Work energy power and efficiency

Work Done by a Force

Key Equations

Work done:

$$\Delta W=F\Delta scos\theta $$

Variables:

• ΔW = work done (J)
• F = magnitude of the applied force (N)
• Δs = magnitude of the displacement (m)
• θ = angle between the force and the displacement

SI unit: joule (J), where 1 J = 1 N m

Rearrangements:

$$F=\frac{\Delta W}{\Delta scos\theta } \Delta s=\frac{\Delta W}{Fcos\theta }$$

ProportionalityWork done is directly proportional to the applied force when displacement is fixed, and directly proportional to displacement when force is fixed.

Work is done on an object whenever a force causes a displacement in the direction of that force.

When the force acts along the line of motion, the equation reduces to ΔW = FΔs because cos 0° = 1. When the force acts at an angle θ to the displacement, only the component F cos θ acts along the motion. The perpendicular component F sin θ does no work, because no displacement occurs in that direction.

If the force is perpendicular to the displacement, cos 90° = 0 and no work is done. This is why the centripetal force in uniform circular motion does zero work on the object. The kinetic energy of the object therefore remains constant.

Work done is a scalar quantity, but its sign carries physical meaning. Positive work means the force has a component in the direction of motion, transferring energy to the object. Negative work means the force opposes the motion, transferring energy away from the object.

MisconceptionA force can be present without doing any work. A waiter carrying a tray horizontally exerts an upward force, but the displacement is horizontal, so cos 90° = 0 and no work is done on the tray.

Exam TipAlways check the angle between force and displacement before applying ΔW = FΔs.

Worked Example: Pulling a Sledge at an Angle

Scenario

A child pulls a sledge along level snow using a rope inclined at 35° above the horizontal. The tension in the rope is 48 N and the sledge moves 22 m. Calculate the work done by the tension.

Equation used

$$\Delta W=F\Delta scos\theta $$

Given

$$F=48\text{ N}$$

$$\Delta s=22\text{ m}$$

$$\theta =35{}^{∘}$$

Substitution:

$$\Delta W=48\times 22\times cos35{}^{∘}$$

$$\Delta W=48\times 22\times 0.8192$$

$$\Delta W=865.0\text{ J}$$

Answer

$$\Delta W\approx 865\text{ J (3 s.f.)}$$

Interpretation

Only the horizontal component of the tension does work on the sledge. The vertical component lifts the rope but produces no horizontal displacement.

Kinetic Energy of a Body

Key Equations

Kinetic energy:

$${E}_{k}=\frac{1}{2}m{v}^{2}$$

Variables:

• $E_{\mathrm{k}}$ = kinetic energy (J)
• m = mass (kg)
• v = speed (m s⁻¹)

SI unit: joule (J)

Rearrangements:

$$v=\sqrt{\frac{2{E}_{k}}{m}} m=\frac{2{E}_{k}}{{v}^{2}}$$

ProportionalityKinetic energy is directly proportional to mass at fixed speed. Kinetic energy is proportional to the square of speed; doubling the speed quadruples the kinetic energy.

The kinetic energy of a body is the energy stored in it because of its motion.

Because v is squared, kinetic energy depends much more strongly on speed than on mass. Tripling the speed multiplies the kinetic energy by nine. Halving the speed reduces the kinetic energy to one quarter of its original value.

When a resultant force does work on a body, the work done equals the change in kinetic energy. This is the work–energy principle. It links the force-based and energy-based descriptions of motion together into one consistent framework.

Worked Example: Speed of a Cyclist Given Kinetic Energy

Scenario

A cyclist and bicycle have a combined mass of 78 kg and a kinetic energy of 1.40 kJ. Calculate the cyclist’s speed.

Conversion of kinetic energy to SI units:

$${E}_{k}=1.40\text{ kJ}\times 1000$$

$${E}_{k}=1400\text{ J}$$

Equation used

$${E}_{k}=\frac{1}{2}m{v}^{2}$$

Rearranging for v:

$$v=\sqrt{\frac{2{E}_{k}}{m}}$$

Given

$${E}_{k}=1400\text{ J}$$

$$m=78\text{ kg}$$

Substitution:

$$v=\sqrt{\frac{2\times 1400}{78}}$$

$$v=\sqrt{35.90}$$

$$v=5.99{\text{ m s}}^{-1}$$

Answer

$$v\approx 6.0{\text{ m s}}^{-1}\text{ (2 s.f.)}$$

Interpretation

The rearrangement is essential here because the unknown sits inside a square term. Squaring the calculated speed and using ½mv² confirms the original kinetic energy.

