Cambridge AS & A Level9702

Work, energy and power

Physics 9702 Chapter Notes

What this chapter covers

Work, energy and power - Energy conservationWork, energy and power - Gravitational potential energy and kinetic energy
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1. Understanding Work Done

In physics, 'work' has a very specific meaning. Work is done whenever a force causes an object to move. It is the transfer of energy from one form to another. For work to be done, two conditions must be met: there must be a force, and the object must move in the direction of the force. The amount of work done is calculated as the product of the force and the distance moved in the direction of the force. If you push against a wall and it doesn't move, you have done no work on the wall, even though you might feel tired! The unit of work is the Joule (J). One Joule is the work done when a force of 1 Newton moves an object by 1 metre. If the force acts at an angle to the direction of motion, we only consider the component of the force that is in the direction of motion. This is calculated using `W = Fs cos(θ)`, where θ is the angle between the force and the displacement.

W = Fs

W = Fs cos(θ)

Key term

Work Done: The energy transferred when a force moves an object through a distance in the direction of the force.

Examiner insight

Examiners award marks for correctly identifying that only the component of the force parallel to the displacement does work. Show your use of trigonometry clearly.

Common pitfall

Forgetting to use the component of the force in the direction of motion. Students often just multiply the total force by the distance, even when the force is at an angle.

Worked example 13 marks

A student pulls a block of mass 5.0 kg along a horizontal surface for a distance of 8.0 m. The student pulls with a rope that exerts a tension of 40 N at an angle of 30° to the horizontal. Calculate the work done by the student on the block.

  1. 1

    Step 1: Identify the formula. The force is at an angle to the displacement, so we use W = Fs cos(θ).

  2. 2

    Step 2: Identify the values. F = 40 N, s = 8.0 m, θ = 30°.

  3. 3

    Step 3: Calculate the horizontal component of the force: F_horizontal = F cos(θ) = 40 cos(30°).

  4. 4

    Step 4: Calculate the work done: W = (40 cos(30°)) × 8.0 or W = 40 × 8.0 × cos(30°).

  5. 5

    Step 5: W = 40 × 8.0 × 0.866 = 277.1 J. Rounding to 2 significant figures gives 280 J.

Recap

  • Work is done when a force causes displacement.
  • The unit of work is the Joule (J).
  • Work done = Force × distance moved in the direction of the force (W = Fs).
  • If the force is at an angle θ to the displacement, use W = Fs cos(θ).
  • No work is done if the object does not move (s=0).
  • No work is done by a force that is perpendicular to the direction of motion (cos(90°) = 0).

Quick check

  1. A crane lifts a 1000 kg load vertically by 15 m. How much work is done by the lifting force? (Take g = 9.81 N/kg)2 marks

2. Gravitational Potential Energy (GPE)

Gravitational Potential Energy (GPE) is the energy an object has stored because of its position in a gravitational field. When you lift an object, you do work against the force of gravity. This work done is converted into GPE and stored in the object. To lift an object of mass 'm' to a vertical height 'h', you must apply a force at least equal to its weight, which is 'mg' (where 'g' is the acceleration of free fall, approximately 9.81 m/s²). The work done is Force × distance, so `W = (mg) × h`. This work done is equal to the gain in GPE. Therefore, the change in gravitational potential energy is given by the formula `ΔEp = mgΔh`. Note that it is the change in vertical height that matters, not the path taken to get there.

ΔEp = mgΔh

Key term

Gravitational Potential Energy: The energy an object possesses due to its position within a gravitational field, relative to a reference point.

Common pitfall

Using mass in grams instead of kilograms, or using the total distance travelled instead of the vertical height change.

Fun fact

The Three Gorges Dam in China, by holding back a massive reservoir of water at a high altitude, slightly slowed the Earth's rotation, increasing the length of a day by 0.06 microseconds due to the change in mass distribution and its stored GPE.

Worked example 12 marks

A hiker of mass 75 kg climbs from sea level to the summit of a mountain that is 1200 m high. Calculate the minimum increase in her gravitational potential energy.

  1. 1

    Step 1: Identify the formula for change in GPE: ΔEp = mgΔh.

  2. 2

    Step 2: Identify the given values. m = 75 kg, Δh = 1200 m. Use g = 9.81 m/s².

  3. 3

    Step 3: Substitute the values into the formula: ΔEp = 75 × 9.81 × 1200.

