(a) Explain (i) work: (ii) power (b) Show that the efficiency E, the force ratio M.A. and the velocity ratio V.R. of a machine are related by the equation: ...

(a)(i) Work is done when a force moves its point of application in the direction of the force; it is the product of the force and the distance moved in that direction, \( W = F \times d \) (unit: joule).

(ii) Power is the rate of doing work (or the rate of transfer of energy), \( P = \dfrac{W}{t} \) (unit: watt).

(b) Relation between E, M.A. and V.R. For a machine:

\[ E = \frac{\text{work output}}{\text{work input}} = \frac{\text{load} \times \text{distance moved by load}}{\text{effort} \times \text{distance moved by effort}} \] \[ E = \left(\frac{\text{load}}{\text{effort}}\right) \times \left(\frac{\text{distance moved by load}}{\text{distance moved by effort}}\right) = \text{M.A.} \times \frac{1}{\text{V.R.}} \]

since \( \text{V.R.} = \dfrac{\text{distance moved by effort}}{\text{distance moved by load}} \). Expressed as a percentage:

\[ E = \frac{\text{M.A.}}{\text{V.R.}} \times 100\% \]

(c) Inclined plane: angle \(=15^\circ\), load \(=4500\,\text{N}\), height \(=2\,\text{m}\), efficiency \(=75\%\).

(i) Velocity ratio of an inclined plane:

\[ \text{V.R.} = \frac{1}{\sin\theta} = \frac{1}{\sin 15^\circ} = \frac{1}{0.2588} = 3.86 \]

(ii) Work done on the load (useful work output):

\[ W_{\text{load}} = \text{load} \times \text{height} = 4500 \times 2 = 9000\,\text{J} \]

(d) Charles' law from the kinetic theory: at constant pressure, raising the temperature increases the average kinetic energy of the gas molecules, so they move faster and strike the walls harder and more often. To keep the pressure constant the gas must expand, so that the number of collisions per unit area per unit time is unchanged. Hence the volume increases in proportion to the absolute temperature, \( V \propto T \).