Mw8 - Power as rate of working

Mw8 - Power as rate of working

Power measures the rate at which a force does work. You calculate average and instantaneous power, connect a vehicle's useful power to its driving force, and find steady-speed limits when driving and resistance balance. Constant power will not usually mean constant force or constant acceleration.

Average power and work over a time interval

A car develops a useful power of 20 kW20\,\mathrm{kW}. Is its driving force larger at 10 m s−110\,\mathrm{m\,s^{-1}} or at 20 m s−120\,\mathrm{m\,s^{-1}}?

The answer will be the slower speed: at the same power, each newton can act through fewer metres each second, so more newtons are needed. First, connect power to the work completed over a time interval.

You already know

Work and energy are measured in joules. A constant force along a displacement does work FsFs; the total external work equals the change in kinetic energy. Vehicle power means the useful power of its driving force, rather than the engine's output before transmission losses.

A machine transferring 1000 J1000\,\mathrm{J} every second has power 1000 W1000\,\mathrm{W}. A larger total energy transfer need not mean greater power if it takes longer. Average power measures the total transfer over the entire elapsed time; instantaneous power describes the rate at one instant.

Average and constant power

If a force does total work WW in elapsed time T>0T>0:

Paverage=WTP_{\mathrm{average}}=\frac WT

If its power is constant at PP, then

W=PTW=PT

The watt is 1 J s−11\,\mathrm{J\,s^{-1}}, and 1 kW=1000 W1\,\mathrm{kW}=1000\,\mathrm{W}. Use seconds and watts when calculating work in joules.

Not in the formula booklet: learn it.

The work in this equation belongs to the force whose power is specified. For a vehicle, the driving work may supply kinetic-energy gain, gravitational potential-energy gain and energy dissipated against resistance.

Worked example 1: time from work and power

Question. A machine must do 120 kJ120\,\mathrm{kJ} of useful work. Its useful power is constant at 8 kW8\,\mathrm{kW}. Calculate the required time.

Plan. Convert to SI units and divide work by power.

W=120000 J,P=8000 Wconvert both prefixesT=WP=1200008000constant power=15 sjoules divided by joules per second\begin{aligned} W&=120000\,\mathrm{J},\qquad P=8000\,\mathrm{W}&&\text{convert both prefixes}\\ T&=\frac WP=\frac{120000}{8000}&&\text{constant power}\\ &=15\,\mathrm{s}&&\text{joules divided by joules per second} \end{aligned}

Why this step? The ratio must use matching energy and power units. Converting both avoids a factor of 10001000 error.

Answer. 15 s15\,\mathrm{s}.

Check. 8000(15)=120000 J8000(15)=120000\,\mathrm{J}, the required useful work.

The next example changes only how the work is supplied in the question: it must be found from a kinetic-energy change.

Worked example 2: time from a change in speed

Question. A car of mass 1000 kg1000\,\mathrm{kg} accelerates from 1010 to 20 m s−120\,\mathrm{m\,s^{-1}} along a horizontal road. Neglect resistance. Its useful power is constant at 8 kW8\,\mathrm{kW}. Use energy to find the time taken.

Plan. Find the kinetic-energy gain, equate it to the driving work PTPT, then solve for time.

ΔK=12(1000)(202−102)final minus initial kinetic energy=150000 Jno other energy destination8000T=150000driving work equals the gainT=18.75 sdivide by the constant power\begin{aligned} \Delta K&=\frac12(1000)(20^2-10^2)&&\text{final minus initial kinetic energy}\\ &=150000\,\mathrm{J}&&\text{no other energy destination}\\ 8000T&=150000&&\text{driving work equals the gain}\\ T&=18.75\,\mathrm{s}&&\text{divide by the constant power} \end{aligned}

Why this step? A speed change specifies an energy change. Constant power supplies that energy at a constant rate, even though the acceleration need not be constant.

Answer. 18.75 s18.75\,\mathrm{s}.

Check. The kinetic energy rises from 50000 J50000\,\mathrm{J} to 200000 J200000\,\mathrm{J}; the difference is 8000(18.75)8000(18.75).