  4. 4

    Step 4: Calculate the result: ΔEp = 882,900 J.

  5. 5

    Step 5: Express the answer in a suitable form, e.g., kilojoules or standard form: ΔEp = 883 kJ (to 3 s.f.) or 8.8 × 10⁵ J.

Recap

  • GPE is the energy stored due to an object's height.
  • The change in GPE depends on mass, gravitational field strength (g), and the change in vertical height (Δh).
  • The formula is ΔEp = mgΔh.
  • The path taken does not affect the change in GPE, only the vertical displacement.
  • Mass must be in kilograms (kg) and height in metres (m).

Quick check

  1. An apple of mass 150 g falls 3.0 m from a tree. What is its loss of GPE? (Take g = 9.81 m/s²)2 marks

3. Kinetic Energy (KE)

Kinetic Energy (KE) is the energy an object possesses because of its motion. Anything that is moving has kinetic energy, from a moving car to a tiny atom. The amount of kinetic energy an object has depends on two things: its mass(m) and its speed (v). The formula for kinetic energy is `Ek = ½mv²`. This formula can be derived from the work-energy principle. The work done on an object by a net force causes a change in its kinetic energy. If a force 'F' acts on a mass 'm' over a distance 's', the work done is `W = Fs`. Using Newton's second law, `F = ma`, so `W = mas`. From the equations of motion, `v² = u² + 2as`. If the object starts from rest (u=0), then `v² = 2as`, so `as = v²/2`. Substituting this into the work equation gives `W = m(v²/2)`, which is `W = ½mv²`. This work done is equal to the kinetic energy gained. Notice the `v²` term: this means that doubling the speed of an object quadruples its kinetic energy, which is why high-speed collisions are so devastating.

Ek = ½mv²

Key term

Kinetic Energy: The energy an object possesses due to its motion.

Examiner insight

Examiners often test the work-energy principle. Be prepared to calculate the change in kinetic energy and equate it to the net work done on the object.

Common pitfall

Forgetting to square the speed (v) in the kinetic energy formula is the most frequent error.

Worked example 14 marks

A car of mass 1200 kg accelerates from 10 m/s to 20 m/s. Calculate:a) its initial kinetic energy,b) its final kinetic energy, andc) the work done by the engine to cause this increase in speed, assuming no resistive forces.

  1. 1

    a) Initial KE: Use Ek = ½mv² with v = 10 m/s. Ek_initial = ½ × 1200 × (10)² = 600 × 100 = 60,000 J or 60 kJ.

  2. 2

    b) Final KE: Use Ek = ½mv² with v = 20 m/s. Ek_final = ½ × 1200 × (20)² = 600 × 400 = 240,000 J or 240 kJ.

  3. 3

    c) Work Done: The work done is equal to the change in kinetic energy (Work-Energy Principle). Work Done = ΔEk = Ek_final - Ek_initial.

  4. 4

    Work Done = 240,000 J - 60,000 J = 180,000 J or 180 kJ.

Recap

  • Kinetic energy is the energy of motion.
  • The formula for kinetic energy is Ek = ½mv².
  • KE depends on mass and the square of the speed.
  • Doubling the speed of an object quadruples its kinetic energy.
  • The net work done on an object equals its change in kinetic energy.

Quick check

  1. By what factor does the kinetic energy of a car change if its speed triples?1 mark

4. The Principle of Conservation of Energy

This is one of the most fundamental principles in all of physics. The Principle of Conservation of Energy states that in an isolated system, energy cannot be created or destroyed; it can only be converted from one form to another. The total amount of energy in the system remains constant. For example, when a ball is dropped, its gravitational potential energy (GPE) is converted into kinetic energy (KE) as it falls and speeds up. If we ignore air resistance, the GPE lost is exactly equal to the KE gained. In a real-world scenario, some energy is always converted into less useful forms, like heat and sound, due to forces like friction and air resistance. The principle still holds true; the initial energy just gets distributed among more forms. So, a more complete equation is: `Initial Energy = Final Useful Energy + Energy dissipated (e.g. as heat)`.

Initial Total Energy = Final Total Energy

GPE_lost = KE_gained

GPE_lost = KE_gained + Work done against friction

Key term

Principle of Conservation of Energy: In an isolated (closed) system, the total energy remains constant; it can be converted from one form to another, but not created or destroyed.

