4.5 - Power and motion | Cambridge A-Level Mathematics

An engine's power tells you how quickly it transfers energy, rather than the size of its driving force alone. You will calculate average power from work and time, use speed to find an instantaneous driving force, and combine that force with resistance and gravity to calculate a vehicle's acceleration on a hill.

Power as a rate

Two hoists lift the same load through the same height, but one takes half the time. Which does more work? Neither: the work is the same. The faster hoist delivers it at twice the average power.

A $100\,\mathrm{kg}$ load raised $8\,\mathrm{m}$ at constant speed, with resistance neglected, needs work $100(10)(8)=8000\,\mathrm{J}$ . Delivering that work in $20\,\mathrm{s}$ requires $400$ joules each second on average. Throughout, take $g=10\,\mathrm{m\,s^{-2}}$ .

Power

Power is the rate at which a force does work. Its unit is the watt, $1\,\mathrm{W}=1\,\mathrm{J\,s^{-1}}$ . A kilowatt is $1000\,\mathrm{W}$ . Average power describes an interval; instantaneous power describes a particular instant.

An interval's work and duration give its average rate of energy transfer. The power can vary within that interval.

Average power

For work $W$ done in time $t>0$ ,

P_{\text{average}}=\frac Wt.

Use joules and seconds to obtain watts. If power is constant, this average equals that constant value.

Not in the formula booklet: learn it.

First calculate the work of the force whose power is requested. Lifting at constant speed with no resistance needs $mgh$ ; lifting with increased speed also needs the kinetic-energy increase, and resistance requires additional work. Do not insert a force in newtons directly into work divided by time.

Worked example 1: a slower lift

Question. A hoist raises a $100\,\mathrm{kg}$ load vertically through $8\,\mathrm{m}$ at constant speed in $20\,\mathrm{s}$ . Resistance is neglected. Find the average power of the lifting force.

Plan. Constant speed means no kinetic-energy change, so the lifting work equals the potential-energy gain.

\begin{aligned} W&=100(10)(8)=8000\,\mathrm{J}\\ P_{\text{average}}&=8000/20=400\,\mathrm{W} \end{aligned}

Answer. Average power $400\,\mathrm{W}=0.4\,\mathrm{kW}$ .

Check. $400\,\mathrm{J}$ each second for $20\,\mathrm{s}$ transfers $8000\,\mathrm{J}$ , the required work. Power has units of energy per time, not force per time.

Halve only the duration, keeping constant speed within each lift and the same load and height. The required work stays fixed.

Worked example 2: the same work in less time

Question. A hoist raises the same load through the same $8\,\mathrm{m}$ , now at constant speed in $10\,\mathrm{s}$ , again with resistance neglected. Find its average power.

Plan. Divide the unchanged work by the new duration.

P_{\text{average}}=8000/10=800\,\mathrm{W}=0.8\,\mathrm{kW}.

Answer. Average power $800\,\mathrm{W}$ , twice the previous power.

Check. $800(10)=8000\,\mathrm{J}$ , so this hoist transfers the same energy. Halving the time doubles work per second.

These models describe the stated constant-speed lifting interval; they do not include a separate startup or stopping interval. If such a stage is included in a question, account for its actual work and duration.

Watch out

Power and work are different quantities. A high power can mean the same work is done in less time. Convert $\mathrm{kW}$ to $\mathrm{W}$ before combining with joules and seconds.

Find the work of lifting before dividing by the journey time.

Instantaneous power

During a short interval, a force $D$ in the direction of motion does approximately $D\Delta s$ of work. Dividing by $\Delta t$ , then considering the rate at that instant, gives $D$ times speed. This lets us turn an engine's power into its driving force at a known speed.

Power, force and speed

For a force in the direction of motion,

P=Dv.

Use $P$ in watts, driving force $D$ in newtons and speed $v$ in $\mathrm{m\,s^{-1}}$ . For $v>0$ ,

D=\frac Pv.

These are instantaneous quantities. Constant power need not mean constant force.

Not in the formula booklet: learn it.

The unit check is $\mathrm{N}\times\mathrm{m\,s^{-1}}=\mathrm{J\,s^{-1}}$ . A force of $1500\,\mathrm{N}$ acting at $12\,\mathrm{m\,s^{-1}}$ transfers $18000\,\mathrm{J}$ each second, or $18\,\mathrm{kW}$ .

Worked example 3: driving force at a lower speed

Question. A car's engine delivers power $18\,\mathrm{kW}$ . Find its driving force when the car travels at $12\,\mathrm{m\,s^{-1}}$ , with the force in the direction of motion.

Plan. Convert power to watts, then divide by the instantaneous speed.

D=\frac{18000}{12}=1500\,\mathrm{N}.

Answer. Driving force $1500\,\mathrm{N}$ .

Check. $1500(12)=18000\,\mathrm{W}$ , recovering the specified power. No acceleration information was needed to find this force.

Increase only the speed while holding the engine power fixed. Each newton now transfers energy faster, so fewer newtons provide the same power.

Worked example 4: driving force at a higher speed

Question. The engine still delivers $18\,\mathrm{kW}$ , but the car now travels at $20\,\mathrm{m\,s^{-1}}$ . Find the driving force.

Plan. Use the same power with the new speed.

D=\frac{18000}{20}=900\,\mathrm{N}.

Answer. Driving force $900\,\mathrm{N}$ .

Check. Speed has increased by factor $20/12=5/3$ , while force has decreased by factor $900/1500=3/5$ . Their product remains $18000\,\mathrm{W}$ .

The expression $P/v$ requires $v>0$ ; it cannot describe starting from rest at fixed positive power with a finite force. Use it within the moving-vehicle interval for which the model is given. If the question supplies power at one instant only, do not assume it stays constant afterwards.

Watch out

Power is not a force. The line $18000-r=ma$ mixes watts and newtons. First calculate $D=P/v$ , then put $D$ , resistance and any weight component into Newton's law.

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4.5 - Power and motion

Power as a rate

Instantaneous power

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