MC7 - Power

MC7 - Power

Two engines can do the same work but take different times. MC7 measures that rate using power and connects it to force and speed. You will distinguish driving force from resultant force, analyse steady and changing speeds, and use a supplied efficiency to connect input power with useful output.

Power as the rate of work

A 600 kg600\,\mathrm{kg} van develops a constant driving power of 6 kW6\,\mathrm{kW} on a horizontal road against a constant 300 N300\,\mathrm{N} resistance. Can it accelerate at 10 m s−110\,\mathrm{m\,s^{-1}}, and can it sustain 20 m s−120\,\mathrm{m\,s^{-1}}? An engine's power does not by itself tell us its force: speed matters too.

First consider a hoist. Lifting a load through a given height transfers the same gravitational energy whether the lift takes three seconds or six. Doing that work in half the time requires twice the average power.

Power

Power is the rate at which work is done, or energy is transferred. Its SI unit is the watt: 1 W=1 J s−11\,\mathrm{W}=1\,\mathrm{J\,s^{-1}}. A kilowatt is 1000 W1000\,\mathrm{W}; it is a unit of power, not energy.

A rate over a whole time interval need not equal the rate at every instant inside it.

Average and instantaneous power

If work ΔW\Delta W is done in time Δt>0\Delta t>0,

Paverage=ΔWΔtP_{\text{average}}=\frac{\Delta W}{\Delta t}

At an instant, P=dW/dtP=\mathrm{d}W/\mathrm{d}t. For constant power, W=PΔtW=P\Delta t; otherwise total work is the area under a power-time graph, W=∫P dtW=\int P\,\mathrm{d}t.

Not in the formula booklet: learn the meaning of power and these rate relations.

A motor transferring 600 J600\,\mathrm{J} in one second and 200 J200\,\mathrm{J} in the next has average power 400 W400\,\mathrm{W} over both seconds. Its power need not be constant at 400 W400\,\mathrm{W}.

Worked example 1: a six-second lift

Question. A hoist lifts an 80 kg80\,\mathrm{kg} load vertically through 3 m3\,\mathrm{m} in 6 s6\,\mathrm{s}. The load starts and ends at rest. Neglect losses. Find the average power transferred to the load. Use g=10 m s−2g=10\,\mathrm{m\,s^{-2}}.

Plan. The work equals the gain in gravitational energy because the endpoint kinetic energies are equal; divide by the time.

W=mgh=80(10)(3)=2400 Jenergy gainedPaverage=24006=400 Wwork per second\begin{aligned} W&=mgh=80(10)(3)=2400\,\mathrm{J} && \text{energy gained}\\ P_{\text{average}}&=\frac{2400}{6}=400\,\mathrm{W} && \text{work per second} \end{aligned}

Answer. 400 W400\,\mathrm{W}.

Check. 400(6)=2400 J400(6)=2400\,\mathrm{J}, recovering the energy transfer. The endpoint condition, rather than constant speed throughout, lets us omit a kinetic-energy change.

Keep the load and height fixed, but halve the lifting time.

Worked example 2: a three-second lift

Question. The same load is lifted through the same height, again starting and ending at rest, in 3 s3\,\mathrm{s}. Find the average useful power, neglecting losses.

Plan. Keep the work at 2400 J2400\,\mathrm{J}, but divide by the shorter time.

Paverage=80(10)(3)3same energy, new time=800 Wevaluate\begin{aligned} P_{\text{average}}&=\frac{80(10)(3)}{3} && \text{same energy, new time}\\ &=800\,\mathrm{W} && \text{evaluate} \end{aligned}

Answer. 800 W800\,\mathrm{W}.

Check. Power has doubled while time has halved: 800(3)=400(6)800(3)=400(6). Both lifts transfer the same energy.

The average tells us nothing about the largest instantaneous power needed during either lift. That requires information about how the motion changes during the interval.

Watch out

Wrong: power is the energy transferred, or P=WΔtP=W\Delta t. Power is a rate: divide work by elapsed time. Convert kilowatts to watts before combining them with joules and seconds.

