6.02k–6.02m - Power and driving force

6.02k–6.02m - Power and driving force

Interpret power as a rate of work, calculate driving force at a given speed and find the greatest steady speed on a level or inclined road. You will also calculate instantaneous power from a variable force and a velocity vector in two dimensions, keeping the sign of the result.

Average and instantaneous power

Can an engine keep the same power while its driving force gets smaller? Yes: if the vehicle moves faster, each newton acts through more metres each second. The same work per second can then be delivered by a smaller force.

First separate an amount of work from the rate at which it is done. Two motors might each transfer 6000 J6000\,\mathrm{J}, but the one that takes half as long has twice the average power.

You already know

Work is measured in joules, 1 J=1 N m1\,\mathrm{J}=1\,\mathrm{N\,m}. The work-energy principle relates signed resultant work to the kinetic-energy change. Newton's second law is Fresultant=maF_{\mathrm{resultant}}=ma.

Power refers to a specified force or energy transfer. Engine power, power against resistance and the resultant rate of kinetic-energy change are different quantities unless the force balance makes them equal.

Power

Power is the rate at which a force does work. Average power is total work divided by the elapsed time; instantaneous power is the work rate at one instant. Its SI unit is the watt, with 1 W=1 J s−11\,\mathrm{W}=1\,\mathrm{J\,s^{-1}} and 1 kW=1000 W1\,\mathrm{kW}=1000\,\mathrm{W}. Work and power are not interchangeable: 600 J600\,\mathrm{J} is an amount of work, while 600 W600\,\mathrm{W} is a rate.

If W(t)W(t) is cumulative work by a specified force, the following expressions distinguish its average and instantaneous rates.

Power as a work rate

Over a time interval of duration Δt>0\Delta t>0,

P‾=ΔWΔtandP=dWdt\overline P=\frac{\Delta W}{\Delta t} \qquad\text{and}\qquad P=\frac{dW}{dt}

If power is constant, work done is W=PtW=Pt. A force doing negative work has negative power.

Not in the formula booklet: learn it.

For example, if the numerical cumulative work in joules is W(t)=3t2W(t)=3t^2, with tt in seconds, average power over the first 2 s2\,\mathrm{s} is 12/2=6 W12/2=6\,\mathrm{W}, while instantaneous power at t=2t=2 is 6t=12 W6t=12\,\mathrm{W}. The rate has increased during the interval.

Worked example 1: average power

Question. A motor does 6000 J6000\,\mathrm{J} of useful work in 20 s20\,\mathrm{s}. Find its average useful power in watts and kilowatts.

Plan. Divide useful work by elapsed time, then convert watts to kilowatts.

P‾=600020=300 Wwork divided by time=0.300 kWdivide by 1000\begin{aligned} \overline P&=\frac{6000}{20}=300\,\mathrm{W} && \text{work divided by time}\\ &=0.300\,\mathrm{kW} && \text{divide by }1000 \end{aligned}

Why this step? The calculation uses the useful work specified, so it gives useful output power; it does not determine the motor's electrical input power.

Answer. The average useful power is 300 W=0.300 kW300\,\mathrm{W}=0.300\,\mathrm{kW}.

Check. (300 J s−1)(20 s)=6000 J(300\,\mathrm{J\,s^{-1}})(20\,\mathrm{s})=6000\,\mathrm{J}.

Worked example 2 changes only the time taken: the same work is now done in half the time.

Worked example 2: the same work more quickly

Question. The motor does the same 6000 J6000\,\mathrm{J} of useful work in 10 s10\,\mathrm{s}. Find its average useful power.

Plan. Divide the unchanged work by the new elapsed time.

P‾=600010same work, shorter time=600 W=0.600 kWevaluate and convert\begin{aligned} \overline P&=\frac{6000}{10} && \text{same work, shorter time}\\ &=600\,\mathrm{W}=0.600\,\mathrm{kW} && \text{evaluate and convert} \end{aligned}

Why this step? Power compares work with time. Halving the time doubles the rate without changing the work total.

Answer. The average useful power is 600 W600\,\mathrm{W}.

Check. 600/300=2600/300=2, the reciprocal of the time factor 10/2010/20.

An average alone does not tell you how power varied within the interval.

Watch out

Wrong: P=WtP=Wt, with units written as watts. The product has units J s\mathrm{J\,s}, which are not power units. Right: average power is W/tW/t, with units J s−1\mathrm{J\,s^{-1}}.

Keep work and elapsed time visible in your next calculation.

Driving force and acceleration

When a driving force acts along the velocity, its instantaneous power is force multiplied by speed. This gives the driving force at the specified instant, not the resultant force on the vehicle.

