# Work, kinetic energy, and potential energy

See attached file.

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## SOLUTION This solution is **FREE** courtesy of BrainMass!

Let's use the notation as depicted in the figure:

F_{i,j} = force exerted on particle i by particle j.

I'll also use the notation:

F_{i} = Total force exerted on particle i. We can write:

F_{i} = Sum over all j different from i of F_{i,j} plus external forces on particle i (if any).

The work-KE theorem says that the total work performed on a particle equals the increase of the kinetic energy of that particle.

The total force exerted on particle 1 is:

F_{1} = F_{1,2} + F_{1,3} + F_{1,4}

The total force exerted on particle 2 is:

F_{2} = F_{2,1} + F_{2,3} + F_{2,4}

The total force exerted on particle 3 is:

F_{3} = F_{3,1} + F_{3,2} + F_{3,4}

The total force exerted on particle 4 is:

F_{4} = F_{4,1} + F_{4,2} + F_{4,3}

The increase in kinetic energy of a particle is the work done on the particle, which is given by the inner product of force and the distance the particle moves. The distance it moves in a time interval dt is its velocity times dt. We can thus write the increase in kinetic energy of particles 1...4 as:

dT1 = F_{1} dot v_{1}dt

dT2 = F_{2} dot v_{2}dt

dT3 = F_{3} dot v_{3}dt

dT4 = F_{4} dot v_{4}dt

The total change in kinetic energy is:

dT = d(T1 + T2 + T3 + T4) = dT1 + dT2 + dT3 + dT4 =

F_{1} dot v_{1}dt + F_{2} dot v_{2}dt + F_{3} dot v_{3}dt + F_{4} dot v_{4}dt

This is cleary the work done by all the forces on all the particles during the time interval dt.

To see that this is minus the change in total potential energy, we take the first term on the right hand side:

F_{1} dot v_{1}dt

v_{1}dt is the displacement of particle 1, dx_{1}

substituting:

F_{1} = F_{1,2} + F_{1,3} + F_{1,4}

gives:

[ F_{1,2} + F_{1,3} + F_{1,4}] dot dx_{1} =

F_{1,2}dot dx_{1} + F_{1,3}dot dx_{1} + F_{1,4}dot dx_{1}

The forces F_{i,j} are derived from some potential energy function:

F_{i,j} = -nabla_{i} Uij(x_{i},x_{j}) .

The potential energy is actually a function of x_{i} - x_{j}, but we don't need that property here. Nabla_{i} is the gradient w.r.t. the variable x_{i}

The first term is thus:

F_{1,2} = - nabla_{1} U12(x_{1},x_{2})

The inner product with dx_{1} is:

F_{1,2}dot dx_{1} = - nabla_{1} U12(x_{1},x_{2}) dot dx_{1}.

In general the change of a function of several variables, say y1, y2, y3 is:

dF = dF/dy1 dy1 + dF/dy2 dy2 + dF/dy3 dy3 + ...=

nabla F dot dy

where the components of nabla F are the partial derivatives w.r.t. the yi and dy is the vector:

dy = (dy1, dy2,dy3,...)

This means that:

nabla_{1} U12(x_{1},x_{2}) dot dx_{1} is the change in the function U12(x_{1},x_{2}) caused by a change of the position vector x_{1} --> x_{1} + dx_{1}

The term F_{1,2}dot dx_{1} is thus minus the increase in the potential energy function U12 caused by the displacement by particle 1. The same is thus also true for all the other terms

F_{i,j} dot x_{i}.We have derived that:

F_{1}dot dx_{1} =

F_{1,2}dot dx_{1} + F_{1,3}dot dx_{1} + F_{1,4}dot dx_{1}

is minus the increase in U12 + U13 + U14 caused by dx_{1}

And since there is nothing special about particle 1 you can do the same for the other particles with the analogous result:

F_{2}dot dx_{2} is minus the increase in U21 + U23 + U24 caused by dx_{2}

F_{3}dot dx_{3} is minus the increase in U31 + U32 + U34 caused by dx_{3}

F_{4}dot dx_{4} is minus the increase in U41 + U42 + U43 caused by dx_{4}

Let's write:

U1 = U12 + U13 + U14

U2 = U21 + U23 + U24

U3 = U31 + U32 + U34

U4 = U41 + U42 + U43

The change in kinetic energy for the individual particles equal the work done on the articles, so:

dT1 = -dU1

dT2 = -dU2

dT3 = -dU3

dT3 = -dU3

Note that dUi is not the total change in the potential energy function, but only the change caused by the displacement of particle i. Let's add up the four terms to find the total increase in kinetic energy:

dT = -dU1 - dU2 - dU3 - dU4 =

-d[U12 + U13 + U14]

-d[U21 + U23 + U24]

-d[U31 + U32 + U34]

-d[U41 + U42 + U43]

Note that the d-operator has a different meaning in each of the lines. In the last line, e.g. it is the change in the function in the square bracket caused by the change in the position of particle 4 only. Now we use that Uij = Uji. The d-operator in front of Uij gives the change caused by the change of the position of particle i, while the d-operator in front of Uji gives the change caused by the change in the position of particle j. The sum of the two is thus the total change of the function Uij. If we add up all the terms above we thus obtain:

dT = -d[U12 + U13 + U14 + U23 + U24 + U34]

where the d-operator on the right now denotes the total change.

U12 + U13 + U14 + U23 + U24 + U34 is the total potential energy of all the particles, therefore:

dT = -dU --->

d[T + U] = 0

Note that the fact that Uij(x_{i},x_{j}) = Uji(x_{j},x_{i}) does not imply that action is minus reaction. For that you must assume that the potential is of the form Uij(|x_{i} - x_{j}|). Still we were able to derive conservation of energy without this. This seems strange, but look at what happens if you consider the system from another frame of reference which moves relative to the original frame of reference. In that frame of reference x_{i} ---> x_{i} - vt where v is the velocity of the reference frame. The potential energy function then becomes time dependent, unless all the mutual potential energies only depend on the differences of the positions of the particles. A time dependent potential energy will lead to violation of energy conservation. So, energy conservation can be violated in other frames even if it is conserved in a particular frame if you don't have action = -reaction (or conservation of momentum).

The reverse is also true. If you demand that energy is conserved in all frames of reference, then you can show that momentum must be conserved, which in turn implies that action is minus reaction. If we forget about a potential energy terms here and only look at collisions.

If you write down that

1/2 m1 v1^2 + 1/2 m2 v2^2 + ... = 1/2 m1 v1'^2 + 1/2 m2 v2'^2 + ... (1)

The primes denote the velocities some time later after collisions have occurred

An observer moving at velocity U sees an energy before the collisions of:

1/2 m1 (v1 - U)^2 + 1/2 m2 (v2 - U)^2 + ... (2)

afterwards it is:

1/2 m1 (v1' - U)^2 + 1/2 m2 (v2' - U)^2 + ... (3)

Equate (2) to (3) and demand that it is an equality for all arbitrary U. You'll then find that the momentum must be conserved.

Â© BrainMass Inc. brainmass.com December 24, 2021, 6:20 pm ad1c9bdddf>https://brainmass.com/physics/conservation-of-energy/work-kinetic-energy-potential-energy-106847