Chapter 5 · Work, Energy and Power

5.1 Work, Kinetic Energy and the Work–Energy Theorem

Work is done when a force moves its point of application: $W=\mathbf{F}\cdot\mathbf{d}$. The net work on a body equals the change in its kinetic energy $K=\tfrac12mv^2$ — for constant and for variable forces alike.

Hold a full bucket of water still at arm's length for five minutes and you will be exhausted — yet a physicist would say you have done no work on the bucket at all. Lift the same bucket onto a shelf and you have done work; carry it along a level corridor at steady speed and, once again, your upward pull on it does none. In physics, work has a precise meaning: a force does work only when the point where it acts moves, and only the part of the force along that motion counts.

Why define such a quantity? Because it is the bridge between forces and speeds. Newton's second law (Chapter 4) tells you how a force changes velocity from moment to moment; work tells you, in one step, how all the forces acting along a path change a body's kinetic energy. A car's braking distance, the depth a bullet penetrates into wood, the speed of a block pushed by a force that varies along the way — all follow from one equation, the work–energy theorem, without ever finding the acceleration or the time. To state it cleanly we first need a new way of multiplying two vectors.

The scalar product of two vectors

In Chapter 3 you added vectors and multiplied them by numbers. Work needs a product of two vectors — force and displacement — whose answer is an ordinary number.

📘 Definition 5.1 — Scalar (dot) product

The scalar product (or dot product) of two vectors $\mathbf{A}$ and $\mathbf{B}$ is the scalar

$$\mathbf{A}\cdot\mathbf{B}=AB\cos\theta,$$

where $A$ and $B$ are their magnitudes and $\theta$ ($0\le\theta\le180^\circ$) is the angle between them when they are drawn tail to tail. Since $B\cos\theta$ is the projection of $\mathbf{B}$ on the direction of $\mathbf{A}$, the scalar product is "$A$ times the part of $\mathbf{B}$ along $\mathbf{A}$" (and equally "$B$ times the part of $\mathbf{A}$ along $\mathbf{B}$").

The sign of $\mathbf{A}\cdot\mathbf{B}$ tells you the angle at a glance: positive when $\theta\lt90^\circ$, zero when the vectors are perpendicular, negative when $\theta\gt90^\circ$. Applying the definition to the unit vectors, which have length 1 and are mutually perpendicular, gives $\hat{\mathbf{i}}\cdot\hat{\mathbf{i}}=1$ and $\hat{\mathbf{i}}\cdot\hat{\mathbf{j}}=0$; multiplying out two vectors written in components then leaves only the "like with like" terms.

📐 Result 5.1 — Properties of the scalar product

For any vectors $\mathbf{A}$, $\mathbf{B}$, $\mathbf{C}$ and any real number $\lambda$:

(i) $\mathbf{A}\cdot\mathbf{B}=\mathbf{B}\cdot\mathbf{A}$ (commutative); (ii) $\mathbf{A}\cdot(\mathbf{B}+\mathbf{C})=\mathbf{A}\cdot\mathbf{B}+\mathbf{A}\cdot\mathbf{C}$ (distributive); (iii) $\mathbf{A}\cdot(\lambda\mathbf{B})=\lambda(\mathbf{A}\cdot\mathbf{B})$;

(iv) $\hat{\mathbf{i}}\cdot\hat{\mathbf{i}}=\hat{\mathbf{j}}\cdot\hat{\mathbf{j}}=\hat{\mathbf{k}}\cdot\hat{\mathbf{k}}=1$ and $\hat{\mathbf{i}}\cdot\hat{\mathbf{j}}=\hat{\mathbf{j}}\cdot\hat{\mathbf{k}}=\hat{\mathbf{k}}\cdot\hat{\mathbf{i}}=0$; (v) $\mathbf{A}\cdot\mathbf{A}=A^2$.

Hence, in components,

$$\mathbf{A}\cdot\mathbf{B}=A_xB_x+A_yB_y+A_zB_z,$$ $$\cos\theta=\frac{\mathbf{A}\cdot\mathbf{B}}{AB}.$$

Two non-zero vectors are perpendicular if and only if $\mathbf{A}\cdot\mathbf{B}=0$.

