Why the Push Dies Away
A car pulls away with its engine delivering a constant $24\ \text{kW}$, against a resistance that grows with speed, $R = 10v$. The driving force still available is $F = \dfrac{P}{v}$ â and it falls every time the speed rises, while the resistance climbs to meet it. Scrub the speed slider and watch the acceleration fade: the journey is heading for the crossing point, where nothing is left to accelerate.
driving force $\dfrac{24\,000}{v}$ = 960 N ¡ resistance $10v$ = 250 N ¡ acceleration $a = \dfrac{F - R}{1000}$ = 0.71 m sâťÂ˛
The violet curve is the driving force; the amber line is the resistance. They cross at $v = 49\ \text{m s}^{-1}$ â the steady speed where the acceleration has run out.
Try each in your head before clicking an answer â this is a test, not a reveal. See how many you get first try. The clock starts on your first answer.
R1. A car engine works at 40 kW while the car travels at 20 m/s. The driving force is:
R2. A ship is slowing down while its engines still work. The resistance must be:
R3. The formula P = Fv rearranged for the driving force is:
When an engine is working at power $P$ while the vehicle moves at speed $v$, the driving force it can exert at that instant is
$F = \dfrac{P}{v}$
No time is involved â this is an instantaneous link. Mini-example: at $30\ \text{kW}$ and $20\ \text{m s}^{-1}$ the driving force is $\dfrac{30\,000}{20} = 1500\ \text{N}$; at $30\ \text{m s}^{-1}$ it has dropped to $1000\ \text{N}$.
The driving force and the resistance combine normally: $F - R = ma$. Two resistance models appear in every question:
| Resistance model | Example |
|---|---|
| constant | $R = 500\ \text{N}$ whatever the speed |
| proportional to speed | $R = kv$, e.g. $R = 20v$ |
Mini-example: a car with a driving force of $900\ \text{N}$ and resistance $300\ \text{N}$ has resultant $600\ \text{N}$; for mass $1200\ \text{kg}$, $a = 0.5\ \text{m s}^{-2}$.
At constant power, as $v$ rises the available force $\dfrac{P}{v}$ falls while the resistance $R$ rises. The resultant shrinks, so the acceleration shrinks with it.
Motion steadies when the available force has fallen to the resistance: $\dfrac{P}{v} = R$. For a resistance $R = kv$ this gives $v^{2} = \dfrac{P}{k}$.
A lorry with $45\ \text{kW}$ against a constant $900\ \text{N}$ steadies at $v = \dfrac{45\,000}{900} = 50\ \text{m s}^{-1}$ â however long the driver waits, it will not go faster.
Groups of 4: two vehicles, both with constant power. Car C faces a flat resistance; ship S faces a resistance that grows with speed. Move the speed slider and read the driving force, the resistance and the acceleration; find where each journey steadies, then settle the three graded challenges.
Work alone. Four claims about cars and ships moving under constant power. For each, decide "It is correct" or "There is an error" and read the explanation. Two of the four are sound â the other two are the classic slips that cost marks, so look at each line carefully.
A supertanker of mass $4 \times 10^{8}\ \text{kg}$ steams at a constant $8\ \text{m s}^{-1}$. The resistance force is $2 \times 10^{6}\ \text{N}$. (i) What power are the ship's engines producing? One of the two engines suddenly fails but the other continues at the same rate. (ii) Find the deceleration of the ship immediately after the failure. The resistance is directly proportional to the speed. (iii) Find the eventual steady speed of the ship under one engine, assuming the single engine maintains a constant power output.
At constant speed $P = Rv$. After the failure the power halves â but the speed has not changed yet, so recompute the driving force at $8\ \text{m s}^{-1}$ with the new power. For part (iii), first find $k$ from $R = kv$ at the original speed, then set driving force equal to resistance at the new steady speed.
$P = Rv = 2 \times 10^{6} \times 8 = 1.6 \times 10^{7}\ \text{W}$, that is $16\ \text{MW}$.
