More practice questions

4. Q4. A bob is fixed to a light rigid rod 1 m long, pivoted at the other end. What is the least speed it needs at the lowest point to go round a full vertical circle? Take g = 9.8 m s⁻².

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   A rod can push as well as pull. The value below is the limiting bottom speed; a full circle in finite time requires a speed strictly greater than it. Equality would approach the top asymptotically with speed tending to zero. It must rise 2L, so its kinetic energy at the bottom must equal mg(2L):

   $$\begin{aligned} v_0 &= \sqrt{4gL} = \sqrt{39.2} \\ &\approx 6.26\ \text{m s}^{-1} \end{aligned}$$

5. Q5. A 2 kg block slides from rest down a rough slope 5 m high and reaches the bottom at 8 m s^⁻¹. How much mechanical energy is lost to friction? Take g = 9.8 m s⁻².

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   Potential energy lost: 2 × 9.8 × 5, that is 98 J. Kinetic energy gained: $\tfrac{1}{2}(2)(8)^{2}$, that is 64 J. The difference, 34 J, is the mechanical energy lost to friction (as heat).

6. Q6. A ball is thrown straight up from the ground at 15 m s^⁻¹. Find the greatest height it reaches, and its speed at half that height. Ignore air resistance and take g = 9.8 m s⁻².

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   At the top all the kinetic energy has become potential energy:

   $$\begin{aligned} h &= \frac{v^{2}}{2g} = \frac{225}{19.6} \\ &\approx 11.5\ \text{m} \end{aligned}$$

   At half that height, half the kinetic energy remains, so the speed is $15/\sqrt{2}$, about 10.6 m s^⁻¹.