If power varies during an interval, W/TW/T still gives the average power. It does not state that this same power was delivered at every instant.

Watch out

Wrong: use constant-acceleration equations just because the power is constant. Right: find the work from the energy changes, then use W=PTW=PT. We will shortly see why a constant-power vehicle's acceleration usually changes with speed.

The next car has a different mass and useful power. The energy method stays the same.

Instantaneous power and the velocity component

A force's instantaneous power depends on the component of velocity in the direction of that force.

For a constant force acting along a short displacement, work is force times displacement. Dividing by the short time gives force times speed. If the directions differ, only the component vcos⁡θv\cos\theta lies along the force.

Power of a force

For force magnitude FF, speed vv, and angle θ\theta between the force and velocity:

P=Fvcos⁡θP=Fv\cos\theta

For a force along the velocity, P=FvP=Fv. A perpendicular force has zero power. A force opposite the velocity has negative power.

Not in the formula booklet: learn it.

This formula applies at an instant even when the force, speed or angle is changing. Use all three values from the same instant. A resistance RR directly opposing the motion has power −Rv-Rv; the positive power needed to overcome it is RvRv.

Worked example 3: power along the motion

Question. A force of 200 N200\,\mathrm{N} acts along the motion of a particle travelling at 3 m s−13\,\mathrm{m\,s^{-1}}. Calculate its power.

Plan. Multiply the force by the speed, since their directions agree.

P=Fv=200(3)θ=0=600 Wnewton metres per second\begin{aligned} P&=Fv=200(3)&&\theta=0\\ &=600\,\mathrm{W}&&\text{newton metres per second} \end{aligned}

Why this step? The particle moves 3 m3\,\mathrm{m} along the force each second, so the force does 600 J600\,\mathrm{J} of work per second at that instant.

Answer. 600 W600\,\mathrm{W}.

Check. A 1 s1\,\mathrm{s} interval with unchanged force and speed gives work 200(3)=600 J200(3)=600\,\mathrm{J}.

The next example changes only the angle. The full speed is no longer the component in the force's direction.

Worked example 4: an oblique force

Question. The 200 N200\,\mathrm{N} force now acts at 60∘60^\circ to the particle's velocity. The speed is still 3 m s−13\,\mathrm{m\,s^{-1}}. Calculate the force's power.

Plan. Resolve the velocity along the force and multiply by the force magnitude.

v∥=3cos⁡60∘=1.5 m s−1velocity component along the forceP=200(1.5)force times its velocity component=300 Wpositive power\begin{aligned} v_{\parallel}&=3\cos60^\circ=1.5\,\mathrm{m\,s^{-1}}&&\text{velocity component along the force}\\ P&=200(1.5)&&\text{force times its velocity component}\\ &=300\,\mathrm{W}&&\text{positive power} \end{aligned}

Why this step? A tension that is oblique to the motion does not act through the full distance travelled in its own direction. The cosine is as necessary here as in the work formula.

Answer. 300 W300\,\mathrm{W}.

Check. The component is half the speed, so the power is half of the parallel-force value.

For the horizontally carried bag, the upward support has zero power because its velocity component in the upward direction is zero, even though the bag is moving.

Watch out

Wrong: always write P=FvP=Fv, including an oblique rope tension. Right: use Fvcos⁡θFv\cos\theta, with θ\theta between the tension and the velocity. Omitting this component overstates the power.

Use the acute angle between the force and velocity in the next question.

Driving force and acceleration at constant power

At constant useful power, a vehicle's driving force decreases as its speed increases.

The driving force points along the velocity, so P=FDvP=F_Dv. Rearranging gives the force available at that speed. To find acceleration, then subtract the resistance and apply Newton's second law.

A constant-power vehicle on a level road

For useful power PP, positive speed vv, mass mm, and opposing resistance RR:

FD=Pv,ma=Pv−RF_D=\frac Pv,\qquad ma=\frac Pv-R

The expression P/vP/v applies for v>0v>0; this ideal constant-power law does not describe starting exactly from rest.

Not in the formula booklet: learn it.

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