Kinetic Energy — Question 3

Question 3

Even a kinetic-energy change depends on the reference frame

Use classical translational kinetic energy $K=\tfrac12 m|\vec v|^2$. A 2 kg cart changes ground velocity from +1 to +3 m/s. An observer moves at constant velocity u along the same axis; the cart velocities in that frame are $v^{\prime}=v-u$.

1. Find $\Delta K$ in the ground frame.

2. Find $\Delta K^{\prime}$ for $u=4$ m/s.

3. Derive $\Delta K^{\prime}$ as a function of u and find the frame in which it vanishes.

Question 3 - Solution

Even a kinetic-energy change depends on the reference frame

(a) Compare the two ground speeds.

The governing relation is $$\boxed{v^\prime=v-u,\qquad\Delta K^\prime=\Delta K-mu(v_f-v_i)}$$

$$\Delta K=\tfrac12(2)(3^2-1^2)=8\ \mathrm J.$$

(b) Transform each endpoint velocity.

The new velocities are $-3$ and $-1\ \mathrm{m/s}$, giving $$\Delta K^{\prime}=1-9=-8\ \mathrm J.$$ The same physical change can increase kinetic energy in one frame and decrease it in another.

(c) Keep the cross term.

In SI numerical values, $$\Delta K^{\prime}=(3-u)^2-(1-u)^2=8-4u\ \mathrm J.$$ It vanishes at $$u=2\ \mathrm{m/s},$$ where the endpoint velocities are $-1$ and $+1\ \mathrm{m/s}$. Classical kinetic energy is frame dependent.