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Work, Energy and Simple Machines - Power

Grade 9CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Power is defined as the rate of doing work or the rate of transfer of energy. If an agent does a work WW in time tt, then P=WtP = \frac{W}{t}.

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The SI unit of power is the Watt (WW), named after James Watt. 11 Watt is defined as the power of an agent which does work at the rate of 11 Joule per second (1W=1Js−11 W = 1 J s^{-1}).

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Larger units of power include the Kilowatt (kWkW), where 1kW=1000W1 kW = 1000 W or 103W10^3 W.

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Average power is used when the rate of doing work varies over time. It is calculated as Pavg=Total work doneTotal time takenP_{avg} = \frac{\text{Total work done}}{\text{Total time taken}}.

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The commercial unit of energy is the kilowatt-hour (kWhkWh). It is the energy consumed in one hour at the rate of 1000J/s1000 J/s (1kW1 kW).

📐Formulae

P=WtP = \frac{W}{t}

P=EtP = \frac{E}{t}

1W=1J/1s1 W = 1 J / 1 s

1kWh=1000W×3600s=3.6×106J1 kWh = 1000 W \times 3600 s = 3.6 \times 10^6 J

💡Examples

Problem 1:

An electric bulb of 60W60 W is used for 66 hours per day. Calculate the 'units' of energy consumed in one day by the bulb.

Solution:

Power of bulb P=60W=601000kW=0.06kWP = 60 W = \frac{60}{1000} kW = 0.06 kW. Time t=6ht = 6 h. Energy E=P×t=0.06kW×6h=0.36kWhE = P \times t = 0.06 kW \times 6 h = 0.36 kWh.

Explanation:

Since 1kWh1 kWh is equal to 11 unit, the energy consumed is 0.360.36 units.

Problem 2:

A girl of mass 40kg40 kg climbs a staircase of 4040 steps, each 15cm15 cm high, in 10s10 s. Calculate her power. (Take g=10m/s2g = 10 m/s^2)

Solution:

Mass m=40kgm = 40 kg. Total height h=40×15100m=6mh = 40 \times \frac{15}{100} m = 6 m. Work done W=mgh=40×10×6=2400JW = mgh = 40 \times 10 \times 6 = 2400 J. Time t=10st = 10 s. Power P=Wt=240010=240WP = \frac{W}{t} = \frac{2400}{10} = 240 W.

Explanation:

The work done against gravity is the potential energy mghmgh. Power is then found by dividing this work by the time taken.

Problem 3:

Calculate the total power consumption of a house that has two appliances running simultaneously with power ratings of 1250W1250 W and 850W850 W.

Solution:

1250+8502100\begin{array}{r} 1250 \\ + 850 \\ \hline 2100 \end{array} Total Power = 2100W2100 W.

Explanation:

Total power is the sum of the individual power ratings of all appliances used at the same time.