Section 13–1 Energy of a falling body

(Rate of change of potential energy / Power expended / Work done by gravity)

In this section, Feynman discusses the rate of change of potential energy, power expended, and work done by gravity.

1. Rate of change of potential energy:

“…the rate of change of the kinetic energy is −mg(dh/dt), which quantity, miracle of miracles, is minus the rate of change of something else! It is minus the time rate of change of mgh! (Feynman et al., 1963, section 13–1 Energy of a falling body).”

Feynman did not begin chapter 13 by discussing the work-kinetic energy theorem. Firstly, he derives how the time rate of change of kinetic energy is related to the time rate of change of potential energy. Next, he specifies that the gravitational force is constant and it is equal to −mg. This is an approximation because we can show that GMm/(R + h)² » mg if the height h is less than 100 km above the sea level. In a sense, this section (13.1) can be titled as “work done by constant gravitational force, whereas the next section (13.2) is about “work done by variable gravitational force.” Furthermore, we assume there is no air resistance (an idealization) such that there is a miracle of miracles (Feynman’s words) in which the rate of change of the kinetic energy is minus the rate of change of potential energy.

In Tips on Physics, Feynman (2006) elaborates that “the value of the potential energy can be zero any place you want. The way we’re going to use potential energy is to talk about its changes - and then, of course, it doesn’t make any difference if you add a constant (p. 47).” In other words, what matters is the change in gravitational potential energy rather than its absolute value. (There are locations, e.g., sea level or somewhere at infinity, where they can be set to zero as a matter of convenience.) In this section, Feynman proves that the work done by gravity results in a change of gravitational potential energy. This is in contrast to conventional physics textbook authors that prove the work done by a force results in a change of kinetic energy. Nevertheless, one should appreciate the beauty of how the time rate of change of kinetic energy, −mg(dh/dt), is the minus time rate of change of mgh (potential energy).

2. Power expended:

“We thus have a marvelous theorem: the rate of change of kinetic energy of an object is equal to the power expended by the forces acting on it (Feynman et al., 1963, section 13–1 Energy of a falling body).”

Using vector analysis, Feynman shows that dT/dt = F_xv_x + F_yv_y + F_zv_z. = F.v. He also expresses F.v as the power being delivered to an object: the dot product of a force acting on an object and the velocity of the object. Moreover, he describes a theorem: the rate of change of kinetic energy of an object is equal to the power expended by the forces acting on it. Weaker students may prefer Feynman adds one more step such that d(½mv²)/dt = d(½mv.v)/dt = ½m(v.dv/dt + dv/dt.v) = m(dv/dt).v = F.v. Conversely, some textbook authors simply state dT/dt = F.v. To be precise, physicists prefer using the term instantaneous power instead of power.

In daily lives, the word power may mean average power. It is worthwhile to distinguish average power and instantaneous power. Essentially, average power (ΔW/Δt) is the amount of work done (ΔW) by a force divided by a period of time (Δt). In general, if the force acting on an object is constant and the average velocity is v_ave, then the average power can be shown to be equal to F.v_ave. If the object’s initial velocity is zero, then it is possible that the average power is equal to F.v_max/2 in which v_max is the maximum velocity of the object. On the other hand, the instantaneous power is numerically equal to the average power as the time taken, Δt, approaches zero. Mathematically, the time rate of change of energy transferred is equal to the dot product of the force at an instant and the instantaneous velocity, F.v.

3. Work done by gravity:

“We also see that it is only a component of force in the direction of motion that contributes to the work done (Feynman et al., 1963, section 13–1 Energy of a falling body).”

The work done by a force on an object can be written as F.s or F_xdx + F_ydy + F_zdz. In a daily life example, the gravitational force can be vertical, and it has only a single component, say F_z, that is equal to –mg. (Based on the convention that upward is positive, the gravitational force is acting downward and thus, it is negative.) Using this example, Feynman shows that ∫²₁F.ds = −mg(z₂ − z₁). In short, this is equivalent to the work-potential energy theorem, DW = DP.E. (or DW = −DP.E. for gravity). For instance, if your hand lifts a book upward, the work done by your hand (external force) is (+mg)(+h) and it is positive. At the same time, the work done by the gravitational force is (–mg)(+h) and thus, negative. Because of Newton’s third law of motion, we have two different works that are equal in magnitude and opposite in sign.

In Tips on Physics, Feynman (2006) explains that “[i]n certain cases, that integral can be calculated easily ahead of time, because the force on the particle depends only on its position in a simple way. Under those circumstances, we can write that the work done on the particle is equal to the change in another quantity called its potential energy, or P.E. Such forces are said to be ‘conservative’: DW = DP.E. (p. 44).” Strictly speaking, we need to idealize the particle moves slower than a snail (its speed is infinitesimally close to zero) such that there is no gain in kinetic energy. Some may prefer a more realistic work-energy theorem that includes both potential energy and kinetic energy: DW = DP.E. + DK.E.

Questions for discussion:

1. How do you explain the time rate of change of kinetic energy, −mg(dh/dt), is the minus time rate of change of mgh (potential energy)?

2. How do you prove that the rate of change of kinetic energy of an object is equal to the power expended by the forces acting on it?

3. How do you show that the work done by a force is positive or negative?

The moral of the lesson: average power is equal to work done per second and it is only a component of force in the direction of motion that contributes to the work done.

References:

1. Feynman, R. P., Gottlieb, & M. A., Leighton, R. (2006). Feynman’s tips on physics: reflections, advice, insights, practice: a problem-solving supplement to the Feynman lectures on physics. San Francisco: Pearson Addison-Wesley.

2. Feynman, R. P., Leighton, R. B., & Sands, M. (1963). The Feynman Lectures on Physics, Vol I: Mainly mechanics, radiation, and heat. Reading, MA: Addison-Wesley.