Gravitational Potential Energy Near Earth’s Surface

Key Equations

Change in gravitational potential energy:

$$\Delta {E}_{grav}=mg\Delta h$$

Variables:

• $\Delta E_{\mathrm{grav}}$ = change in gravitational potential energy (J)
• m = mass (kg)
• g = gravitational field strength (N kg⁻¹), taken as 9.81 N kg⁻¹ near Earth’s surface
• Δh = change in vertical height (m)

SI unit: joule (J)

Rearrangements:

$$\Delta h=\frac{\Delta {E}_{grav}}{mg} m=\frac{\Delta {E}_{grav}}{g\Delta h}$$

Proportionality$\Delta E_{\mathrm{grav}}$ is directly proportional to each of m, g and Δh.

The change in gravitational potential energy is the work done against the gravitational force when an object is raised through a vertical height Δh.

Only the vertical component of any displacement contributes to $\Delta E_{\mathrm{grav}}$. Moving an object horizontally changes no gravitational potential energy, because no work is done against gravity. A box pushed along a flat floor stays at the same height, so Δh = 0.

The equation $\Delta E_{\mathrm{grav}}$ = mgΔh is valid only when g is approximately constant. Near the Earth’s surface, g varies negligibly with height, so this approximation is excellent for everyday problems. For motion through large changes in altitude, a more general expression for gravitational potential energy is required.

Examiner InsightMany students plug in the diagonal distance travelled rather than the vertical height. For an inclined ramp of length L at angle α, the vertical height gained is L sin α.

Exam TipIdentify Δh as the vertical change only.

Worked Example: Skier on a Slope

Scenario

A skier of mass 72 kg descends a slope of length 250 m inclined at 18° to the horizontal. Calculate the loss in gravitational potential energy.

Finding the vertical height change:

$$\Delta h=Lsin\alpha $$

$$\Delta h=250\times sin18{}^{∘}$$

$$\Delta h=77.25\text{ m}$$

Equation used

$$\Delta {E}_{grav}=mg\Delta h$$

Given

$$m=72\text{ kg}$$

$$g=9.81{\text{ N kg}}^{-1}$$

$$\Delta h=77.25\text{ m}$$

Substitution:

$$\Delta {E}_{grav}=72\times 9.81\times 77.25$$

$$\Delta {E}_{grav}=54562\text{ J}$$

Answer

$$\Delta {E}_{grav}\approx 5.5\times {10}^{4}\text{ J (2 s.f.)}$$

Interpretation

This energy is transferred from the gravitational potential store of the skier–Earth system. It becomes kinetic energy and thermal energy as the skier descends the slope.

Conservation of Energy

The principle of conservation of energy states that the total energy of a closed system is constant; energy is only transferred between stores or by transfer pathways.

Energy can be stored as kinetic, gravitational potential, elastic, thermal, chemical, nuclear, electrostatic or magnetic. Energy is transferred mechanically (by a force doing work), electrically, by heating, or by radiation. No process creates or destroys energy.

For a body falling freely under gravity, gravitational potential energy is transferred to the kinetic energy store. If air resistance is negligible, mgΔh = ½mv², which gives v = √(2gΔh). When air resistance is significant, some energy is also transferred to the thermal store of the air and the object.

A practical strategy in conservation problems is to identify the initial and final states. List the energy stores at each, then equate the total before with the total after. Any work done by external forces (such as friction) appears as a transfer to the thermal store.

MisconceptionEnergy is not “used up” or “lost” when an object slows down due to friction. Work is done against friction, transferring energy from the kinetic store to the thermal store of the object and surroundings.

Exam TipAlways name the destination store rather than saying energy disappears.

Worked Example: Pendulum Swing

Scenario

A pendulum bob of mass 0.30 kg is released from rest at a height of 0.18 m above its lowest point. Calculate its speed at the lowest point, assuming air resistance is negligible.

Applying conservation of energy:

$$\Delta {E}_{grav}\text{ lost}={E}_{k}\text{ gained}$$

$$mg\Delta h=\frac{1}{2}m{v}^{2}$$

Rearranging for v (mass cancels):

$$v=\sqrt{2g\Delta h}$$

Given

$$g=9.81{\text{ N kg}}^{-1}$$

$$\Delta h=0.18\text{ m}$$

Substitution:

$$v=\sqrt{2\times 9.81\times 0.18}$$

$$v=\sqrt{3.532}$$

$$v=1.879{\text{ m s}}^{-1}$$

Answer

$$v\approx 1.9{\text{ m s}}^{-1}\text{ (2 s.f.)}$$

Interpretation

The speed depends only on the height drop, not on the mass. A heavier bob would reach the same speed because both the energy gained and the kinetic energy term scale with mass.