After the failure $P = 8 \times 10^{6}\ \text{W}$. At $8\ \text{m s}^{-1}$ the driving force is $\dfrac{8 \times 10^{6}}{8} = 1 \times 10^{6}\ \text{N}$. Resultant $= 1 \times 10^{6} - 2 \times 10^{6} = -1 \times 10^{6}\ \text{N}$, so $a = \dfrac{-1 \times 10^{6}}{4 \times 10^{8}} = -0.0025\ \text{m s}^{-2}$ â a deceleration of $0.0025\ \text{m s}^{-2}$.
From the original data, $k = \dfrac{2 \times 10^{6}}{8} = 2.5 \times 10^{5}$, so $R = 2.5 \times 10^{5}v$. Steady speed: $\dfrac{8 \times 10^{6}}{v} = 2.5 \times 10^{5}v$, giving $v^{2} = 32$ and $v = 5.66\ \text{m s}^{-1}$.
A car of mass $850\ \text{kg}$ has a maximum speed of $50\ \text{m s}^{-1}$ and a maximum power output of $40\ \text{kW}$. The resistance force at speed $v\ \text{m s}^{-1}$ is modelled by $R = kv$. (i) Find the value of $k$. (ii) Find the resistance force when the car's speed is $20\ \text{m s}^{-1}$. (iii) Find the power needed to travel at a constant $20\ \text{m s}^{-1}$ along a level road. (iv) Find the maximum acceleration of the car when it is travelling at $20\ \text{m s}^{-1}$ (a) along a level road, (b) up a hill at $5^{\circ}$ to the horizontal.
At the maximum speed the driving force from the 40 kW has fallen to equal the resistance. Part (iii) is a constant-speed question â the engine only has to beat the resistance. For part (iv) remember to add the weight component $mg\sin 5^{\circ}$ to the resistance when climbing.
At $50\ \text{m s}^{-1}$ the driving force is $\dfrac{40\,000}{50} = 800\ \text{N}$ and it balances the resistance, so $800 = 50k$ and $k = 16$.
$R = 16 \times 20 = 320\ \text{N}$. At constant speed the driving force is also 320 N, so $P = Fv = 320 \times 20 = 6400\ \text{W} = 6.4\ \text{kW}$.
On full power the driving force at $20\ \text{m s}^{-1}$ is $\dfrac{40\,000}{20} = 2000\ \text{N}$. (a) Level road: resultant $= 2000 - 320 = 1680\ \text{N}$, so $a = \dfrac{1680}{850} = 1.98\ \text{m s}^{-2}$. (b) Up the hill add $850 \times 9.8 \times \sin 5^{\circ} = 726\ \text{N}$ of weight component: resultant $= 2000 - 320 - 726 = 954\ \text{N}$, so $a = \dfrac{954}{850} = 1.12\ \text{m s}^{-2}$.
Answer, then check. Options are shuffled every load. The clock starts on your first answer.
Q1. A car engine works at 20 kW while the car travels at 25 m/s. The driving force is:
Q2. A 1500 kg car has a driving force of 1200 N and a resistance of 300 N. Its acceleration is:
Q3. A boat's resistance is modelled as R = 40v. At 5 m/s the resistance is:
Q4. A car with constant power 12 kW faces a constant resistance of 400 N. Its steady speed is:
Work each on paper, then reveal the solution to check your working.
A mouse of mass $15\ \text{g}$ is stationary $2\ \text{m}$ below its hole when it sees a cat. It runs to the hole, arriving $1.5\ \text{s}$ later with a speed of $3\ \text{m s}^{-1}$. (i) Show that the acceleration of the mouse is not constant. (ii) Calculate the average power of the mouse.
