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SingaporeCombined ScienceSyllabus dot point

How is energy stored, transferred and conserved, and how do we measure work and power?

State the principle of conservation of energy, describe energy stores and transfers, and apply the equations for work, kinetic and potential energy, power and efficiency

A focused answer to the O-Level Combined Science outcome on energy. Conservation of energy, energy stores and transfers, work done, kinetic and gravitational potential energy, power and efficiency.

Generated by Claude Opus 4.89 min answer

Reviewed by: AI editorial process; not yet individually human-reviewed

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  1. What this dot point is asking
  2. The answer
  3. Examples in context
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What this dot point is asking

SEAB wants you to state the principle of conservation of energy, to describe how energy is stored and transferred between stores, and to apply the simple equations for work done, kinetic energy, gravitational potential energy, power and efficiency. The calculations are short; the marks come from picking the right equation, keeping units consistent, and understanding that energy is never created or destroyed.

The answer

Conservation of energy

The principle of conservation of energy states that energy cannot be created or destroyed, only transferred from one store to another or one form to another. The total amount of energy stays the same. The unit of energy is the joule (J\text{J}).

Energy stores and transfers

Common energy stores include kinetic (movement), gravitational potential (height), chemical (fuel and food), elastic (stretched or compressed), thermal (internal), and electrical. Energy is transferred between stores by doing work, by heating, or electrically. For example, a falling ball transfers gravitational potential energy to kinetic energy.

Work done

Work is done when a force moves its point of application through a distance in the direction of the force:

W=F×dW = F \times d

Work done equals the energy transferred, both measured in joules.

Kinetic and potential energy

The kinetic energy of a moving object and the gravitational potential energy of a raised object are:

Ek=12mv2,Ep=mghE_k = \frac{1}{2}mv^2, \qquad E_p = mgh

where mm is mass, vv is speed, gg is the gravitational field strength and hh is the height raised.

Power and efficiency

Power is the rate of transferring energy (or doing work):

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

measured in watts (W\text{W}), where 1 W=1 J/s1\ \text{W} = 1\ \text{J/s}. Efficiency compares useful output to total input:

efficiency=useful energy outputtotal energy input×100%\text{efficiency} = \frac{\text{useful energy output}}{\text{total energy input}} \times 100\%

No real machine is 100% efficient, because some energy is always transferred to less useful forms such as heat.

Examples in context

Example 1. A roller coaster. At the top of the first hill the car has maximum gravitational potential energy. As it descends, that store transfers to kinetic energy so the car speeds up, then back to potential as it climbs the next hill. Friction gradually transfers some energy to heat, so each hill is lower than the last.

Example 2. An efficient kettle. A kettle puts most of its electrical energy into heating the water (useful), but some heats the casing and escapes as steam. Comparing the energy that warms the water with the electrical energy supplied gives an efficiency a little below 100%.

Try this

Q1. State the principle of conservation of energy. [2 marks]

  • Cue. Energy cannot be created or destroyed; it can only be transferred from one store or form to another, and the total stays constant.

Q2. A 3.0 kg3.0\ \text{kg} object moves at 4.0 m/s4.0\ \text{m/s}. Calculate its kinetic energy. [2 marks]

  • Cue. Ek=12mv2=12×3.0×4.02=12×3.0×16=24 JE_k = \tfrac{1}{2}mv^2 = \tfrac{1}{2} \times 3.0 \times 4.0^2 = \tfrac{1}{2} \times 3.0 \times 16 = 24\ \text{J}.

Q3. Explain why a machine can never be 100% efficient. [2 marks]

  • Cue. Some energy is always transferred to less useful forms, mainly heat to the surroundings through friction, so the useful output is always less than the total input.

Exam-style practice questions

Practice questions written in the style of SEAB exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Original4 marksA crane lifts a load of mass 200 kg200\ \text{kg} through a height of 15 m15\ \text{m}. Take g=10 N/kgg = 10\ \text{N/kg}. (a) Calculate the gain in gravitational potential energy. (b) If the lift takes 30 s30\ \text{s}, calculate the useful power output.
Show worked answer →

(a) Gain in potential energy: Ep=mgh=200×10×15=30000 J=3.0×104 JE_p = mgh = 200 \times 10 \times 15 = 30000\ \text{J} = 3.0 \times 10^4\ \text{J}.

(b) Power: P=Et=3000030=1000 WP = \dfrac{E}{t} = \dfrac{30000}{30} = 1000\ \text{W}.

Markers reward the use of Ep=mghE_p = mgh with the correct unit joule, and P=E/tP = E/t giving the power in watts. A common slip is to leave out the height or to divide by the wrong time.

Original3 marksA motor is supplied with 500 J500\ \text{J} of electrical energy and does 350 J350\ \text{J} of useful work lifting a load. (a) Calculate the efficiency. (b) State what happens to the rest of the energy.
Show worked answer →

(a) Efficiency =useful energy outputtotal energy input×100%=350500×100%=70%= \dfrac{\text{useful energy output}}{\text{total energy input}} \times 100\% = \dfrac{350}{500} \times 100\% = 70\%.

(b) The remaining 150 J150\ \text{J} is transferred to the surroundings as heat (and a little sound), mainly through friction in the motor.

Markers reward the efficiency formula and answer, and identifying the wasted energy as heat to the surroundings.

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