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The electrical power of a vacuum cleaner is 1 kW. It expels 50 litre of air per second with a speed of 80 m/s. What percent of the total power is in the form of kinetic energy of air? (Density of the flowing air is $$1.2 \times 10^{-3}\text{ g cm}^{-3}$$)
Correct Answer: 19.2
The air density is $$1.2\text{ kg m}^{-3}$$, so the mass expelled each second is $$0.05 \times 1.2 = 0.06\text{ kg}$$. Its kinetic energy per second is $$\frac{1}{2} \times 0.06 \times 80^2 = 192\text{ W}$$. As a percentage of 1000 W this is $$19.2\%$$.
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