Free Pearson Edexcel International A Level Physics lesson

Work, energy and power

Account for the transfer, then ask how quickly and how usefully it happens.

What you’ll learn

  • Calculate work with force components and interpret its sign.
  • Use kinetic and gravitational energy in a complete conservation account.
  • Distinguish power from energy and calculate useful-output efficiency.

A force transfers energy through displacement

Work done by a constant force equals its component along a displacement multiplied by that displacement. If a force F makes angle θ to the displacement, W = Fs cos θ. Here the point of application must move, and the force must have a component along that movement. A person supporting a stationary case may use metabolic energy, but the support force does no mechanical work on the case while its point of application is stationary.

The sign of work describes the transfer by that force relative to the motion. A driving force in the direction of displacement does positive work; a resisting force opposite the displacement does negative work on the moving object. A perpendicular force does zero work in this ideal description. In a straight-line constant-mass translation, the work of the resultant force equals change in kinetic energy. Do not mistake work by one applied force for net work when resistance also acts.

On a graph of force component along the motion against displacement, area gives work. A constant component makes a rectangle; a changing component requires the actual area or a justified numerical estimate. The triangular force–extension area for an ideal spring is one special example. Keep the distinction between force and energy visible: a large force through a tiny distance can transfer less energy than a smaller force over a longer distance.

W = FΔs cos θ Net work = ΔEk Area under force-along-motion versus displacement = work

Symbols and units

  • W is work in J; F is constant force magnitude in N; Δs is displacement magnitude in m.
  • θ is the angle between force and displacement; Ek is kinetic energy in J.

How the equation explains the idea

  1. The component along displacement is F cos θ. Multiplying it by Δs gives W = FΔs cos θ.
  2. A perpendicular force does no work; an opposing force does negative work. Summing the work of all forces gives ΔEk, not just the work by the driving force.
  3. For changing force, accumulate small contributions: the signed area under the along-motion force against displacement replaces a single rectangle.

Worked example

Pulling an equipment sled

  1. A constant 50.0 N force pulls a sled 4.0 m horizontally at 30.0° above horizontal.
  2. Work by the pull = 50.0 × 4.0 × cos30.0° = 173 J.
  3. If a constant 20.0 N resistance opposes motion, its work is −20.0 × 4.0 = −80.0 J.
  4. The vertical forces do no work over horizontal displacement; net work is approximately 93.2 J.

The pull transfers about 173 J, while the sled’s kinetic energy increases by about 93 J.

Keep this idea. Specify which force does the work, resolve along displacement and keep the sign of the transfer.

Balance the complete energy account

Kinetic energy is Ek = ½mv². It is scalar and depends on speed squared, so reversing velocity without changing speed leaves kinetic energy unchanged. Near Earth in a uniform gravitational field, a change in the Earth–object system’s gravitational potential energy is ΔEgrav = mgΔh. Height is a vertical difference relative to a chosen reference, not distance along a ramp. A reference changes the numerical potential energy but not its physically useful changes.

Conservation of energy means a complete account includes all initial stores, transfers into or out of the chosen system and final stores. In an ideal frictionless descent, lost gravitational potential energy becomes kinetic energy. With friction or drag, include increased internal/thermal energy and any other relevant transfers, so mechanical energy alone may decrease even though total energy is conserved. Do not count gravitational work and the same gravitational potential-energy loss as two separate inputs.

Write an energy balance before substituting. For a mass starting from rest and dropping through height h with energy D transferred away from mechanical stores, mgh = ½mv² + D. If it already moves, include its initial kinetic energy. Energy often finds a speed without needing time or detailed force variation, but it does not give the velocity direction by itself. A negative value calculated for v² or a demanded useful output above total input signals incompatible assumptions or data rather than a new kind of speed.

Ek = ½mv² ΔEgrav = mgΔh Initial energy + transfers in = final energy + transfers out

Symbols and units

  • Ek is kinetic energy in J; m is mass in kg; v is speed in m s⁻¹.
  • ΔEgrav is gravitational potential-energy change in J; g is field magnitude in m s⁻²; Δh is upward height change in m.

