Pushing the Stalled Car
A crate is pushed along the floor by a steady force of $60\ \text{N}$. Nothing dramatic happens — the crate simply slides — yet energy is moving from the pusher into the crate, metre by metre. The shaded area below is that transfer: $W = Fd$. Scrub the distance slider and watch the work grow in step with $d$, then ask what the picture must become when the force itself changes along the journey.
work done $W = Fd = 60 \times d$ = 240 J
One joule is one newton metre: $1\ \text{J} = 1\ \text{N m}$. Doubling the distance doubles the work — for as long as the force stays at $60\ \text{N}$.
Try each in your head before clicking an answer — this is a test, not a reveal. See how many you get first try. The clock starts on your first answer.
R1. A resultant force of 12 N acts on a body of mass 3 kg. Its acceleration is:
R2. A crate is dragged along level ground at constant speed. The driving force:
R3. A block slides to rest on a rough floor. The friction force on the block acts:
A force does work when it moves its point of application. For a constant force acting along the motion:
| Relation | Meaning |
|---|---|
| $W = Fd$ | force along the motion, distance $d$ |
| $W = Fd\cos\theta$ | force at angle $\theta$ to the motion |
Mini-example: a push of $60\ \text{N}$ over $4\ \text{m}$ does $60 \times 4 = 240\ \text{J}$ of work. The joule is the newton metre: $1\ \text{J} = 1\ \text{N m}$. Only the component along the motion works — a force perpendicular to the motion does no work at all.
Each force has its own work done, and the sign says which way the energy flows.
| Situation | Work done by the force |
|---|---|
| Lifting a load steadily | lifting force $+mgh$, gravity $-mgh$ |
| Braking a car | brakes do negative work — energy leaves the car |
Mini-example: raising a hod of bricks of mass $25\ \text{kg}$ through $10\ \text{m}$ at constant speed takes work $25 \times 9.8 \times 10 = 2450\ \text{J}$ against gravity — and the speed never changes, so the resultant work is zero.
When the resultant force does work $W$ on a body of mass $m$, the kinetic energy changes by exactly that amount: $W = \dfrac{1}{2}mv^{2} - \dfrac{1}{2}mu^{2}$. This holds whether or not the force is constant.
For a force $F(x)$ that changes as the body moves, split the journey into thin strips of width $\delta x$; the strip at $x$ contributes $F\,\delta x$. The total work is the area under the force–displacement graph.
A spring of stiffness $8\ \text{N m}^{-1}$ obeys $F = 8x$. Stretching it $2\ \text{m}$ takes the area of a triangle: $\dfrac{1}{2} \times 8 \times 2^{2} = 16\ \text{J}$.
Groups of 4: decide for each situation whether work is being done by the force described. Tap a card to pick it up, then tap the zone you want to file it in. Press Check cards to see which ones are misfiled, and Reset to start again. Two of the six are the traps — discuss them before you tap.
Work alone on the two force laws. Scrub the distance slider and predict the readout before you move it: for the constant push the shaded region is a rectangle, for the spring it is a triangle whose height grows as the stretch grows. Then answer the three graded challenges.
A sledge of mass $10\ \text{kg}$ is pulled across level ground by a rope making $20^{\circ}$ with the horizontal. The tension is $80\ \text{N}$ and there is a resistance force of $14\ \text{N}$. (i) Find the work done while the sledge moves $20\ \text{m}$ by (a) the tension (b) the resistance force. (ii) Find the speed of the sledge after it has moved $20\ \text{m}$, starting from rest.
Only the component of the tension along the motion does work — resolve the $80\ \text{N}$ horizontally: $80\cos 20^{\circ}$. The resistance acts backwards, so its work is negative. For part (ii), the resultant work changes the kinetic energy.
$W_{T} = Fd\cos\theta = 80 \times 20 \times \cos 20^{\circ} = 1503.5\ \text{J} \approx 1.5\ \text{kJ}$.
The resistance opposes the motion, so its work is negative: $W_{R} = -14 \times 20 = -280\ \text{J}$.
Resultant work $= 1503.5 - 280 = 1223.5\ \text{J}$. The sledge starts from rest, so $\dfrac{1}{2} \times 10 \times v^{2} = 1223.5$, giving $v^{2} = 244.7$ and $v = 15.6\ \text{m s}^{-1}$ (3 s.f.).
A car of mass $0.9\ \text{tonnes}$ is driven $200\ \text{m}$ up a slope inclined at $5^{\circ}$ to the horizontal, against a resistance force of $100\ \text{N}$. (i) Find the work done by the car against gravity. (ii) Find the work done against the resistance. (iii) If the car slows from $12\ \text{m s}^{-1}$ to $8\ \text{m s}^{-1}$, find the total work done by the engine.
Work against gravity needs the vertical rise, not the slope length — use the sine component: rise $= 200\sin 5^{\circ}$. Part (iii) is an energy audit: the engine supplies the rise in potential energy, the work against resistance, and the (negative) change in kinetic energy.
Rise $= 200\sin 5^{\circ} = 17.43\ \text{m}$, so $W = mgh = 900 \times 9.8 \times 17.43 = 153\,743\ \text{J} \approx 154\ \text{kJ}$.
$W = 100 \times 200 = 20\,000\ \text{J} = 20\ \text{kJ}$.
