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Physics - Energy Transfers and Conservation

Grade 8IGCSE

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

🔑Concepts

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Energy Stores: Energy can be stored in different ways, including Kinetic Energy (KEKE), Gravitational Potential Energy (GPEGPE), Chemical, Elastic Potential, Nuclear, and Thermal stores.

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Energy Transfers: Energy is moved between stores through four main pathways: Mechanically (by forces), Electrically (by current), by Heating (conduction, convection), and by Radiation (waves like light or sound).

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Law of Conservation of Energy: Energy cannot be created or destroyed; it can only be transferred from one store to another. The total energy in a closed system remains constant.

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Efficiency: This is a measure of how much 'useful' energy is produced compared to the 'total' energy input. It is often represented as a percentage. Energy that is not useful is usually 'dissipated' (wasted) to the surroundings, typically as thermal energy.

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Sankey Diagrams: Visual representations used to show energy transfers. The width of the arrows is proportional to the amount of energy in Joules (JJ).

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Work Done: Work is done when a force (FF) moves an object through a distance (dd). Work done is equivalent to energy transferred.

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Power: The rate at which energy is transferred or the rate at which work is done, measured in Watts (WW), where 1W=1J/s1 W = 1 J/s.

📐Formulae

W=F×dW = F \times d

ΔGPE=m×g×Δh\Delta GPE = m \times g \times \Delta h

KE=12mv2KE = \frac{1}{2} m v^2

Efficiency=Useful energy outputTotal energy input×100%\text{Efficiency} = \frac{\text{Useful energy output}}{\text{Total energy input}} \times 100\%

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

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

💡Examples

Problem 1:

A ball with a mass of 0.5 kg0.5\text{ kg} is dropped from a height of 10 m10\text{ m}. Calculate its Gravitational Potential Energy (GPEGPE) at the start. (Assume g=9.8 m/s2g = 9.8\text{ m/s}^2)

Solution:

ΔGPE=m×g×Δh\Delta GPE = m \times g \times \Delta h ΔGPE=0.5 kg×9.8 m/s2×10 m=49 J\Delta GPE = 0.5\text{ kg} \times 9.8\text{ m/s}^2 \times 10\text{ m} = 49\text{ J}

Explanation:

The energy is calculated by multiplying the mass, the gravitational field strength, and the change in vertical height.

Problem 2:

An electric motor uses 200 J200\text{ J} of electrical energy to lift a weight. If 150 J150\text{ J} of that energy is converted into useful GPEGPE, what is the efficiency of the motor?

Solution:

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

Explanation:

Efficiency is the ratio of useful energy output to total energy input, expressed as a percentage.

Problem 3:

A car of mass 1000 kg1000\text{ kg} is traveling at a velocity of 20 m/s20\text{ m/s}. Calculate its Kinetic Energy (KEKE).

Solution:

KE=12mv2KE = \frac{1}{2} m v^2 KE=0.5×1000 kg×(20 m/s)2KE = 0.5 \times 1000\text{ kg} \times (20\text{ m/s})^2 KE=500×400=200,000 J (or 200 kJ)KE = 500 \times 400 = 200,000\text{ J (or } 200\text{ kJ)}

Explanation:

The kinetic energy depends on the mass and the square of the velocity. Doubling the speed quadruples the kinetic energy.