Work and Kinetic Energy — Question 9

Question 9

Equal work increments do not produce equal speed increments

Use the work–kinetic-energy theorem $W_{\mathrm{net}}=\Delta K$ for the specified particle in an inertial frame. A 4 kg cart receives +8 J net work in each of two separate trials. It starts from rest in trial A and at speed 6 m/s in trial B.

1. Find each final speed.

2. Compare the speed increases.

3. For arbitrary initial speed v and positive fixed work W, derive the speed increase and its high-speed trend.

Question 9 - Solution

Equal work increments do not produce equal speed increments

(a) Add work to the squared-speed relation.

The governing relation is $$\boxed{\Delta v=\sqrt{v_i^2+2W/m}-v_i}$$

$$v_f^2=v_i^2+2W/m=v_i^2+4.$$ Thus A ends at 2 m/s; B ends at $$\sqrt{40}=2\sqrt{10}\approx6.325\ \mathrm{m/s}.$$

(b) Subtract the respective initial speeds.

The increases are 2 m/s and approximately 0.325 m/s. Equal kinetic-energy gains need not imply equal speed gains.

(c) Rationalize the square-root difference.

$$\Delta v=\sqrt{v^2+2W/m}-v=\dfrac{2W/m}{\sqrt{v^2+2W/m}+v}.$$ At large positive v it approaches $W/(mv)$, so the same work gives a smaller speed increment.