In a short time Δt\Delta t, a vehicle moving at speed vv travels approximately vΔtv\Delta t. A forward force FF then does approximately FvΔtFv\Delta t work. Dividing by Δt\Delta t and taking the instantaneous limit gives P=FvP=Fv.

Power and tractive force

For a tractive force FF acting in the direction of velocity of speed vv,

P=Fvand, for v>0,F=PvP=Fv \qquad\text{and, for }v>0,\qquad F=\frac Pv

Use PP in watts and vv in m s−1\mathrm{m\,s^{-1}} to obtain FF in newtons. The resultant force must still include resistance and any component of weight.

Not in the formula booklet: learn it.

At fixed positive power, doubling speed halves the available driving force. The ideal constant-power formula cannot describe the instant of starting from rest: P/vP/v is undefined at v=0v=0, and real engines have force limits.

Worked example 3: the driving force is not the resultant

Question. A vehicle of mass 1200 kg1200\,\mathrm{kg} moves along a straight horizontal road against constant resistance 400 N400\,\mathrm{N}. Its engine supplies 24 kW24\,\mathrm{kW}. Find its acceleration when its speed is 20 m s−120\,\mathrm{m\,s^{-1}}.

Plan. Use power and speed to find driving force, then subtract resistance before applying Newton's second law.

F=2400020=1200 Nconvert kilowatts to watts1200−400=1200ahorizontal resultant =maa=23 m s−2solve\begin{aligned} F&=\frac{24000}{20}=1200\,\mathrm{N} && \text{convert kilowatts to watts}\\ 1200-400&=1200a && \text{horizontal resultant }=ma\\ a&=\frac23\,\mathrm{m\,s^{-2}} && \text{solve} \end{aligned}

Why this step? The engine's 1200 N1200\,\mathrm{N} is only one force. Resistance reduces the resultant to 800 N800\,\mathrm{N}.

Answer. The acceleration is 0.667 m s−20.667\,\mathrm{m\,s^{-2}}, to 3 significant figures.

Check. Power used against resistance is (400)(20)=8000 W(400)(20)=8000\,\mathrm{W}. The remaining 16000 W16000\,\mathrm{W} equals mav=(1200)(2/3)(20)mav=(1200)(2/3)(20).

Worked example 4 changes only the speed. With the same engine power, the available driving force decreases.

Worked example 4: faster motion at the same power

Question. For the same vehicle, resistance and engine power, find the acceleration when the speed is 30 m s−130\,\mathrm{m\,s^{-1}}.

Plan. Recalculate driving force at the new speed before resolving forces.

F=2400030=800 Nsame power, new speed800−400=1200aresultant driving forcea=13 m s−2solve\begin{aligned} F&=\frac{24000}{30}=800\,\mathrm{N} && \text{same power, new speed}\\ 800-400&=1200a && \text{resultant driving force}\\ a&=\frac13\,\mathrm{m\,s^{-2}} && \text{solve} \end{aligned}

Why this step? Constant power does not mean constant driving force or constant acceleration. You must use the speed at the instant asked for.

Answer. The acceleration is 0.333 m s−20.333\,\mathrm{m\,s^{-2}}, to 3 significant figures.

Check. The remaining power is 24000−(400)(30)=12000 W24000-(400)(30)=12000\,\mathrm{W}, agreeing with (1200)(1/3)(30)(1200)(1/3)(30).

That declining acceleration leads to a greatest steady speed when the available forward force just balances the opposing forces.

Watch out

Wrong: P=mavP=mav using engine power when resistance is present. Right: engine power is FvFv, while ma=F−Rma=F-R on this level road. The difference P−RvP-Rv is the rate of kinetic-energy increase.

Change the instantaneous speed once more.

Greatest steady speed on a slope

A greatest steady speed is found from zero acceleration at the maximum available power, with all forces resolved along the road.

For a road inclined at angle α\alpha to the horizontal, weight has component mgsin⁡αmg\sin\alpha down the road and mgcos⁡αmg\cos\alpha perpendicular to it. For a vehicle moving straight along the road without perpendicular acceleration, the normal contact force is mgcos⁡αmg\cos\alpha, if there are no other perpendicular forces. These resolving results are assumed mechanics and are not in the formula booklet.

[DIAGRAM: asset_name: Driving force and resistance on an inclined road; asset_slug: h245_6_02_power_incline; caption: For uphill motion, both resistance and the component mg sin alpha oppose the driving force; description: An inclined straight road rises to the right at angle alpha above the horizontal. A vehicle is shown with blue driving force F up the road, clay resistance R down the road, vertical weight mg downward, and a sage normal reaction perpendicular to the road. A separate component sketch resolves weight into mg sin alpha down the road and mg cos alpha into the road. A panel states at greatest steady uphill speed, F equals R plus mg sin alpha, and maximum power equals Fv.]
For uphill motion, both resistance and the component mg sin alpha oppose the driving force

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