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 $300$ $\text{kg}$ of water through a vertical height of $4$ $\text{m}$ in $10$ $\text{sec}$. (Take $g = 10$ $\text{ms}^{-2}$)

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


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

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, \[\text{Total Opposing Force}= mg\sin θ+\text{Resistance}\] \[=10000×10×\frac{1}{100}+1000\] \[=2000\text{ N}\] and, \[\text{Velocity}=27 \text{ km/hr}\] \[=7.5 \text{ m/s}\]

Therefore, \[\text{Power of the engine}=\text{Force}×\text{Velocity}\] \[=2000×7.5\] \[=15000\text{ W}\] \[=15 \text{ kW}\]


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