Common pitfall

Ignoring the effects of friction or air resistance in real-world problems. Unless a question states a situation is 'frictionless' or 'in a vacuum', you should consider that some mechanical energy is likely converted to heat.

Worked example 13 marks

A child on a sledge with a combined mass of 40 kg starts from rest at the top of a frictionless slope of vertical height 5.0 m. What is their speed at the bottom of the slope? (Take g = 9.81 m/s²)

  1. 1

    Step 1: Apply the principle of conservation of energy. Since the slope is frictionless, all the GPE lost is converted into KE.

  2. 2

    Step 2: State the energy conservation equation: GPE lost = KE gained.

  3. 3

    Step 3: Substitute the formulas: mgΔh = ½mv².

  4. 4

    Step 4: Notice that mass 'm' cancels out on both sides: gΔh = ½v².

  5. 5

    Step 5: Rearrange to make v the subject: v² = 2gΔh, so v = √(2gΔh).

  6. 6

    Step 6: Substitute the values: v = √(2 × 9.81 × 5.0) = √98.1 = 9.90 m/s (to 3 s.f.).

Worked example 24 marks

Now assume the same child on the 40 kg sledge reaches the bottom of the 5.0 m high slope with a speed of 8.0 m/s. How much energy was converted to heat due to friction?

  1. 1

    Step 1: Calculate the initial GPE at the top. GPE = mgΔh = 40 × 9.81 × 5.0 = 1962 J.

  2. 2

    Step 2: The final energy is a combination of KE and heat. Calculate the final KE. KE = ½mv² = ½ × 40 × (8.0)² = 20 × 64 = 1280 J.

  3. 3

    Step 3: Apply the conservation of energy principle: Initial GPE = Final KE + Energy lost to friction.

  4. 4

    Step 4: Rearrange to find the energy lost: Energy lost = Initial GPE - Final KE.

  5. 5

    Step 5: Calculate the value: Energy lost = 1962 J - 1280 J = 682 J.

Recap

  • Energy cannot be created or destroyed, only transferred between different forms.
  • In a closed system, the total energy is always constant.
  • In mechanical problems, GPE is often converted to KE and vice versa.
  • Friction and air resistance convert mechanical energy into thermal energy (heat).
  • Always account for all forms of energy at the start and end of a process.

Quick check

  1. Describe the main energy transformations for a pendulum swinging from its highest point to its lowest point.2 marks

5. System Efficiency

While energy is always conserved, it is not always converted into the form we want. When we use a machine, like a light bulb or a car engine, we provide it with an energy input (e.g., electrical energy, chemical energy from fuel). The machine converts this into a useful energy output (e.g., light, kinetic energy) but also produces wasted energy, usually as heat and sound. Efficiency is a measure of how good a device is at converting the total energy input into useful energy output. It is calculated as a ratio, often expressed as a percentage. A perfectly efficient machine would have an efficiency of 100%, but this is impossible in practice due to the second law of thermodynamics. Improving efficiency is a major goal in engineering to save energy and reduce costs.

Efficiency = (Useful Energy Output / Total Energy Input)

Efficiency = (Useful Power Output / Total Power Input)

Efficiency (%) = (Useful Output / Total Input) × 100%

Key term

Efficiency: A measure of how well a system converts total input energy into useful output energy, calculated as the ratio of useful output to total input.

Examiner insight

Examiners reward clear identification of what constitutes 'useful output' and 'total input' in a given scenario. Always state these explicitly in your working.

Fun fact

An incandescent light bulb is only about 5% efficient at producing light; the other 95% of the electrical energy is converted directly to heat. An LED bulb can be over 50% efficient.

Worked example 14 marks

An electric motor is used to lift a 200 kg load through a vertical height of 15 m. The motor is supplied with 40,000 J of electrical energy. Calculate the efficiency of the motor. (Take g = 9.81 m/s²).

  1. 1

    Step 1: Identify the total energy input. Total Input Energy = 40,000 J.

  2. 2

    Step 2: Identify and calculate the useful energy output. The useful work is lifting the load, which increases its GPE. Useful Output Energy = GPE gained = mgΔh.

  3. 3

    Step 3: Calculate the GPE: GPE = 200 kg × 9.81 m/s² × 15 m = 29,430 J.

  4. 4

    Step 4: Use the efficiency formula: Efficiency = (Useful Energy Output / Total Energy Input).

  5. 5

    Step 5: Substitute the values: Efficiency = 29,430 J / 40,000 J = 0.73575.