Reveal Solution
(i) If the acceleration were constant from rest, then in $1.5\ \text{s}$ the distance would be $\dfrac{1}{2} \times 3 \times 1.5 = 2.25\ \text{m}$, not $2\ \text{m}$ â so the acceleration is not constant. (ii) The mouse gains kinetic energy and potential energy: $\dfrac{1}{2} \times 0.015 \times 3^{2} + 0.015 \times 9.8 \times 2 = 0.0675 + 0.294 = 0.36\ \text{J}$, so the average power is $\dfrac{0.36}{1.5} = 0.24\ \text{W}$.
A car of mass $1150\ \text{kg}$ travels up a straight hill inclined at $1.2^{\circ}$ to the horizontal. The resistance to motion of the car is $975\ \text{N}$. Find the acceleration of the car at an instant when it is moving with speed $16\ \text{m s}^{-1}$ and the engine is working at a power of $35\ \text{kW}$.
Reveal Solution
Driving force $= \dfrac{35\,000}{16} = 2187.5\ \text{N}$. Weight component up the hill $= 1150 \times 9.8 \times \sin 1.2^{\circ} = 236\ \text{N}$. Resultant $= 2187.5 - 975 - 236 = 976.5\ \text{N}$, so $a = \dfrac{976.5}{1150} = 0.849\ \text{m s}^{-2}$.
A boat of mass $1200\ \text{kg}$ is winched a distance $30\ \text{m}$ up a flat beach inclined at $10^{\circ}$ to the horizontal. (i) Neglecting all resistances, calculate the work done. (ii) Given that the process takes 2 minutes at a constant speed, calculate the power of the winch motor. A better model takes account of the resistance of the beach. The winch motor develops a constant $4.5\ \text{kW}$, the resistance is a constant $5\ \text{kN}$ and the boat moves at constant speed. (iii) Calculate how long the winching will take. (iv) Show that if the cable broke off at the boat the boat would come to rest in about 35 mm.
Reveal Solution
(i) The boat rises $30\sin 10^{\circ} = 5.21\ \text{m}$, so $W = 1200 \times 9.8 \times 5.21 = 61\,270\ \text{J} \approx 61.3\ \text{kJ}$. (ii) $P = \dfrac{61\,270}{120} = 511\ \text{W}$. (iii) Total resistance along the slope $= 1200 \times 9.8 \times \sin 10^{\circ} + 5000 = 2042 + 5000 = 7042\ \text{N}$, so the speed is $\dfrac{4500}{7042} = 0.639\ \text{m s}^{-1}$ and the time is $\dfrac{30}{0.639} = 47\ \text{s}$ (2 s.f.). (iv) If the cable breaks, the resultant force is the whole $7042\ \text{N}$ backwards, so $a = \dfrac{7042}{1200} = 5.87\ \text{m s}^{-2}$ and the stopping distance is $\dfrac{0.639^{2}}{2 \times 5.87} = 0.035\ \text{m}$ â about $35\ \text{mm}$.
A car of mass $1200\ \text{kg}$ travels along a horizontal straight road. The power provided by the engine is constant and equal to $20\ \text{kW}$. The resistance to the car's motion is constant and equal to $500\ \text{N}$. The car passes through the points $A$ and $B$ with speeds $10\ \text{m s}^{-1}$ and $25\ \text{m s}^{-1}$ respectively, taking $30.5\ \text{s}$ to travel from $A$ to $B$. (i) Find the acceleration of the car at $A$. (ii) By considering work and energy, find the distance $AB$.
Reveal Solution
(i) At $A$: $F = \dfrac{20\,000}{10} = 2000\ \text{N}$, so $a = \dfrac{2000 - 500}{1200} = 1.25\ \text{m s}^{-2}$. (ii) Work done by the engine in $30.5\ \text{s}$: $20\,000 \times 30.5 = 610\,000\ \text{J}$. Energy audit: engine work â work against resistance = gain in kinetic energy: $610\,000 - 500s = \dfrac{1}{2} \times 1200 \times (25^{2} - 10^{2}) = 315\,000$, so $500s = 295\,000$ and $s = 590\ \text{m}$.
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