Rate of Doing Work: Physics Explained

Work is said to be done when the application of the force causes displacement of the object. In other words, we can say that the rate of transfer of emergy is called power. In automobile companies, the power rating of different engines shows the speed of vehicles. Power is a scalar quantity.

Fundamentals of Rate of Doing Work

The rate of work done is known as power. We can calculate the power by dividing the work done in joules by the time taken in seconds to complete the work.

Mathematical Formula of Power

If an object does a work $W$ in time $t$, then

$\text{Power}= \dfrac{\text{Work}}{\text{Time}}\\ \therefore P = \dfrac{W}{t}$

The SI unit of power is $\dfrac{{Joule}}{{\sec }}$ which is expressed as Watt, which is denoted by W. This SI unit was named after Scottish inventor James Watt.

Power is 1 watt when 1 joule of work is done in 1 second or we can say that power is 1 watt when 1 joule of energy is consumed in 1 second.

$1Watt = \dfrac{{1\,Joule}}{{1\sec }}$

1 Watt = 1 Joule per second

The bigger unit of power is Kilowatt which is denoted as kW.

$1\text{ kilowatt}= 1000\,watt = 1000\,Joule/\sec $

$1 \text{ Megawatt} = 1000\,kW$

Understanding of Power with Examples

If we have a bulb of 60 watt, then we can say that the bulb consumes 60 joules of energy in 1 second or in simple words, we can say that the bulb converts 60 joules of energy into the light every second. Similarly, if we have a bulb of 100 watt, then we can say that it consumes 100 joules of energy per second.

Suppose, we have two bulbs. First bulb uses 1200 joules over 20 seconds and the second bulb uses 1600 joules over 40 seconds.

Case I:

$P = \dfrac{{1200}}{{20}} = 60\,W$

Case II:

$P = \dfrac{{1600}}{{40}} = 40\,W$

So, from the above discussion, we can say that the first bulb has a power of 60 watts and the second bulb has a power of 40 watts. So, the first bulb is more powerful than the second bulb.

Power may vary with time. That means the rate of doing work may be different at different time intervals. Thus, the concept of average power is required. Average Power is equal to total work done divided by total time taken.

Commercial Unit of Power

The commercial unit of power is kilowatt. The commercial unit of energy is kilowatt hour. One kilowatt hour is the energy consumed in 1 hour at the rate of 1000 Joule per second.

$1\,kWh = 1\,kW \times 1\,h = 1000\,W \times 3600\,s = 3600000\,J = 3.6 \times {10^6}\,J$

Solved Examples

1. Calculate the amount of energy required to glow a bulb of 15 watts for 2 hours.

Ans: Given, Power $P = 15\,Watt$

Time $t = 2hors = 3600 \times 2 = 7200\sec $

Here, we will use the relation $P = \dfrac{W}{t}$

$15 = \dfrac{W}{{7200}}$

$\Rightarrow W = 15 \times 7200 = 108000J$

$Therefore, W = 108\,kJ$

Hence, the amount of energy required is $108\,kJ$.

2. What will be the power of an electrical fan if the current drawn by it is $0.5A$ when connected to $200V$ supply.

Ans: Given, Current $I = 0.5A$

$V = 200Volt$

Here, we will use the relation $P = V \times I$

$Therefore, P = (200) \times (0.5) = 100Watt$

So, the power of an electrical fan would be $100\,Watt$.

Interesting Facts

• CFL stands for compact fluorescent lamp. In a CFL bulb, the electric current passes through a CFL tube which contains argon and a small amount of mercury vapour. This generates ultraviolet light that excites the fluorescent coating inside the tube which generates visible light.

• LED bulbs are made using light emitting diodes. They are semiconductors. When electrons pass through these types of semiconductors, they emit light.

Conclusion

So, we can conclude that power is also expressed as energy consumed divided by time. It is the ratio of the amount of work done and the time it takes to do that work. So, power is the rate at which a certain amount of work is done. It is the measure of the speed with which work is done on an object.