How the equation explains the idea

  1. Kinetic energy Ek = mv²/2 grows with speed squared; doubling speed quadruples this energy at fixed mass.
  2. Near Earth with constant g, raising mass through Δh requires work mgΔh. If only gravity transfers energy between these stores, their sum stays constant.
  3. A fall through height h from rest then gives mgh = mv²/2 and v = √(2gh). Include drag or other transfers before using that result.
Kinetic energy depends quadratically on speed

Conditions. Computed Ek = ½mv² for a constant mass m = 2 kg in a nonrelativistic model. The graph compares states, not a specified time history.

Speed / m s⁻¹; Kinetic energy / J.

At 4 m s⁻¹, Ek is 16 J; at 8 m s⁻¹ it is 64 J. The fourfold increase comes from squaring speed. A doubling of kinetic energy increases speed by √2, rather than by two.

Open the interactive lesson to view the plotted curves and their coordinate tables.

Worked example

A descent with dissipated energy

  1. A 2.0 kg cart starts from rest and drops 1.5 m vertically. Use g = 10 m s⁻²; 6.0 J transfers to thermal energy.
  2. Initial available gravitational energy = 2.0 × 10 × 1.5 = 30 J.
  3. Final kinetic energy = 30 − 6.0 = 24 J.
  4. ½ × 2.0 × v² = 24, so speed v = √24 = 4.90 m s⁻¹.

The final speed is lower than the frictionless value because 6 J leaves the mechanical energy account.

Keep this idea. State the system, use vertical height changes and include every relevant energy store exactly once.

Rate and efficiency answer different questions

Power is the rate of energy transfer or work done, measured in watts, where 1 W = 1 J s⁻¹. Average power is total energy transferred divided by elapsed time. Two motors can transfer the same energy but have different average powers if one completes the task sooner. Instantaneous power can vary through a lift as force and speed change; a whole-task average does not describe every instant.

Efficiency compares useful output with total input over matching boundaries and time intervals. Use either energy ratio or power ratio and multiply by 100 for a percentage. The useful output depends on the task: in a hoist it might be the gravitational energy gained by the load, while thermal energy from motor resistance is not useful for lifting. State whether an input is electrical, chemical or mechanical, and do not compare one stage’s output with a different stage’s input. A passive device cannot yield more total useful energy than the total supplied within a complete account.

At steady motion with a constant force along velocity, P = Fv follows from W = Fs and v = s/t. It is a useful extension of the rate equation, with the force component along motion if inclined. Greater power does not automatically mean greater efficiency. A powerful motor may waste a larger fraction while still doing useful work faster. For sequential stages with no other inputs, multiply fractional efficiencies; percentages cannot simply be added. Always check that the final fraction is between zero and one for the stated energy-conversion device.

Pavg = ΔE/Δt = W/Δt Efficiency = useful output/total input Percentage efficiency = 100 × fraction

Symbols and units

  • Pavg is average power in W = J s⁻¹; ΔE or W is transferred energy or work in J; Δt is duration in s.
  • Efficiency is a dimensionless useful-output/input ratio, expressed as a fraction or percentage.

How the equation explains the idea

  1. Divide transferred energy by its duration: Pavg = ΔE/Δt. Doubling transfer time for the same energy halves average power.
  2. Use either energies for the same event or powers for the same operating condition in the efficiency ratio. An 80 W useful output from 100 W input gives 0.80 = 80%.
  3. The remaining 20 W is an energy-transfer rate to other destinations, not destroyed energy.

Worked example

A hoist’s useful power and efficiency

  1. A hoist lifts a 12.0 kg load through 2.50 m in 5.00 s at unchanged initial/final speed. Use g = 10.0 m s⁻².
  2. Useful energy = mgh = 12.0 × 10.0 × 2.50 = 300 J, so average useful power = 300/5.00 = 60.0 W.
  3. If average electrical input power is 100 W over the same lift, input energy is 500 J.
  4. Efficiency = 60.0/100 = 0.600 = 60.0%.

Power measures transfer rate; efficiency measures the useful fraction of the matching input.

Keep this idea. Match the system and time interval before taking an efficiency ratio; more power and better efficiency are different claims.

Open the interactive lesson for free practice and reviewed three-question samples when available. New full question sets and timed unit checks are in preparation. Existing complete Physics revision packs retain paid access; existing free teaching PDFs remain free.

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