More practice questions

4. Q4. A 0.5 kg particle moving at 2 m s^⁻¹ at x = 1 m is acted on by a net force F = 3/x² N (with x in metres) along its motion. Find its speed at x = 3 m.

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   $$\begin{aligned} W &= \int_1^{3} \frac{3}{x^{2}}\,dx \\ &= \Big[-\frac{3}{x}\Big]_1^{3} \\ &= -1 + 3 = 2\ \text{J} \end{aligned}$$

   The initial kinetic energy is $\tfrac{1}{2}(0.5)(2)^{2}$, that is 1 J, so the final kinetic energy is 3 J, and

   $$\begin{aligned} v &= \sqrt{\frac{2 \times 3}{0.5}} = \sqrt{12} \\ &\approx 3.46\ \text{m s}^{-1} \end{aligned}$$

5. Q5. Starting from $K = \tfrac{1}{2}mv^{2}$, show that the rate at which the net force does work on a body of constant mass, dK/dt, equals Fv.

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   Differentiate: dK/dt = mv (dv/dt). By the second law, m dv/dt = F, so dK/dt = Fv. This rate of doing work is the power delivered by the net force, met later in this chapter.