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Work, kinetic energy, and potential energy

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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, ...

Solution Summary

The solution provides details on kinetic and potential energy.