25-30 - Work, Energy, Power and Efficiency

Work as Energy Transfer

Imagine pulling a heavy trolley across a laboratory floor with a rope. Your hand pulls up and forwards, but the trolley moves horizontally. The whole force is real, but only the forward component transfers energy into the trolley's horizontal motion.

In physics, work done is an energy transfer caused by a force acting through a displacement:

Since $1\,\mathrm{J} = 1\,\mathrm{N\,m}$, work is the bridge between a force story and an energy story.

If the force is not along the line of motion, use the component of the force along the displacement. For a force $F$ at angle $\theta$ to the displacement:

so

That is not a new idea to memorise separately. It is just $\Delta W = F\Delta s$ after choosing the part of the force that actually acts along the displacement.

The sign also has a physical meaning. A force in the direction of motion does positive work and transfers energy into the object's mechanical stores. A force opposite the motion does negative work and removes mechanical energy. A sideways or perpendicular force does no work on that displacement, even if the force is large.

Worked example: a trolley is pulled $12.0\,\mathrm{m}$ along a level floor by a rope tension of $180\,\mathrm{N}$ at $35^{\circ}$ above the horizontal. Calculate the work done by the rope on the trolley.

Decision point 1: the trolley moves horizontally, so the vertical part of the rope force is not along the displacement.

Decision point 2: the force component is in the direction of motion, so the work done by the rope is positive.

A common trap is to use $180 \times 12.0$. That would treat the upward component as if it moved the trolley horizontally. Another common trap is everyday language: holding a heavy bag can feel like hard work, but if the force on the bag is vertical and the bag's displacement is zero, the work done on the bag by that force is zero.

In an Edexcel calculate question, show the component step, substitution and unit. In a show that question, keep more digits than the rounded value in the question before you state the final comparison.

Kinetic Energy

Now suppose the trolley speeds up. The energy transferred by work done has appeared as kinetic energy, the energy store associated with motion.

For a body of mass $m$ moving at speed $v$:

Kinetic energy is a scalar: it does not have a direction. Velocity is a vector in mechanics, but this equation uses speed, so either velocity sign gives the same positive squared value. The energy itself has no direction.

The square matters. Doubling the speed does not double the kinetic energy; it makes it four times as large. That is why stopping a fast object is so demanding: the work needed to remove its kinetic energy grows with $v^{2}$.

Worked example: a $0.42\,\mathrm{kg}$ ball leaves a launcher at $18\,\mathrm{m\,s^{-1}}$. Calculate its kinetic energy.

Decision point 1: use kilograms and metres per second, not grams or kilometres per hour.

Decision point 2: square the speed before multiplying by $\frac{1}{2}m$.

If the speed were doubled to $36\,\mathrm{m\,s^{-1}}$, the kinetic energy would be $\frac{1}{2} \times 0.42 \times 36^{2} = 272\,\mathrm{J}$, four times larger. The ball is not "twice as energetic" because speed is not the whole story; the equation says the motion energy depends on speed squared.

This is also where work and kinetic energy begin to talk to each other. If the ball gained $68\,\mathrm{J}$ of kinetic energy from rest, the net work done by all forces on its translational motion was $68\,\mathrm{J}$. Work by one chosen force alone need not equal the kinetic-energy gain, because other forces may also do work.

The work–energy theorem connects the equations: for constant resultant force along the motion, $W_{\rm net}=Fs=mas$. Since $v^2-u^2=2as$, this gives $W_{\rm net}=\tfrac12m(v^2-u^2)=\Delta E_k$. For braking, the work on the object is negative and the energy lost is a positive magnitude. Calculate the difference of the two squared speeds, not the square of their difference.

Gravitational Potential Energy and Conservation

The same trolley can be pulled up a ramp. The rope may act along the sloping path, but gravity cares about vertical height. Near the Earth's surface, the change in gravitational potential energy is:

The phrase "vertical height" is doing important work. A long shallow ramp and a short steep ramp can lead to the same change in gravitational potential energy if the vertical rise is the same. The path length matters for work done by a force acting along the path; the vertical height matters for gravitational potential energy.

Conservation of energy then becomes an accounting rule:

For many mechanics problems in this lesson, that means tracking work done, kinetic energy and gravitational potential energy. If resistive forces are negligible, a loss in gravitational potential energy can become a gain in kinetic energy. If resistive forces are present, some energy is dissipated to thermal stores in the surroundings.

Worked example: a $65\,\mathrm{kg}$ rider rolls from rest down a hill with a vertical drop of $2.40\,\mathrm{m}$. Air resistance and friction are negligible. Determine the rider's speed at the bottom.

Decision point 1: use vertical drop, not the distance along the slope.

Loss of gravitational potential energy:

Decision point 2: the rider starts from rest and the lost gravitational potential energy becomes kinetic energy.

The mass appears on both sides, so it cancels. This is not because mass is irrelevant to energy; it is because both the lost gravitational potential energy and the gained kinetic energy are proportional to mass in this ideal model.

Common misconception contrast: conservation of energy does not mean kinetic energy is constant. It means the total energy transfer account balances. In this example, gravitational potential energy decreases while kinetic energy increases.

