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Work, Energy and Simple Machines - Define power and calculate it using appropriate formulae

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. It measures how fast work is being done.

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The S.I. unit of power is the Watt (WW), where 1 W=1 J/s1 \text{ W} = 1 \text{ J/s}. A body is said to have a power of 1 W1 \text{ W} if it does 1 Joule1 \text{ Joule} of work in 1 second1 \text{ second}.

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Larger units of power include the kilowatt (kWkW), where 1 kW=1000 W=103 W1 \text{ kW} = 1000 \text{ W} = 10^3 \text{ W}, and the megawatt (MWMW), where 1 MW=106 W1 \text{ MW} = 10^6 \text{ 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 electrical energy is the kilowatt-hour (kWhkWh), commonly known as a 'unit'.

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Relationship between commercial unit and S.I. unit: 1 kWh=3.6×106 J1 \text{ kWh} = 3.6 \times 10^6 \text{ J}. This is derived as 1 kW×1 hour=1000 W×3600 s1 \text{ kW} \times 1 \text{ hour} = 1000 \text{ W} \times 3600 \text{ s}.

📐Formulae

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

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

P=mghtP = \frac{mgh}{t}

1 Watt=1 Joule second−11 \text{ Watt} = 1 \text{ Joule} \text{ second}^{-1}

1 kWh=3.6×106 J1 \text{ kWh} = 3.6 \times 10^6 \text{ J}

💡Examples

Problem 1:

A boy of mass 50 kg50 \text{ kg} runs up a staircase of 4545 steps in 9 s9 \text{ s}. If the height of each step is 15 cm15 \text{ cm}, find his power. (Take g=10 m/s2g = 10 \text{ m/s}^2)

Solution:

Weight of the boy (mgmg) = 50 kg×10 m/s2=500 N50 \text{ kg} \times 10 \text{ m/s}^2 = 500 \text{ N}. Total height (hh) = 45×15100 m=6.75 m45 \times \frac{15}{100} \text{ m} = 6.75 \text{ m}. Time taken (tt) = 9 s9 \text{ s}. Power P=mght=500×6.759=375 WP = \frac{mgh}{t} = \frac{500 \times 6.75}{9} = 375 \text{ W}.

Explanation:

Power is calculated by dividing the total work done (which is the potential energy mghmgh gained) by the time taken.

Problem 2:

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

Solution:

Power of bulb = 60 W=601000 kW=0.06 kW60 \text{ W} = \frac{60}{1000} \text{ kW} = 0.06 \text{ kW}. Time (tt) = 6 h6 \text{ h}. Energy (EE) = P×t=0.06 kW×6 h=0.36 kWh=0.36 unitsP \times t = 0.06 \text{ kW} \times 6 \text{ h} = 0.36 \text{ kWh} = 0.36 \text{ units}.

Explanation:

To find energy in commercial units, convert power to kilowatts and multiply by time in hours.