How Fast Can the Job Be Done?
A car cruises at a steady speed on a level road. Two dials decide everything: the driving force $F$ the engine is supplying and the speed $v$ it is travelling at. The product of the two is the power $P = Fv$ â the rate at which the engine does work. Move both sliders and watch: the same force at double the speed needs double the power.
power $P = Fv$ = 6000 W = 6.0 kW
At a steady speed the driving force exactly balances the resistance â so this same product also tells the engine what it must deliver to hold that speed.
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 travels at constant speed while the driving force is 400 N. The resistance is:
R2. A load of mass 20 kg is raised 5 m. The work done against gravity is:
R3. A cyclist travels 300 m in 60 s at a steady speed. The speed is:
Two cranes can do the same work in very different times; power measures how fast the work is delivered.
$P = \dfrac{W}{t}$
The unit is the watt: $1\ \text{W} = 1\ \text{J s}^{-1}$. Mini-example: raising a load of weight $600\ \text{N}$ through $10\ \text{m}$ takes $600 \times 10 = 6000\ \text{J}$; doing it in $12\ \text{s}$ needs $\dfrac{6000}{12} = 500\ \text{W}$.
In time $t$ a force along the motion moves the body $vt$ and does work $F \times vt$, so
$P = Fv$
| At constant speed | Consequence |
|---|---|
| driving force = resistance | $P = Rv$ â the engine pays for every metre per second |
| speed doubles at fixed power | the driving force available halves |
Mini-example: a driving force of $500\ \text{N}$ at $12\ \text{m s}^{-1}$ develops $500 \times 12 = 6000\ \text{W} = 6\ \text{kW}$.
An engine has a maximum power $P_{max}$. The top speed of the vehicle is where the driving force it can still deliver equals the resistance: $\dfrac{P_{max}}{v} = R$.
A car with maximum power $30\ \text{kW}$ facing a constant resistance of $750\ \text{N}$ tops out at $v = \dfrac{30\,000}{750} = 40\ \text{m s}^{-1}$ â faster needs more power or less drag.
Groups of 4: both cars have a fixed maximum power. The steep curve is the driving force still available, F = P / v; the other curve is the resistance. Drag the speed slider along and read the resultant force and acceleration â then find the crossing point, where the acceleration has died away. Answer the three graded challenges together.
Work alone. Only the current stage can be answered: a wrong choice turns red, locks and shows a hint, so think before you tap. Stages cleared on the first attempt score on the HUD; stages cleared after a hint still unlock the next one. Stage 5 is a reasoning stage â read it carefully.
A car of mass $1\ \text{tonne}$ moves at a constant $60\ \text{km h}^{-1}$ up a road inclined at $6^{\circ}$ to the horizontal. (i) Find the weight of the car and the normal reaction from the road. The non-gravitational resistance down the slope is 2000 N; find (ii) the tractive force propelling the car up the slope, (iii) the rate at which the tractive force does work. The engine has maximum power 80 kW. (iv) Find the maximum speed of the car up the same slope, assuming the resistances are unchanged.
At constant velocity the tractive force must balance everything acting down the slope: the component of weight $mg\sin 6^{\circ}$ plus the 2000 N resistance. Convert $60\ \text{km h}^{-1}$ to $\text{m s}^{-1}$ before using $P = Tv$.
Weight $W = 1000 \times 9.8 = 9800\ \text{N}$. Perpendicular to the road: $R = 9800\cos 6^{\circ} = 9746\ \text{N} \approx 9750\ \text{N}$.
Down the slope: $9800\sin 6^{\circ} = 1024\ \text{N}$ of weight component plus $2000\ \text{N}$ of resistance, so $T = 1024 + 2000 = 3024\ \text{N}$.
$60\ \text{km h}^{-1} = 16.67\ \text{m s}^{-1}$, so $P = Tv = 3024 \times 16.67 = 50\,400\ \text{W} \approx 50.4\ \text{kW}$.
At the maximum speed the engine's 80 kW only just balances the same forces: $v = \dfrac{80\,000}{3024} = 26.5\ \text{m s}^{-1}$ (about $95\ \text{km h}^{-1}$).
A diesel shunter of mass $100\ \text{tonnes}$ pulls a truck of mass $25\ \text{tonnes}$ along a level track, the engine working at $125\ \text{kW}$. The resistance to motion is $50\ \text{N}$ per tonne. (i) Calculate the constant speed of the train. While travelling at this speed the truck becomes uncoupled, but the engine continues to produce the same power. (ii) Find the acceleration of the shunter immediately afterwards. (iii) Find the greatest speed the shunter can now reach.
Total resistance is $50\ \text{N}$ for every tonne â first for $125\ \text{tonnes}$, then for the shunter's $100\ \text{tonnes}$ alone. At the constant speed the driving force equals the resistance; immediately after uncoupling the driving force is still the old value, because the speed has not changed yet.
