Kinetic Energy — Question 8

Question 8

Equal endpoint kinetic energies can hide a complete stop

Use classical translational kinetic energy $K=\tfrac12 m|\vec v|^2$. A 2 kg particle has $v(t)=(2-2t)$ m/s for numerical time $0\le t\le2$ in seconds.

1. Find $K(t)$ and the endpoint energies.

2. Find when kinetic energy is smallest and interpret the motion then.

3. Find the energy change over each one-second half and explain why the full-interval change does not describe the whole history.

Question 8 - Solution

Equal endpoint kinetic energies can hide a complete stop

(a) Square the signed velocity.

The governing relation is $$\boxed{K(t)=\tfrac12m[v(t)]^2,\qquad\frac{dK}{dt}=mv\frac{dv}{dt}}$$

$$K(t)=(2-2t)^2\ \mathrm J.$$ Therefore $$K(0)=K(2)=4\ \mathrm J.$$

(b) Locate zero speed.

$v=0$ at $$t=1\ \mathrm s,$$ where $K=0$. Velocity changes from positive to negative, so this is a stop and reversal.

(c) Keep the interval information.

$$\Delta K_{0\to1}=-4\ \mathrm J,\qquad \Delta K_{1\to2}=+4\ \mathrm J.$$ Their sum is zero, but kinetic energy was not constant. A zero net change can conceal large intermediate changes.