Paper 3 can turn this into a practical reasoning question. If you were checking this model with a ramp and trolley, measuring the vertical height with a metre rule and timing or using light gates for speed would both matter. Repeating readings helps judge random variation; a poorly measured height gives a percentage uncertainty in $mg\Delta h$.

Power as Rate of Transfer

Work and energy tell you how much has been transferred. Power tells you how quickly the transfer happens.

A powerful motor does not necessarily transfer more energy overall than a weaker motor. It transfers energy at a greater rate.

There is a notation warning hiding in plain sight: in $P = W/t$, the $W$ in the numerator means work done, while $W$ after a number such as $370\,\mathrm{W}$ means the unit watt. The surrounding words and units tell you which meaning is being used.

Worked example: a motor lifts a $45\,\mathrm{kg}$ crate through a vertical height of $2.50\,\mathrm{m}$ in $3.0\,\mathrm{s}$ at constant speed. Calculate the useful power output.

Decision point 1: constant speed means the crate's kinetic energy does not change. The useful energy output is the gain in gravitational potential energy.

Decision point 2: power is the rate of this energy transfer.

So the useful power output is about $370\,\mathrm{W}$.

There is a useful graph representation here. On an energy-time graph, the gradient is power:

A straight section gives a constant power. A steeper section means a greater rate of transfer. If the graph is curved, the average power over an interval is the gradient of a chord, while the instantaneous power at one moment is the gradient of a tangent.

This is where Edexcel wording matters. Calculate power if the energy and time are directly given. Determine power if you first have to work out the energy transfer from the situation, such as $mg\Delta h$.

For a constant force component along the motion, $P=W/t=Fs/t=Fv$. More generally the instantaneous mechanical power is $Fv\cos\theta$. A motor pulling at $150\,\mathrm N$ along the motion at $2.0\,\mathrm{m\,s^{-1}}$ transfers $300\,\mathrm W$. This relation lets you connect a steady lifting speed or resistive force to a power rating.

Efficiency and Useful Output

An energy transfer can be complete without being useful. A winch lifting a crate may transfer energy electrically, mechanically, thermally and by sound. Energy has not disappeared, but only part of the input may end up in the store we wanted.

Efficiency measures the useful fraction of the input:

Efficiency has no unit. It can be written as a decimal or as a percentage. For a percentage, multiply the decimal by $100$.

The word useful must be defined by the task. For a lamp, useful output may be light. For a winch, useful output may be gravitational potential energy gained by the load. For brakes, useful output might be a controlled reduction in kinetic energy, even though the energy ends up mostly as thermal energy.

Worked example: a pump transfers $1.80\,\mathrm{kJ}$ of energy from its power supply while raising water. The useful gain in gravitational potential energy of the water is $1.35\,\mathrm{kJ}$. Calculate the efficiency of the pump.

Decision point 1: both quantities are energies and both are in $\mathrm{kJ}$, so the ratio is allowed without converting. If one were a power and one were an energy, that would not be allowed.

As a percentage:

The remaining $25.0\%$ was not destroyed. It was transferred to less useful stores, such as thermal energy of the pump and surroundings, and sound.

Common misconception contrast: a $75\%$ efficient device does not "lose" $25\%$ of the energy from physics. It means $25\%$ of the input was not transferred to the useful output named in the question.

Energy Transfers in a Motor and Load

The hardest part of this topic is often not the algebra. It is choosing the right energy route from the words.

Use this decision sequence:

1. Is there a force acting through a distance? Use work done. If the force is angled, use the component along the displacement.
2. Is a speed involved? Use kinetic energy.
3. Is there a vertical height change near Earth's surface? Use gravitational potential energy.
4. Is time involved? Use power as a rate of energy transfer or work done.
5. Is the question asking how effective a device is? Define the useful output, then use efficiency.

Worked synthesis: a motor pulls a $120\,\mathrm{kg}$ crate $18\,\mathrm{m}$ up a ramp. The ramp raises the crate vertically by $4.0\,\mathrm{m}$. The mean cable force over the path is $410\,\mathrm{N}$ along the ramp. The crate starts and ends at rest. The motor takes $9.0\,\mathrm{s}$, and its total electrical input power is $920\,\mathrm{W}$.

Determine the work done by the cable, the energy dissipated as the crate moves, and the overall efficiency if the useful output is the gain in gravitational potential energy.

Decision point 1: cable work uses the ramp distance because the cable force is along the ramp.

Decision point 2: gravitational potential energy uses the vertical height.

Decision point 3: the crate starts and ends at rest, so there is no net change in kinetic energy. The difference between cable work and gained gravitational potential energy is dissipated by resistive forces while the crate moves.

Decision point 4: efficiency uses the total input energy and the defined useful output. The total input energy is found from input power and time.

Overall efficiency $= 57\%$.

Notice the command-word transfer. "Determine" required choosing a route from the context. A "calculate" version would probably give the relevant quantities more directly. An "explain" version would need linked reasoning, such as "height is used for gravitational potential energy because the gravitational force acts vertically". A "show that" version would need enough working and precision to justify the stated value.