Energy conservation law question with capacitor

[annamal]

Homework Statement

We wish to use the energy in a capacitor to accelerate a car of mass 1000 kg from 0-15 km/hr; if the voltage is initially 500 V; what is the area if the dielectric constant = 3.4 and d = 0.0005 inches between the capacitor plates?

Relevant Equations

Energy of capacitor = change in kinetic energy

I was wondering why energy of capacitor does not equal change in kinetic energy PLUS change in potential energy where potential energy is the change in the potential energy of the charges. I believe that should be so because energy conservation = change in kinetic energy plus change in potential energy. So why in this problem would we leave potential energy out?

 

[TSny]

energy conservation = change in kinetic energy plus change in potential energy.

Energy conservation: Total energy remains constant.

So, final kinetic energy + final potential energy= initial kinetic energy plus initial potential energy

or, change in kinetic energy plus change in potential energy = 0.

So why in this problem would we leave potential energy out?

The energy of the charges in the capacitor is the potential energy of the system.

 

[Steve4Physics]

The energy stored in a capacitor IS electrical potential energy!

When a charged capacitor is fully discharged, all of the capacitor's electrical potential energy is converted to some other form(s) of energy.

If we assume all the capacitor's potential energy is converted to kinetic energy, then the law of conservation of energy tells us:

(change in capacitor's energy) + (change in kinetic energy) = 0

Remember, a change is positive if a quantity increases; and a change is negative if a quantity decreases.

Edit: Ahah! @TSny just beat me to it.

 

[annamal]

Ok, I had a follow up question. How come in the work energy theorem, work = change in kinetic energy; but in the energy conservation law work = change in kinetic energy + change in potential energy?

 

[kuruman]

The work done by a conservative force is the negative of the change in potential energy. If an object is displaced under the influence of a conservative force from A to B, $W_{AB}=\int_A^B \mathbf{F}\cdot d\mathbf{s}.$ The change in the potential energy from which this force is derived is $\Delta U=U_B-U_A$. Since $\mathbf{F}=-\mathbf{\nabla}U$, $$W_{AB}=\int_A^B \mathbf{F}\cdot d\mathbf{s}=\int_A^B(- \mathbf{\nabla}U)\cdot d\mathbf{s}=-(U_B-U_A)=-\Delta U.$$So starting with the work-energy theorem, you get energy conservation: $$\begin{align} & \Delta K=W_{AB}\nonumber \\ & \Delta K=-\Delta U\nonumber
\\ & \Delta K+\Delta U=0. \nonumber \\ \end{align}
$$

 

[Steve4Physics]

work = change in kinetic energy + change in potential energy?

Your equation isn’t always correct (but conservation of energy is!). You have to be very careful about what you mean by ‘work’.

What's always correct is the 'work-energy' theorem: work done by the resultant force = Δ(KE).

E.g. you push a block uphill (no friction) from rest, giving the block a kinetic energy (KE) of 1000J and increasing its gravitational potential energy (GPE) by 2000J. You have done 3000J of work.

Here are 2 ways to write the energy-conservation equation:

1) Work done by you = Δ(KE) + Δ(GPE)
3000J = 1000J + 2000J

2) (Work done by you) + (work done by gravity) = Δ(KE)
3000 + (-2000) = 1000J
(Because work done by gravity = -Δ(GPE) = -2000J.)

For a block sliding (no friction) downhill - with no force applied by you - you could write
1) Δ(KE) + Δ(GPE) = 0 or
2) Work done by gravity = Δ(KE)
They mean the same because work done by gravity = -Δ(GPE).

 

[annamal]

Does the work energy theorem: delta W = delta KE apply with external and nonconservative forces as well or does that formula only work if there are only conservative forces and external forces = 0?

 

[nasu]

First, there is no "delta W". The work (not change in work) is equal to the change in kinetic energy. This is because the work is a quantity that describes what happens in a process, in the transition from initital to final state. The kinetic energy, on the other hand, is a quantity that describes the state and has different values in the initial and final states. So the change in the state (the "delta") depends on what happens in the transition from initial to final (the work done by the forces).
And second, the change in kinetic energy of a system is equal to the work done by all forces, internal and external to the system. It does not matter if they are or not conservative.
The conservative character is important only for defining a potential energy. It is not relevant to the work-energy theorem.

 

[kuruman]

Does the work energy theorem: delta W = delta KE apply with external and nonconservative forces as well or does that formula only work if there are only conservative forces and external forces = 0?

It works with non-conservative forces as well. The general form of the W-E theorem is
$$\Delta K = W_{Net}$$ where $W_{Net}=\sum_i\int\mathbf{F}_i\cdot d\mathbf{s}$ is the sum of all the forces, conservative and non-conservative. Textbooks often split $W_{Net}$ in two parts, one that includes the work $W_C$ done on the object by all the conservative forces and one that includes the work $W_{NC}$ done by all the non-conservative forces. Then the W-E theorem becomes, $$\Delta K = W_{C}+W_{NC}.$$ The work done by the conservative forces can then appear (with a sign change) on the other side of the equation as the change in potential energy $$\Delta K +\Delta U= W_{NC}.$$It seems to me that you are using the term "external force" incorrectly. As the name implies, external forces are outside the system and cross the system boundary to do work on it. You cannot say that a force is external if you have not defined the system and its boundaries.

For example, you drop a book of mass $m$ on the floor from height $h$. If you choose the book only as your system, the force of gravity $mg$ is external to the book because it exerted by the Earth and that is not part of the system. The force of gravity does work $W= mgh$ and the W-E theorem says $$\Delta K = mgh.$$ Note that a single-component system does not have potential energy. Potential energy requires at least two components and depends on the relative configuration of these, i.e. their relative position in space.

If you choose the Earth and the book as your system, there are no external forces acting on the system to do external work. (We ignore the gravitational attraction of the Moon, the Sun, Jupiter, etc.) Therefore, $$W_{Net} = 0$$ However, the potential energy of the system changes because the book moves closer to the center of the Earth. What is work done by gravity in the one-component system is change in potential energy on the other side of the equation in the two-component system $$0=\Delta K +\Delta U=\Delta K - mgh.$$Whatever system you use, the W-E theorem gives the same result.