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Energy - Energy Stores and Transfers

Grade 6IGCSE

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

🔑Concepts

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Energy is a property that must be transferred to an object in order to perform work on, or to heat, the object. It is measured in Joules (JJ).

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The Law of Conservation of Energy states that energy cannot be created or destroyed, only transferred from one store to another. The total energy in a closed system remains constant.

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Kinetic Energy (EkE_k) is the energy stored in moving objects. It depends on the mass (mm) and the square of the velocity (v2v^2).

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Gravitational Potential Energy (EpE_p) is the energy stored in an object due to its height (hh) above the ground in a gravitational field (gg).

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Chemical Energy is stored in the bonds of chemical compounds, such as food, batteries, and fuels like coal or gas.

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Elastic Potential Energy is the energy stored when an object (like a spring or rubber band) is stretched or compressed.

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Thermal Energy is the internal energy of an object due to the kinetic energy of its particles; hotter objects have more thermal energy.

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Energy Transfer Pathways include: Mechanical (by a force), Electrical (by a current), Heating (by conduction/convection), and Radiation (by light or sound waves).

📐Formulae

W=FimesdW = F imes d

Ep=mimesgimeshE_p = m imes g imes h

Ek=12mv2E_k = \frac{1}{2}mv^2

Efficiency=Useful Energy OutputTotal Energy Input×100%\text{Efficiency} = \frac{\text{Useful Energy Output}}{\text{Total Energy Input}} \times 100\%

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

💡Examples

Problem 1:

A ball with a mass of 2 kg2\text{ kg} is held at a height of 5 m5\text{ m}. Calculate its Gravitational Potential Energy (EpE_p). Assume g=9.8 m/s2g = 9.8\text{ m/s}^2.

Solution:

Ep=2 kg×9.8 m/s2×5 m=98 JE_p = 2\text{ kg} \times 9.8\text{ m/s}^2 \times 5\text{ m} = 98\text{ J}

Explanation:

We use the formula for GPE, which is Ep=mghE_p = mgh. By multiplying the mass, the gravitational field strength, and the height, we find the energy stored in the gravitational store.

Problem 2:

An electric motor uses 200 J200\text{ J} of total energy. If 150 J150\text{ J} is transferred usefully to lift a weight, what is the efficiency of the motor?

Solution:

Efficiency=150 J200 J×100=75%\text{Efficiency} = \frac{150\text{ J}}{200\text{ J}} \times 100 = 75\%

Explanation:

Efficiency is the ratio of useful energy output to the total energy input. Here, 50 J50\text{ J} is likely wasted as thermal energy due to friction.

Problem 3:

How much work is done when a force of 10 N10\text{ N} moves a box 3 m3\text{ m} across a floor?

Solution:

W=10 N×3 m=30 JW = 10\text{ N} \times 3\text{ m} = 30\text{ J}

Explanation:

Work done (WW) is the energy transferred when a force (FF) moves an object over a distance (dd). Using W=F×dW = F \times d, we calculate the energy transferred mechanically.

Energy Stores and Transfers - Revision Notes & Key Formulas | IGCSE Grade 6 Science