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)

Question (1) - Power

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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