Use the endpoint energy change in this new lift.

Driving power, force and speed

A driving force of 200 N200\,\mathrm{N} does 200 J200\,\mathrm{J} of work for every metre travelled along its direction. At 10 m s−110\,\mathrm{m\,s^{-1}}, it transfers 2000 J2000\,\mathrm{J} each second. The force multiplied by speed gives power.

For aligned force and motion, a short displacement gives dW=F ds\mathrm{d}W=F\,\mathrm{d}s. Divide by the short time interval to obtain dW/dt=F ds/dt\mathrm{d}W/\mathrm{d}t=F\,\mathrm{d}s/\mathrm{d}t.

Power of a force

For a force of magnitude FF acting along the velocity, with speed vv,

P=FvP=Fv

More generally, at angle θ\theta between force and velocity, P=Fvcos⁡θP=Fv\cos\theta. The power is positive for a force aiding motion and negative for a force opposing it. Apply the formula to the particular force whose power is required.

Not in the formula booklet: learn P=FvP=Fv and its direction condition.

Driving power uses the driving force. The net power of all forces is instead the rate of change of kinetic energy; these differ when resistance acts. At steady speed a vehicle can use substantial driving power while its kinetic energy stays constant.

Worked example 3: force at one speed

Question. A vehicle's engine supplies a constant driving force of 200 N200\,\mathrm{N}, aligned with its motion. Find the driving power at 10 m s−110\,\mathrm{m\,s^{-1}}.

Plan. Multiply the driving force by the instantaneous speed.

P=Fvforce aligned with velocity=200(10)=2000 W=2 kWSI units\begin{aligned} P&=Fv && \text{force aligned with velocity}\\ &=200(10)=2000\,\mathrm{W}=2\,\mathrm{kW} && \text{SI units} \end{aligned}

Answer. 2 kW2\,\mathrm{kW}.

Check. If that speed were maintained for one second, the force would do 200(10)=2000 J200(10)=2000\,\mathrm{J} of work in that second.

Keep the driving force fixed and double only the speed. The engine must transfer energy twice as fast.

Worked example 4: the same force at twice the speed

Question. The same driving force acts when the vehicle travels at 20 m s−120\,\mathrm{m\,s^{-1}}. Find the driving power.

Plan. Substitute the new speed into the same relation.

P=200(20)=4000 Wsame force, twice the speed=4 kWconvert units\begin{aligned} P&=200(20)=4000\,\mathrm{W} && \text{same force, twice the speed}\\ &=4\,\mathrm{kW} && \text{convert units} \end{aligned}

Answer. 4 kW4\,\mathrm{kW}.

Check. P/v=4000/20=200 NP/v=4000/20=200\,\mathrm{N}, recovering the specified driving force. Doubling vv at fixed FF doubles PP.

If power is fixed instead, rearrange to F=P/vF=P/v for v>0v>0. A higher speed then means a smaller available driving force. That is a different assumption from fixed driving force.

Watch out

Wrong: P=ma vP=ma\,v for the engine when resistance acts. The engine power is driving force times speed. Use Newton’s law to connect that force, resistance and acceleration.

Rearrange the relation and convert kilowatts carefully.

Steady and changing speeds

At constant power, a vehicle can have a large driving force at a modest speed but less force available at a high speed. Compare that force with resistance to decide whether it speeds up, slows down or maintains its speed.

For horizontal straight-line motion with driving power PP, speed v>0v>0, mass mm and opposing resistance RR, Newton’s second law gives

Pv−R=ma\frac{P}{v}-R=ma

This is an instantaneous equation. If PP and RR are constant while vv changes, then aa changes too.

Power and steady speed

On a horizontal road at steady speed,

Fdrive=R,P=RvF_{\text{drive}}=R,\qquad P=Rv

If PP is a constant available power and R>0R>0 is constant, the balance speed is v=P/Rv=P/R. Below it the resultant force is forwards; above it the resultant force is backwards. A maximum sustainable steady speed uses the maximum available power under the stated resistance model.

Not in the formula booklet: derive this balance using P=FvP=Fv and Newton’s law.

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