Kinetic Energy — Question 10

Question 10

Averaging speed before squaring loses kinetic-energy information

Use classical translational kinetic energy $K=\tfrac12 m|\vec v|^2$. A 2 kg cart spends 3 s at speed 2 m/s and 1 s at speed 6 m/s. Ignore the durations of transitions when taking time averages over this idealized record.

1. Find the mean speed and the kinetic energy evaluated at that speed.

2. Find the actual time-average kinetic energy.

3. Derive the difference between these two expressions in terms of the speed variance $\langle(v-\langle v\rangle)^2\rangle$.

Question 10 - Solution

Averaging speed before squaring loses kinetic-energy information

(a) Weight the speed by time.

The governing relation is $$\boxed{\langle K\rangle=\tfrac12m\langle v^2\rangle,\qquad K(\langle v\rangle)=\tfrac12m\langle v\rangle^2}$$

$$\langle v\rangle=(3\cdot2+1\cdot6)/4=3\ \mathrm{m/s}.$$ Kinetic energy at that speed is $$\tfrac12(2)3^2=9\ \mathrm J.$$

(b) Average the squared speeds instead.

$$\langle K\rangle=[3(4)+1(36)]/4=12\ \mathrm J.$$ This is 3 J larger.

(c) Expand the nonnegative square.

$$\langle(v-\langle v\rangle)^2\rangle=\langle v^2\rangle-\langle v\rangle^2.$$ Thus $$\langle K\rangle-K(\langle v\rangle)=\tfrac12m\,\mathrm{Var}(v)\ge0.$$ Equality requires constant speed almost everywhere in the averaging interval.