  6. 6

    Step 6: Express as a percentage: Efficiency = 0.736 × 100% = 73.6% (to 3 s.f.).

Recap

  • Efficiency measures how well energy is converted into a useful form.
  • It's the ratio of useful energy (or power) output to total energy (or power) input.
  • No real-world device is 100% efficient.
  • Wasted energy is usually dissipated as heat or sound.
  • Efficiency is a dimensionless quantity, but is often expressed as a percentage.

Quick check

  1. A kettle uses 100 kJ of electrical energy to produce 85 kJ of heat in the water. What is its efficiency?2 marks

6. Power: The Rate of Energy Transfer

In physics, power is not about strength, but about how quickly energy is used or work is done. A powerful machine is one that can transfer a large amount of energy in a short amount of time. Power is defined as the rate of work done or the rate of energy transfer. The standard unit for power is the Watt (W), named after the Scottish engineer James Watt. One Watt is defined as a rate of energy transfer of one Joule per second (1 W = 1 J/s). So, a 100 W light bulb converts 100 Joules of electrical energy into light and heat every second. The main formula for power is `P = W/t` (Power = Work done / time taken) or `P = ΔE/t` (Power = Energy transferred / time taken).

P = W/t

P = ΔE/t

Key term

Power: The rate at which work is done or energy is transferred, measured in Watts (W).

Common pitfall

Confusing energy and power. Remember that energy is a quantity (Joules), while power is the rate at which that quantity is used (Joules per second).

Worked example 12 marks

A lift motor does 3.0 × 10⁵ J of work to raise a lift car and its passengers. This takes 25 seconds. What is the average power output of the motor?

  1. 1

    Step 1: Identify the formula for power: P = W/t.

  2. 2

    Step 2: Identify the given values: Work done W = 3.0 × 10⁵ J, time taken t = 25 s.

  3. 3

    Step 3: Substitute the values into the formula: P = (3.0 × 10⁵) / 25.

  4. 4

    Step 4: Calculate the result: P = 12,000 W.

  5. 5

    Step 5: Express the answer in a suitable form, e.g., kilowatts (kW). 12,000 W = 12 kW.

Worked example 24 marks

A 4.0 kW electric motor runs for 1 minute.a) How much work does it do?b) If it is 70% efficient, how much energy is supplied to it?

  1. 1

    a) Step 1: Rearrange the power formula to find work: W = P × t. Convert power to Watts and time to seconds. P = 4000 W, t = 1 × 60 = 60 s.

  2. 2

    a) Step 2: Calculate work done: W = 4000 W × 60 s = 240,000 J or 240 kJ. This is the useful work output.

  3. 3

    b) Step 1: Use the efficiency formula. Efficiency = Useful Output / Total Input. We need to find the Total Input.

  4. 4

    b) Step 2: Rearrange the formula: Total Input = Useful Output / Efficiency.

  5. 5

    b) Step 3: Substitute values. Efficiency = 70% = 0.70. Total Input = 240,000 J / 0.70.

  6. 6

    b) Step 4: Calculate the total energy supplied: Total Input = 342,857 J ≈ 343 kJ (to 3 s.f.).

Recap

  • Power is the rate of doing work or transferring energy.
  • The unit of power is the Watt (W), where 1 W = 1 J/s.
  • The formula for power is P = W/t or P = ΔE/t.
  • A high power rating means a large amount of energy is transferred per second.

Quick check

  1. Define the Watt.1 mark
  2. How much energy does a 60 W lamp use in 2 minutes?2 marks

7. Power, Force, and Velocity

There is a very useful relationship between power, force, and velocity, especially for vehicles like cars, trains, and planes. We can derive it from the definition of power. We know that `P = W/t` and `W = Fs`. Substituting the work formula into the power formula gives `P = (Fs)/t`. We can group the terms as `P = F × (s/t)`. Since displacement divided by time (`s/t`) is velocity (`v`), we arrive at the equation `P = Fv`. This equation tells us the power `P` required to provide a driving force `F` to an object moving at a constant velocity `v`. In this context, `F` is the driving force that is overcoming resistive forces like friction and air resistance. At a constant velocity, the driving force is equal to the total resistive forces.

P = Fv

Examiner insight

This formula is often tested in the context of an object at its terminal velocity or constant top speed, where examiners expect you to state that the driving force equals the total resistive forces.

