Changes in Mechanical Energy for Nonconservative Forces?

Question

At time ti, the kinetic energy of a particle is 30.0 J and the potential energy of the system to which it belongs is 10.0 J. At some later time tf, the kinetic energy of the particle is 18.0 J. (a) If only conservative forces act on the particle, what are the potential energy and the total energy at time tf? (b) If the potential energy of the system at time tf is 5.00 J, are there any nonconservative forces acting on the particle? Explain.

Answer 1

As energy can be neither created nor destroyed, the changes in KE mean there must also be changes in PE for a system with TE = KE + PE total energy; where KE = 1/2 mv^2 kinetic energy and PE = mgh potential energy for a system in a gravity field. m is your particle mass, v is the current velocity of the mass, g is acceleration due to gravity, and h is the height of the particle in a gravity field relative to some reference point (e.g., the ground).

a) Assuming no losses (e.g., due to friction heat), when KE --> 18 J then PE --> 22 J from TE = 40 J = KE + PE An example would be a mass traveling upward against the pull of gravity (mg = weight) and slowing down as it gains height (h); so it gains PE while losing KE.

b) Assuming friction loss (for example), we have TE = KE + PE + FE; where FE is the friction energy = Fh = kNh or 1/2 rho Cd A v^2 depending of if the FE is due to sliding friction or drag friction for example. k is the coefficient of friction, N is the normal weight, rho is air density, Cd is coefficient of drag, and A is cross sectional area. In which case, given your supposition TE = 40 = KE + PE + FE = 18 + 5 + FE; so that FE = 40 - 23 = 17 J.

The lesson in b) is that the total energy can be neither created nor destroyed, but it can be converted. And some of that converted energy may be unrecoverable useless energy, which we call entropy. So, PE is reduced from what it would otherwise be and subsequent KE when that PE is converted will be reduced from it would have been.