# Power

Power is the time rate of doing work. It is the work done per second or energy spent or used per second. $\text{Power}=\frac{\text{Work done}}{\text{Time taken}}$ $=\frac{\text{Force × Displacement}} {\text{Time taken}}$ $=\text{Force × Velocity}$ $∴P=F×v$ The SI unit of power is Watt (W). $1 \text{ }W=1 \text{ }Js^{-1}=10^7 \text{ Erg }s^{-1}$ Larger Units of Power $1 \text{ Horsepower (HP)}=746 \text{ Watts}$ $1 \text{ kW}=1000 \text{ W (10}^3 \text{ W)}$ $1 \text{ } MW = 1000000 \text{ W (10}^6 \text{ W)}$ If one joule of work is done in the time of one second, then, $P=\frac{1 \text{ }J}{1\text{ } s}$ $P=1\text{ }W$ Thus, if a body can do one joule of work in one second then the body is said to have 1 watt of power.
Different electrical appliances are labelled with different watt of power. A bulb labelled with 100 watt means that the bulb can convert 100 joule of electrical energy into heat and light energy in 1 second.

Q. Calculate the power of a pump which can lift 300 kg of water through a vertical height of 4 m in 10 sec. (Take g = 10 ms-2)

$\text{Work done}=mgh$ $=300×10×4$ $=12000 \text{ }J$ $\text{and, Time taken}=10 \text{ secs}$ $\text{Thus, Power of the pump}=\frac{\text{Work done}}{\text{Time taken}}$ $=\frac{12000}{10}$ $=1200 \text{ } W$

Q. Find the power of an engine which can travel at the rate of 27 km/hr up an incline of 1 in 100, the mass of the engine and the load being 10 metric tonnes and the resistance due to friction, etc. being 10 kg wt. per metric tonne. (Take g = 10 ms-2)

Mass of the Engine and load (m) $=10 \text{ metric tonnes}$ $=10000 \text{ } kg$ Resistance due to friction, etc. $=10 \text{ kg wt. per metric tonnes}$ Resistance due to friction, etc. for the mass of the engine and load $=100 \text{ kg wt.}$ $=100×10 \text{ } N$ $=1000 \text{ }N$ Thus, $\text{Total Opposing Force}= mgsinθ+\text{Resistance}$ $=10000×10×\frac{1}{100}+1000$ $=2000\text{ }N$ $\text{and, Velocity}=27 km/hr$ $=7.5 \text{ } m/s$ Therefore, $\text{Power of the engine}=\text{Force}×\text{Velocity}$ $=2000×7.5$ $=15000\text{ } W$ $=15 \text{ } kW$

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