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O-Level Physics · Secondary 3 and 4

Energy, Work and Power

Energy is the idea that connects almost every part of physics. In this topic you meet the main energy stores and the ways energy is transferred between them, the principle of conservation of energy, work done by a force, kinetic and gravitational potential energy, power and efficiency. The calculations are short, but questions often combine several of them in one situation, such as a roller coaster or a motor lifting a load. The same ideas return in thermal physics and in electrical power.

What the syllabus expects

  • Describe energy stores, such as kinetic, gravitational potential, chemical potential, elastic potential, internal and nuclear, and the ways energy is transferred between them
  • Apply the principle of conservation of energy to simple situations
  • Calculate kinetic energy using Ek = ½mv² and changes in gravitational potential energy using Ep = mgh
  • Calculate work done using W = Fs, where s is the distance moved in the direction of the force
  • Define power as the rate of work done or energy transferred and use P = W/t = E/t
  • Calculate efficiency as useful energy output ÷ total energy input × 100%

Energy stores and transfers

Energy is measured in joules (J). It can be thought of as held in stores: kinetic (moving objects), gravitational potential (objects raised up), chemical potential (fuels, food, batteries), elastic potential (stretched or squashed springs), internal (the particles of a hot object) and nuclear. Energy moves between stores by mechanical work done by forces, by electrical work done when a current flows, by heating, and by waves such as light and sound. The principle of conservation of energy states that energy cannot be created or destroyed. It can only be transferred from one store to another, so the total stays constant. When a ball falls, gravitational potential energy is transferred to kinetic energy, and a little is transferred to the internal energy of the air and ball through air resistance.

Work done

Work is done when a force moves an object in the direction of the force. Work done W = F × s, where s is the distance moved in the direction of the force. One joule is the work done when a force of 1 N moves its point of application through 1 m in the direction of the force. Pushing a trolley with a steady 40 N for 15 m does 40 × 15 = 600 J of work, so 600 J of energy is transferred. If there is no movement in the direction of the force, no work is done on the object. Holding a heavy bag still may tire your arms, but no work is done on the bag. Work done against friction is transferred to the internal energy of the surfaces, which is why rubbing your hands warms them.

Kinetic and gravitational potential energy

Kinetic energy is Ek = ½mv², with m in kg and v in m/s. Because the speed is squared, doubling the speed makes the kinetic energy four times as large, while doubling the mass only doubles it. The change in gravitational potential energy near the Earth's surface is Ep = mgh, where h is the vertical height gained or lost. Lifting a 2.0 kg book through 1.5 m increases its gravitational potential energy by 2.0 × 10 × 1.5 = 30 J. When an object falls from rest with no air resistance, all the potential energy lost becomes kinetic energy, so mgh = ½mv². The mass cancels, giving v = √(2gh). An object dropped from 5.0 m therefore reaches √(2 × 10 × 5.0) = √100 = 10 m/s, whatever its mass.

Power and efficiency

Power is the rate of work done or energy transferred: P = W/t or P = E/t. The unit is the watt, where 1 W = 1 J/s. A motor that lifts a 50 kg load through 6.0 m in 12 s does 50 × 10 × 6.0 = 3000 J of useful work, so its useful power output is 3000 ÷ 12 = 250 W. No real machine transfers all its input energy usefully. Some is always dissipated, mostly as internal energy of the machine and its surroundings. Efficiency = useful energy output ÷ total energy input × 100%, or the same ratio using powers. If the motor above takes in 400 W of electrical power, its efficiency is 250 ÷ 400 × 100% = 62.5%. Efficiency can never be more than 100%, so a larger answer means the ratio has been turned upside down.

Worked example

A roller coaster car of mass 500 kg starts from rest at the top of a slope 20 m high. Take g = 10 N/kg. (a) Calculate the gravitational potential energy lost as it reaches the bottom. (b) Calculate its speed at the bottom if no energy is dissipated. (c) Its actual speed at the bottom is 18 m/s. How much energy was transferred to internal energy by friction and air resistance?

  1. Potential energy lost: Ep = mgh = 500 × 10 × 20 = 100 000 J.
  2. With no energy dissipated, all of this becomes kinetic energy: ½ × 500 × v² = 100 000.
  3. So 250 × v² = 100 000, giving v² = 400 and v = 20 m/s.
  4. Actual kinetic energy at 18 m/s: Ek = ½ × 500 × 18² = 250 × 324 = 81 000 J.
  5. Energy dissipated = energy lost from the potential store − energy gained in the kinetic store = 100 000 − 81 000 = 19 000 J.

Answer: (a) 100 000 J (1.0 × 10⁵ J) (b) 20 m/s (c) 19 000 J (1.9 × 10⁴ J)

Common mistakes

  • Forgetting to square the speed in ½mv², or squaring the whole of ½mv. Only v is squared.
  • Using a mass in grams, or using the weight in newtons in place of the mass in ½mv². Convert to kilograms and use mass.
  • Using the distance along a slope as h in mgh. The gravitational potential energy change depends only on the vertical height.
  • Writing that energy is used up or lost. Energy is transferred or dissipated, usually to the internal energy of the surroundings, and the total is conserved.
  • Working out an efficiency greater than 100% by dividing input by output. The useful output always goes on top.
  • Counting work done when the object does not move in the direction of the force, such as carrying a bag horizontally at constant height against gravity.

Exam tips

  • For energy conversion problems, write the stores at the start and at the end, then write the energy equation in words before putting in numbers.
  • State any assumption you make, such as no energy dissipated by friction or air resistance. It often earns a mark and makes your method clear.
  • In efficiency questions, make sure useful output and total input are in the same unit, both in joules or both in watts, before dividing.

Practice questions

1. Easy: A runner of mass 60 kg is moving at 5.0 m/s. Calculate her kinetic energy.

Show answer

Ek = ½mv² = ½ × 60 × 5.0² = 30 × 25 = 750 J.

2. Medium: A crane lifts a crate of mass 800 kg through a vertical height of 15 m in 20 s. Taking g = 10 N/kg, calculate the work done on the crate and the useful power of the crane.

Show answer

Work done = gain in potential energy = mgh = 800 × 10 × 15 = 120 000 J. Power = W ÷ t = 120 000 ÷ 20 = 6000 W (6.0 kW).

3. Harder: An electric motor with an input power of 2.0 kW is 75% efficient. How long does it take to lift a load of mass 300 kg through a height of 10 m? Take g = 10 N/kg.

Show answer

20 s. Useful power output = 0.75 × 2000 = 1500 W. Useful work needed = mgh = 300 × 10 × 10 = 30 000 J. Time = work ÷ useful power = 30 000 ÷ 1500 = 20 s.

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