Common pitfall

Using the P=Fv formula when velocity is not constant without specifying it is for instantaneous power. It is most commonly applied in the case where driving force equals resistive forces at a constant velocity.

Worked example 12 marks

A car travels at a constant speed of 30 m/s on a level road. The engine is providing a driving force of 750 N to overcome air resistance and friction. Calculate the output power of the engine.

  1. 1

    Step 1: Identify the formula that links power, force, and velocity: P = Fv.

  2. 2

    Step 2: Identify the given values. The car is at a constant speed, so the driving force equals the resistive forces. F = 750 N, v = 30 m/s.

  3. 3

    Step 3: Substitute the values into the formula: P = 750 × 30.

  4. 4

    Step 4: Calculate the result: P = 22,500 W.

  5. 5

    Step 5: Express the answer in kilowatts: P = 22.5 kW.

Worked example 24 marks

The engine of a motorbike has a maximum output power of 40 kW. The total resistive forces acting on the bike and rider are given by 0.8v², where v is the speed in m/s. Calculate the maximum possible speed of the motorbike on a level road.

  1. 1

    Step 1: At maximum speed, the bike travels at a constant velocity. The engine's driving force F must equal the total resistive forces.

  2. 2

    Step 2: At maximum speed, the engine is producing its maximum power. P_max = 40 kW = 40,000 W.

  3. 3

    Step 3: Use the power formula P = Fv. At top speed, P_max = F_driving × v_max.

  4. 4

    Step 4: Since F_driving = F_resistive, we have F_driving = 0.8v_max².

  5. 5

    Step 5: Substitute the expression for F_driving into the power equation: P_max = (0.8v_max²) × v_max.

  6. 6

    Step 6: Simplify the equation: 40,000 = 0.8v_max³.

  7. 7

    Step 7: Solve for v_max: v_max³ = 40,000 / 0.8 = 50,000.

  8. 8

    Step 8: v_max = ³√50,000 = 36.8 m/s.

Recap

  • Power can be calculated from force and velocity using P = Fv.
  • This formula is particularly useful for objects moving at a constant velocity.
  • When an object moves at a constant velocity, the driving force equals the total resistive forces.
  • 'F' in the equation is the driving force applied to maintain the velocity 'v'.
  • This gives the instantaneous power, or the power required to maintain a constant speed.

Quick check

  1. A cyclist pedals with a force of 60 N to maintain a constant speed of 8.0 m/s. What is her power output?2 marks

End-of-chapter exercise

Test yourself on the whole chapter. Work through these before moving on.

  1. A 1500 kg car accelerates from rest to a speed of 25 m/s. Calculate the work done to achieve this speed, assuming all energy is converted to kinetic energy.3 marks
  2. Define power and state its SI unit.2 marks
  3. A book of mass 800 g is lifted from the floor and placed on a shelf 2.5 m high. Calculate the gain in its gravitational potential energy. (Take g = 9.81 m/s²).3 marks
  4. A 60 kg skier slides from rest down a 200 m long slope inclined at 30° to the horizontal. If her speed at the bottom is 30 m/s, calculate the average frictional force acting on her. (Take g = 9.81 m/s²).5 marks
  5. An electric pump with a power input of 2.0 kW raises water through a vertical height of 10 m. In one minute, 900 kg of water is raised. Calculate the efficiency of the pump.4 marks
  6. A train has a maximum speed of 50 m/s on a level track when its engines are working at a constant rate of 2.5 MW. Calculate the total resistive force on the train at this speed.3 marks
  7. A crane lifts a 500 kg container vertically upwards with a constant speed of 0.40 m/s. Calculate (a) the tension in the lifting cable, and (b) the rate at which the crane is doing work on the container.4 marks
  8. A box is pulled across a rough horizontal floor by a rope inclined at 20° to the horizontal. The tension in the rope is 150 N and the box moves a distance of 10 m at a constant speed. The work done against friction is 1350 J. Calculate (a) the work done by the rope and (b) the efficiency of the pulling process, considering the work done against friction as 'wasted'.5 marks
  9. Starting from the equations of motion, show that the kinetic energy of an object of mass m moving with velocity v is given by Ek = ½mv². You may assume the object starts from rest.4 marks
  10. A 1.0 kg ball is dropped from a height of 10 m. On its first bounce, it reaches a height of 6.4 m. Calculate (a) the speed of the ball just before it hits the ground for the first time, and (b) the percentage of its mechanical energy lost during the bounce.6 marks

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