Work, Energy And Power

# 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 $g = 10$ $\text{ms}^{-2}$)

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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