✏️ Example 1 — The angle between two vectors
Let $\mathbf{A}=2\hat{\mathbf{i}}+2\hat{\mathbf{j}}-\hat{\mathbf{k}}$ and $\mathbf{B}=6\hat{\mathbf{i}}-3\hat{\mathbf{j}}+2\hat{\mathbf{k}}$. Find (a) $\mathbf{A}\cdot\mathbf{B}$, (b) the angle between $\mathbf{A}$ and $\mathbf{B}$, (c) the component of $\mathbf{B}$ along $\mathbf{A}$, (d) the value of $c$ for which $\mathbf{C}=\hat{\mathbf{i}}+c\,\hat{\mathbf{j}}+4\hat{\mathbf{k}}$ is perpendicular to $\mathbf{A}$.
(a) $\mathbf{A}\cdot\mathbf{B}=(2)(6)+(2)(-3)+(-1)(2)=12-6-2=4$.
(b) $A=\sqrt{4+4+1}=3$, $\ B=\sqrt{36+9+4}=7$. $\cos\theta=\dfrac{4}{3\times7}=0.190$, so $\theta=79.0^\circ$.
(c) $B\cos\theta=\dfrac{\mathbf{A}\cdot\mathbf{B}}{A}=\dfrac43=1.33$ (in the units of $\mathbf{B}$).
(d) $\mathbf{A}\cdot\mathbf{C}=2+2c-4=0$ gives $c=1$.
Check $\theta$ is just under $90^\circ$, consistent with a small positive dot product compared with $AB=21$ ✓.

Work done by a constant force

📘 Definition 5.2 — Work done by a constant force

If a constant force $\mathbf{F}$ acts on a body while its point of application undergoes a displacement $\mathbf{d}$, the work done by the force is

$$W=\mathbf{F}\cdot\mathbf{d}=Fd\cos\theta,$$

where $\theta$ is the angle between $\mathbf{F}$ and $\mathbf{d}$. Work is a scalar. Its SI unit is the joule: $1\ \text{J}=1\ \text{N m}=1\ \text{kg m}^2\text{s}^{-2}$, and its dimensional formula is $[\text{M}\text{L}^2\text{T}^{-2}]$.

Only the component of the force along the displacement, $F\cos\theta$, does work. That one fact explains the bucket puzzle and fixes the sign of work:

"Work done by which force?" is always the first question. Several forces usually act; each does its own work, and the net work is their sum, which equals the work done by the resultant force.

✏️ Example 2 — Pulling a suitcase
A traveller pulls a wheeled suitcase $25\ \text{m}$ along a level airport floor with a force of $40\ \text{N}$ along the handle, which makes $30^\circ$ with the horizontal. A resistive force of $20\ \text{N}$ (rolling friction and air) opposes the motion. Find the work done (a) by the traveller, (b) by the resistive force, (c) by gravity and by the normal force, (d) by all the forces together.
(a) $W_F=Fd\cos30^\circ=40\times25\times0.866=866\ \text{J}$ (about $8.7\times10^2\ \text{J}$).
(b) The resistive force points opposite to $\mathbf{d}$ ($\theta=180^\circ$): $W_r=20\times25\times(-1)=-500\ \text{J}$.
(c) Gravity (down) and the normal force (up) are both perpendicular to the horizontal displacement: each does zero work.
(d) $W_\text{net}=866-500+0+0=366\ \text{J}\approx3.7\times10^2\ \text{J}$. It is positive, so (as we shall see) the suitcase speeds up.
Check Only the horizontal part of the pull, $40\cos30^\circ=34.6\ \text{N}$, moves the suitcase; $34.6\times25=866\ \text{J}$ ✓.
✏️ Example 3 — Work from components; work by gravity on a slope
(a) A constant force $\mathbf{F}=(3\hat{\mathbf{i}}+4\hat{\mathbf{j}}-2\hat{\mathbf{k}})\ \text{N}$ acts on a particle that moves from the point $(1,\ 2,\ 0)\ \text{m}$ to the point $(4,\ -1,\ 3)\ \text{m}$. Find the work done by $\mathbf{F}$. (b) A $4.0\ \text{kg}$ box slides $5.0\ \text{m}$ down a slope inclined at $30^\circ$ to the horizontal. How much work does gravity do on it?
(a) displacement $\mathbf{d}=\mathbf{r}_2-\mathbf{r}_1=(3\hat{\mathbf{i}}-3\hat{\mathbf{j}}+3\hat{\mathbf{k}})\ \text{m}$.
(a) work $W=\mathbf{F}\cdot\mathbf{d}=(3)(3)+(4)(-3)+(-2)(3)=9-12-6=-9\ \text{J}$. The force, on balance, opposes this displacement.
(b) The weight $mg$ is vertical; the displacement makes $60^\circ$ with the vertical. $W_g=mgd\cos60^\circ=4.0\times9.8\times5.0\times0.5=98\ \text{J}$.
Check The box descends $h=d\sin30^\circ=2.5\ \text{m}$ and $mgh=4.0\times9.8\times2.5=98\ \text{J}$ ✓ — gravity's work depends only on the drop in height, a fact we use in Section 5.2.

