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Energy - Units of Work and Energy

Grade 7ICSE

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

🔑Concepts

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Work is defined as the product of the force applied (FF) and the displacement (ss) moved in the direction of the force.

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Energy is the capacity or ability to do work. Since energy is measured by the amount of work an object can perform, the units for work and energy are the same.

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The S.I. unit of work and energy is the Joule (JJ). One Joule is the work done when a force of 1 N1 \text{ N} displaces a body through a distance of 1 m1 \text{ m} in its own direction.

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The C.G.S. unit of work and energy is the erg. One erg is the work done when a force of 1 dyne1 \text{ dyne} displaces a body through a distance of 1 cm1 \text{ cm} in its own direction.

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Relationship between Joule and Erg: 1 Joule=107 ergs1 \text{ Joule} = 10^7 \text{ ergs}.

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Larger units of energy include the Kilojoule (kJkJ) where 1 kJ=103 J1 \text{ kJ} = 10^3 \text{ J}, and the Megajoule (MJMJ) where 1 MJ=106 J1 \text{ MJ} = 10^6 \text{ J}.

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Commercial unit of energy is the Kilowatt-hour (kWhkWh), commonly used for electrical energy. 1 kWh=3.6×106 J1 \text{ kWh} = 3.6 \times 10^6 \text{ J}.

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Heat energy is often measured in calories (calcal). The relationship with Joules is 1 cal≈4.2 J1 \text{ cal} \approx 4.2 \text{ J} (precisely 4.186 J4.186 \text{ J}).

📐Formulae

W=F×sW = F \times s

1 Joule=1 Newton×1 metre1 \text{ Joule} = 1 \text{ Newton} \times 1 \text{ metre}

1 erg=1 dyne×1 cm1 \text{ erg} = 1 \text{ dyne} \times 1 \text{ cm}

1 J=107 ergs1 \text{ J} = 10^7 \text{ ergs}

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

1 cal=4.186 J1 \text{ cal} = 4.186 \text{ J}

💡Examples

Problem 1:

A force of 15 N15 \text{ N} is applied to pull a cart through a distance of 10 m10 \text{ m} in the direction of the force. Calculate the work done.

Solution:

Given: Force F=15 NF = 15 \text{ N}, Displacement s=10 ms = 10 \text{ m}. Using the formula W=F×sW = F \times s, we get W=15×10=150 JW = 15 \times 10 = 150 \text{ J}.

Explanation:

Work is calculated by multiplying the magnitude of force by the displacement. Since both units are in S.I. (NN and mm), the result is in Joules (JJ).

Problem 2:

An engine performs 2×105 J2 \times 10^5 \text{ J} of work. Express this work in ergs.

Solution:

We know that 1 J=107 ergs1 \text{ J} = 10^7 \text{ ergs}. Therefore, 2×105 J=2×105×107 ergs=2×1012 ergs2 \times 10^5 \text{ J} = 2 \times 10^5 \times 10^7 \text{ ergs} = 2 \times 10^{12} \text{ ergs}.

Explanation:

To convert Joules to ergs, we multiply the value by the conversion factor 10710^7.

Problem 3:

How much energy in Joules is contained in a snack that provides 100 calories100 \text{ calories}?

Solution:

Given: Energy in calories = 100 cal100 \text{ cal}. Since 1 cal=4.186 J1 \text{ cal} = 4.186 \text{ J}, the energy in Joules is 100×4.186=418.6 J100 \times 4.186 = 418.6 \text{ J}.

Explanation:

The calorie is a non-S.I. unit of energy used for heat and food. To convert to the S.I. unit (Joule), we multiply by approximately 4.24.2.