Resistance $= 50 \times 125 = 6250\ \text{N}$. At constant speed the driving force equals it, so $125\,000 = 6250v$ and $v = 20\ \text{m s}^{-1}$.
The shunter alone faces $50 \times 100 = 5000\ \text{N}$ of resistance. At $20\ \text{m s}^{-1}$ the driving force is still $6250\ \text{N}$, so $a = \dfrac{6250 - 5000}{100\,000} = 0.0125\ \text{m s}^{-2}$.
As the speed rises the driving force falls. The greatest speed comes when it has dropped to the shunter's own resistance: $125\,000 = 5000v$, so $v = 25\ \text{m s}^{-1}$.
Answer, then check. Options are shuffled every load. The clock starts on your first answer.
Q1. A machine does 3600 J of work in 12 s. Its average power is:
Q2. A tug pulls a barge with a steady force of 2500 N at a constant 2 m/s. The power developed is:
Q3. A car engine works at 30 kW while the car travels at 12 m/s. The driving force is:
Q4. A motorcycle has maximum power 20 kW and faces a constant resistance of 500 N. Its top speed is:
Work each on paper, then reveal the solution to check your working.
A motorcycle has a maximum power output of $26.5\ \text{kW}$ and a top speed of $46\ \text{m s}^{-1}$. Find the force exerted by the engine when the motorcycle is travelling at top speed.
Reveal Solution
At top speed the driving force balances the resistance and the engine is at maximum power: $F = \dfrac{P}{v} = \dfrac{26\,500}{46} = 576\ \text{N}$.
A crane raises a load of $500\ \text{tonnes}$ at a steady rate of $5\ \text{cm s}^{-1}$. What power is the engine of the crane producing? (Ignore friction and air resistance.)
Reveal Solution
The force the crane exerts balances the weight: $F = 500\,000 \times 9.8 = 4.9 \times 10^{6}\ \text{N}$. With $v = 0.05\ \text{m s}^{-1}$, $P = Fv = 4.9 \times 10^{6} \times 0.05 = 245\,000\ \text{W} = 245\ \text{kW}$.
A builder hoists bricks to the top of the house he is building. Each brick has mass $3.5\ \text{kg}$ and the house is $9\ \text{m}$ high. In the course of one hour he raises 120 bricks from ground level to the top of the house, where they are unloaded. (i) Find the increase in gravitational potential energy of one brick. (ii) Find the total work done by the builder in one hour. (iii) Find the average power with which he is working.
Reveal Solution
(i) $3.5 \times 9.8 \times 9 = 308.7\ \text{J}$. (ii) $308.7 \times 120 = 37\,044\ \text{J}$. (iii) $\dfrac{37\,044}{3600} = 10.3\ \text{W}$ â a slower rate than a light bulb.
A winch pulls a crate of mass $180\ \text{kg}$ up a rough slope of angle $30^{\circ}$ against a frictional force of $450\ \text{N}$. The crate moves at a steady speed of $1.2\ \text{m s}^{-1}$. (i) Calculate the gravitational potential energy given to the crate during 30 s. (ii) Calculate the work done against friction during this time. (iii) Calculate the total work done per second by the winch. (iv) Calculate the tension in the cable, which runs parallel to the slope. (v) What information is given by $T \times v$?
Reveal Solution
In 30 s the crate moves $1.2 \times 30 = 36\ \text{m}$ and rises $36\sin 30^{\circ} = 18\ \text{m}$. (i) $mgh = 180 \times 9.8 \times 18 = 31\,752\ \text{J}$. (ii) $450 \times 36 = 16\,200\ \text{J}$. (iii) $\dfrac{31\,752 + 16\,200}{30} = 1598\ \text{W}$ (about $1.6\ \text{kW}$). (iv) $T = 180 \times 9.8 \times 0.5 + 450 = 1332\ \text{N}$. (v) $T \times v = 1332 \times 1.2 = 1598\ \text{W}$ â exactly the power delivered by the winch, as P = Fv promises.
A car of mass $1000\ \text{kg}$ moves along a horizontal straight road, passing through points $A$ and $B$. The power of its engine is constant and equal to $15\,000\ \text{W}$. The driving force exerted by the engine is $750\ \text{N}$ at $A$ and $500\ \text{N}$ at $B$. Find the speed of the car at $A$ and at $B$, and hence find the increase in the car's kinetic energy as it moves from $A$ to $B$.
Reveal Solution
At $A$: $v = \dfrac{15\,000}{750} = 20\ \text{m s}^{-1}$. At $B$: $v = \dfrac{15\,000}{500} = 30\ \text{m s}^{-1}$. Increase in kinetic energy $= \dfrac{1}{2} \times 1000 \times (30^{2} - 20^{2}) = 250\,000\ \text{J} = 250\ \text{kJ}$.
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