M2 3.1 - Power and motion

M2 3.1 - Power and motion

Use power as the rate of doing work and connect it to a driving force. Combine power with equations of motion to analyse vehicles on horizontal roads and slopes, including constant-speed and accelerating motion.

Power as a rate

An engine works at constant power: does that mean the vehicle travels at constant speed? To answer, distinguish an energy transfer rate from the force that causes acceleration.

Moving the same load through the same distance may require the same work, but completing it in half the time requires twice the average power.

Power

Power is the rate of doing work, or transferring energy. Its SI unit is the watt: 1 W=1 J s−11\,\mathrm W=1\,\mathrm{J\,s^{-1}}. Also, 1 kW=1000 W1\,\mathrm{kW}=1000\,\mathrm W.

Average and instantaneous power

Paverage=ΔWΔt,P=dWdt.P_{\rm average}=\frac{\Delta W}{\Delta t},\qquad P=\frac{\mathrm dW}{\mathrm dt}.

Here WW denotes work in joules; the upright unit W\mathrm W denotes watts. Average power over an interval need not equal the power at either end of it.

Not in the formula booklet: learn these relations.

Worked example 1: average and final power

Question. An engine has done work W=400t+20t2W=400t+20t^2 joules by time tt seconds, for t≥0t\ge0. Find its average power over the first 5 s5\,\mathrm s and its power at t=5t=5.

Plan. Divide the work change by the interval for the average, then differentiate for the instantaneous value.

Paverage=W(5)−W(0)5=25005=500 W.P_{\rm average}=\frac{W(5)-W(0)}5 =\frac{2500}5=500\,\mathrm W. P=400+40t⟹P(5)=600 W.P=400+40t\quad\Longrightarrow\quad P(5)=600\,\mathrm W.

Answer. Average 500 W500\,\mathrm W; power at 5 s5\,\mathrm s is 600 W600\,\mathrm W.

Check. Power rises linearly from 400400 to 600 W600\,\mathrm W, so its mean is (400+600)/2=500 W(400+600)/2=500\,\mathrm W.

Change only the interval: the engine's work function stays the same.

Worked example 2: a later interval

Question. For the same engine, find the average power between t=5t=5 and t=10t=10, and its power at t=10t=10.

Plan. Subtract the work already done before the interval begins.

Paverage=W(10)−W(5)10−5=6000−25005=700 W.P_{\rm average}=\frac{W(10)-W(5)}{10-5} =\frac{6000-2500}5=700\,\mathrm W. P(10)=400+40(10)=800 W.P(10)=400+40(10)=800\,\mathrm W.

Answer. Average 700 W700\,\mathrm W; final power 800 W800\,\mathrm W.

Check. The mean of the endpoint powers is (600+800)/2=700 W(600+800)/2=700\,\mathrm W.

Watch out

For a later interval, W(10)/5W(10)/5 includes work done before the interval. Use the change in work, not the final total.

Power equals force times speed

An engine's power gives its driving force at a particular speed, even when the speed is changing.

If a force FF acts along a small displacement ds\mathrm ds, its work is F dsF\,\mathrm ds. Dividing by the small time interval gives P=F ds/dt=FvP=F\,\mathrm ds/\mathrm dt=Fv. More generally it is the force component along the velocity multiplied by the speed.

Power of a driving force

P=Fv,F=Pv(v>0).P=Fv,\qquad F=\frac Pv\quad(v>0).

For power in watts and speed in metres per second, the force is in newtons. Use the driving force here; subtract resistances separately in Newton's second law.

Not in the formula booklet: learn it.

For a van towing a trailer, both share speed and acceleration. The towbar force is internal to the combined system, so it cancels from the combined equation. Resolve for the trailer alone to find its towbar force.

Worked example 3: finding a towbar force

Question. A van of mass 800 kg800\,\mathrm{kg} tows a 400 kg400\,\mathrm{kg} trailer on a horizontal road using a light rigid towbar. Resistances on the van and trailer are 600600 and 200 N200\,\mathrm N, respectively. The engine supplies 18 kW18\,\mathrm{kW}. At speed 10 m s−110\,\mathrm{m\,s^{-1}}, find the acceleration and towbar tension.

Plan. Find the driving force from power, use the whole system for acceleration, then the trailer for tension.

F=1800010=1800 N,1800−600−200=1200a.F=\frac{18000}{10}=1800\,\mathrm N, \qquad 1800-600-200=1200a. a=56 m s−2,T−200=400(56).a=\frac56\,\mathrm{m\,s^{-2}},\qquad T-200=400\left(\frac56\right).

Answer. a=0.833 m s−2a=0.833\,\mathrm{m\,s^{-2}} and T=533 NT=533\,\mathrm N, to three significant figures.

Check. The van's equation gives 1800−600−1600/3=2000/3=800(5/6)1800-600-1600/3=2000/3=800(5/6), agreeing with the trailer.

Change only the instantaneous speed. Constant power then gives a different driving force.

Worked example 4: the same power at higher speed

Question. For the same system at 20 m s−120\,\mathrm{m\,s^{-1}}, with the same power and resistances, find the acceleration and towbar tension.

Plan. Recalculate FF; do not reuse the earlier driving force.

F=1800020=900 N,900−800=1200a.F=\frac{18000}{20}=900\,\mathrm N, \qquad 900-800=1200a. a=112 m s−2,T=200+400(112)=7003 N.a=\frac1{12}\,\mathrm{m\,s^{-2}},\qquad T=200+400\left(\frac1{12}\right)=\frac{700}3\,\mathrm N.

Answer. a=0.0833 m s−2a=0.0833\,\mathrm{m\,s^{-2}} and T=233 NT=233\,\mathrm N, to three significant figures.

Check. Doubling speed halves the driving force, but the resistance stays 800 N800\,\mathrm N; it is the resultant force that determines acceleration.

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