Change in kinetic energy: $\dfrac{1}{2} \times 900 \times (8^{2} - 12^{2}) = -36\,000\ \text{J}$. Energy audit: $W_{E} - 153\,743 - 20\,000 = -36\,000$, so $W_{E} = 137\,743\ \text{J} \approx 138\ \text{kJ}$.
Answer, then check. Options are shuffled every load. The clock starts on your first answer.
Q1. Work is done on a body when:
Q2. A force of 40 N acts at 60° to the direction of motion of a body that moves 5 m. The work done by the force is:
Q3. A crate of mass 20 kg is raised 3 m at constant speed. The work done by the lifting force is:
Q4. The resultant force on a 2 kg body, starting from rest, does 36 J of work. The speed reached is:
Work each on paper, then reveal the solution to check your working.
A man pushes a packing case of mass $35\ \text{kg}$ a distance of $5\ \text{m}$ across a rough floor against a resistance force of $200\ \text{N}$. Find the work he does in each case. (i) The case starts and finishes at rest. (ii) The case starts at rest and finishes with speed $2\ \text{m s}^{-1}$. (iii) The case starts with speed $2\ \text{m s}^{-1}$ and ends at rest. (iv) He holds the case stationary for $20\ \text{s}$.
Reveal Solution
Work against the resistance is $200 \times 5 = 1000\ \text{J}$ in every case. (i) No change in speed, so the man does $1000\ \text{J}$. (ii) He must also supply the kinetic energy gained: $1000 + \dfrac{1}{2} \times 35 \times 2^{2} = 1000 + 70 = 1070\ \text{J}$. (iii) The case loses kinetic energy to the resistance, so he supplies $1000 - 70 = 930\ \text{J}$. (iv) No displacement, so $0\ \text{J}$.
A sports car of mass $1.2\ \text{tonnes}$ accelerates from rest to $30\ \text{m s}^{-1}$ in a distance of $150\ \text{m}$. (i) Calculate the work done in accelerating the car. Does your answer depend on the assumption that the driving force is constant? (ii) If the driving force is in fact constant, what is its magnitude?
Reveal Solution
(i) $W = \dfrac{1}{2}mv^{2} = \dfrac{1}{2} \times 1200 \times 30^{2} = 540\,000\ \text{J} = 540\ \text{kJ}$. No — the work–energy principle works whatever the force does, so the assumption is not needed. (ii) $F = \dfrac{W}{d} = \dfrac{540\,000}{150} = 3600\ \text{N}$.
A crate of mass $50\ \text{kg}$ is dragged along a horizontal floor by a constant force of magnitude $400\ \text{N}$ acting at an angle $\alpha^{\circ}$ upwards from the horizontal. The total resistance to motion has magnitude $250\ \text{N}$. The crate starts from rest at $O$ and passes the point $P$ with speed $2\ \text{m s}^{-1}$; the distance $OP$ is $20\ \text{m}$. Find (i) the increase in kinetic energy, (ii) the work done against the resistance, (iii) the value of $\alpha$.
Reveal Solution
(i) $\dfrac{1}{2} \times 50 \times 2^{2} = 100\ \text{J}$. (ii) $250 \times 20 = 5000\ \text{J}$. (iii) Work done by the 400 N force $= 100 + 5000 = 5100\ \text{J}$, and it equals $400 \times 20 \times \cos\alpha$, so $\cos\alpha = \dfrac{5100}{8000} = 0.6375$ and $\alpha = 50.4^{\circ}$ (1 d.p.).
A load of mass $160\ \text{kg}$ is lifted vertically by a crane with constant acceleration. It starts from rest at $O$. After $7\ \text{s}$ it passes through $A$ with speed $0.5\ \text{m s}^{-1}$. By considering energy, find the work done by the crane in moving the load from $O$ to $A$.
Reveal Solution
Acceleration $a = \dfrac{0.5}{7}\ \text{m s}^{-2}$, so the rise is $s = \dfrac{1}{2}at^{2} = \dfrac{1}{2} \times \dfrac{0.5}{7} \times 49 = 1.75\ \text{m}$. The crane's work supplies the kinetic energy gained and the potential energy gained: $\dfrac{1}{2} \times 160 \times 0.5^{2} + 160 \times 9.8 \times 1.75 = 20 + 2744 = 2764\ \text{J}$.
A block of mass $50\ \text{kg}$ is pulled up a straight hill, passing points $A$ and $B$ with speeds $7\ \text{m s}^{-1}$ and $3\ \text{m s}^{-1}$. The distance $AB$ is $200\ \text{m}$ and $B$ is $15\ \text{m}$ higher than $A$. The resistance to motion has magnitude $7.5\ \text{N}$. Find (i) the loss in kinetic energy, (ii) the gain in potential energy, (iii) the work done by the pulling force.
Reveal Solution
(i) $\dfrac{1}{2} \times 50 \times (7^{2} - 3^{2}) = 1000\ \text{J}$ lost. (ii) $50 \times 9.8 \times 15 = 7350\ \text{J}$ gained. (iii) The pulling force must replace the kinetic energy lost, add the potential energy gained and cover the work against resistance: $1000 + 7350 + 7.5 \times 200 = 9850\ \text{J}$.
Exit Ticket
Check the learning objectives you feel confident about: