', i.e. bonds are being broken, and the energy that is released by this process is used to perform work. There's also losses from friction etc, but the total energy is conserved. – Wouter Jun 27 '13 at 8:46
actually, I wanted to know when both equations are applicable. So, thank you that you are all posting something that has to do with my example but that was just to make clear, that I am not sure that both equations are always valid. – Xin Wang Jun 27 '13 at 8:55
I've turned my comment into an answer with more emphasis on the actual validity of the statements in your question. Let me know if you're still unsure. – Wouter Jun 27 '13 at 9:22
thank you. it is perfect now – Xin Wang Jun 27 '13 at 9:23
You're welcome ;-) – Wouter Jun 27 '13 at 9:24
up vote 2 down vote accepted

Both your equations are valid as long as you're dealing with conservative forces. They - pretty much by definition - express conservation of mechanical energy. There are other kinds of energy as well and in most realistic situations you need to take them into account as well. The total energy is always conserved and for conservative forces in classical mechanics, the only relevant types of energy (the ones that can change) are potential and kinetic energy. So the sum of those is conserved.

In the situation of the car the forces are not conservative and energy is lost through e.g. heat from friction in the engine. The potential energy that has been decreased is the chemical bond energy of the fuel. Fuel is being 'burned', i.e. bonds are being broken, and the energy that is released by this process is used to perform work. However, not all the energy is put to good use, there's losses from the system heating up (and that thermal energy is mostly released into the environment). There's also losses from friction of the wheels etc. but the total energy, minus the losses, is conserved.

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Just to add, as much as 85% of the energy used to make a car go is lost due to heat/frictionl i.e. just about 15% of the energy is used to accelerate the car.… – Greg Jun 27 '13 at 11:06

It depends on how you define potential energy. If gravitational potential energy is all you consider, then it would appear that the car suddenly gained kinetic energy without losing potential energy, but really, it's the gasoline in the car that is being converted from a form of higher potential energy to something of lower potential energy and the resulting heat is used to do work, which in turn makes the car go faster and contributes to its increased kinetic energy.

To my knowledge, Potential energy + Kinetic Energy + Losses = Constant will likely be valid for everything you encounter in mechanics. You probably will cover conservative forces as well and how nonconservative forces do not guarantee that conversions between potential energy and kinetic energy will have 0 losses.

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protected by ACuriousMind Apr 1 at 14:08

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