Power: Rate of Energy Transfer

Key Equations

Power:

$$P=\frac{E}{t} P=\frac{W}{t}$$

Variables:

• P = power (W)
• E = energy transferred (J)
• W = work done (J)
• t = time (s)

SI unit: watt (W), where 1 W = 1 J s⁻¹

Rearrangements:

$$E=Pt t=\frac{E}{P}$$

ProportionalityFor fixed energy, power is inversely proportional to time. Halving the time doubles the power. For a constant force moving an object at constant velocity, P = Fv (combining W = Fs and P = W/t).

The power developed by a device is the rate at which it transfers energy or does work.

A 1 W device transfers 1 J of energy every second. Multiplying power by time gives the total energy transferred. This is why electrical energy is often quoted in kilowatt-hours rather than joules. One kilowatt-hour equals 3.6 × 10⁶ J.

For an object moving at constant velocity v under a driving force F, the work done per second is Fv. Therefore the mechanical power output is P = Fv. This combination is useful for vehicles, lifts, and any machine moving at steady speed.

Worked Example: Lift Motor Power

Scenario

A lift motor raises a fully loaded cabin of mass 850 kg through 24 m at constant speed in 18 s. Calculate the useful output power of the motor.

Step 1: Finding the work done against gravity at constant speed.

Equation used

$$W=mg\Delta h$$

Given

$$m=850\text{ kg}$$

$$g=9.81{\text{ N kg}}^{-1}$$

$$\Delta h=24\text{ m}$$

Substitution:

$$W=850\times 9.81\times 24$$

$$W=200124\text{ J}$$

Step 2: Finding the useful output power.

Equation used

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

Given

$$W=200124\text{ J}$$

$$t=18\text{ s}$$

Substitution:

$$P=\frac{200124}{18}$$

$$P=11118\text{ W}$$

Answer

$$P\approx 1.1\times {10}^{4}\text{ W (2 s.f.)}$$

Interpretation

At constant speed, the upward driving force equals the weight, so all useful work goes into raising gravitational potential energy. A real motor would draw more electrical power because of internal losses.

Efficiency of Energy Transfer

Key Equations

Efficiency in terms of energy:

$$\text{efficiency}=\frac{\text{useful energy output}}{\text{total energy input}}$$

Efficiency in terms of power:

$$\text{efficiency}=\frac{\text{useful power output}}{\text{total power input}}$$

Variables: efficiency is dimensionless. It can be expressed as a decimal (between 0 and 1) or as a percentage by multiplying by 100.

SI unit: none (dimensionless ratio)

ProportionalityEfficiency is directly proportional to useful output for a fixed input. Increasing the wasted output reduces the efficiency.

The efficiency of a device is the fraction of the input energy or power that is transferred to a useful output store.

No real device is 100% efficient, because some energy is always transferred to non-useful stores. In an electric motor, electrical input becomes useful kinetic and gravitational potential energy, plus thermal energy in the windings due to resistance. In an incandescent lamp, most electrical input becomes thermal radiation rather than visible light.

A Sankey diagram is a useful visual representation. The width of each arrow is proportional to the energy or power it represents. The useful output arrow continues forward, while wasted energy branches off.

Examiner InsightEfficiency cannot exceed 1 (or 100%). If a calculated value exceeds this, the candidate has either inverted the ratio or confused useful output with input.

Exam TipAlways check that the useful output is smaller than the total input.

Worked Example: Efficiency of a Pump

Scenario

An electric pump uses 4.5 kW of electrical power to lift water against gravity at a useful rate of 3.1 kW. Calculate its efficiency as a percentage.

Equation used

$$\text{efficiency}=\frac{\text{useful power output}}{\text{total power input}}$$

Given

$${P}_{\text{useful}}=3.1\text{ kW}$$

$${P}_{\text{input}}=4.5\text{ kW}$$

Substitution (units cancel, so no conversion needed):

$$\text{efficiency}=\frac{3.1}{4.5}$$

$$\text{efficiency}=0.6889$$

Converting to a percentage:

$$\text{efficiency}=0.6889\times 100$$

Answer

$$\text{efficiency}\approx 69\%\text{ (2 s.f.)}$$

Interpretation

About 31% of the input power is transferred to non-useful stores, mainly thermal energy in the motor and pipework. Improving insulation and reducing friction would raise the efficiency.