Kinetic energy and the work–energy theorem

📘 Definition 5.3 — Kinetic energy

The kinetic energy of a body of mass $m$ moving with speed $v$ is

$$K=\tfrac12mv^2 .$$

It is a scalar, never negative, measured in joules. It depends on the speed, not on the direction of motion. In terms of the momentum $p=mv$ (Chapter 4), $K=\dfrac{p^2}{2m}$.

Where does $\tfrac12mv^2$ come from? Let a constant net force $F$ act on a body along its line of motion through a displacement $d$. Its acceleration $a=F/m$ is constant, so the Chapter 2 equation $v^2=u^2+2ad$ applies. Multiply by $\tfrac12m$:

$$\tfrac12mv^2-\tfrac12mu^2=mad=Fd .$$

The right-hand side is the net work; the left-hand side is the change in the quantity $\tfrac12mv^2$. This is no accident of constant forces — it holds in general, as the next subsection shows.

📐 Result 5.2 — Work–energy theorem

For a body that moves as a particle (every part has the same velocity), the net work done on it by all the forces acting on it equals the change in its kinetic energy:

$$W_\text{net}=K_f-K_i=\tfrac12mv^2-\tfrac12mu^2 .$$

It holds for constant and variable forces and along any path. Positive net work speeds the body up; negative net work slows it down; zero net work leaves its speed unchanged.

🧭 Problem-solving strategy — using the work–energy theorem
  1. Choose the body and the two positions (start and finish) you care about.
  2. List every force on the body (a free-body diagram, as in Chapter 4) and find the work done by each over the path — with its sign. Forces perpendicular to the motion do none.
  3. Add them to get $W_\text{net}$, write $K_i$ and $K_f$, and set $W_\text{net}=K_f-K_i$.
  4. Solve for the unknown (a speed, a distance or an average force). Time never appears — if you need time, use kinematics as well.
✏️ Example 4 — Braking distance grows with the square of speed
A $1200\ \text{kg}$ car travelling at $20\ \text{m/s}$ ($72\ \text{km/h}$) brakes with a constant braking force of $6000\ \text{N}$. (a) Find its braking distance. (b) Find the braking distance from $30\ \text{m/s}$ with the same force. (c) What does this mean for road safety?
Theorem The braking force opposes the motion: $W_\text{net}=-Fs$. Gravity and the normal force do no work on level road. $K_f=0$, so $-Fs=0-\tfrac12mu^2$ and $s=\dfrac{mu^2}{2F}$.
(a) $s=\dfrac{1200\times20^2}{2\times6000}=40\ \text{m}$.
(b) $s=\dfrac{1200\times30^2}{2\times6000}=90\ \text{m}$.
(c) $s\propto u^2$: $1.5$ times the speed needs $1.5^2=2.25$ times the distance. Speed limits near schools are set with this in mind.
Check Kinematics: $a=-5.0\ \text{m/s}^2$, $s=\dfrac{u^2}{2|a|}=\dfrac{400}{10}=40\ \text{m}$ ✓.
✏️ Example 5 — A bullet through a plank
A $20\ \text{g}$ bullet enters a wooden plank $5.0\ \text{cm}$ thick at $400\ \text{m/s}$ and emerges at $150\ \text{m/s}$. (a) Find the average resistive force of the wood. (b) The bullet then enters a second, identical plank. Does it pass through? If not, how deep does it go?
Energies $K_i=\tfrac12(0.020)(400)^2=1600\ \text{J}$; $\ K_f=\tfrac12(0.020)(150)^2=225\ \text{J}$.
(a) $-\bar F d=K_f-K_i=-1375\ \text{J}$, so $\bar F=\dfrac{1375}{0.050}=2.75\times10^4\ \text{N}\approx2.8\times10^4\ \text{N}$ (gravity's work over $5\ \text{cm}$ is negligible).
(b) Passing through a plank needs $1375\ \text{J}$, but only $225\ \text{J}$ is left. The bullet stops after $x=\dfrac{225}{2.75\times10^4}=8.2\times10^{-3}\ \text{m}$, i.e. about $0.82\ \text{cm}$ into the second plank.
Check The force is huge — about $1.4\times10^5$ times the bullet's weight — yet the theorem needs no knowledge of how long the bullet takes ✓.
Interactive 1 — Who Does the Work? A Crate on a Rough Floor

A $10\ \text{kg}$ crate is already sliding at $2.0\ \text{m/s}$. You pull it with a force $F$ at angle $\theta$ to the horizontal (negative $\theta$ = pushing down) for $5.0\ \text{m}$ across a floor with coefficient of kinetic friction $\mu_k$. The bars show the work done by each force: the pull (purple), friction (red), gravity and the normal force (zero — they are perpendicular to the motion). Watch the net work bar match the change in kinetic energy. A steeper pull does less work itself but also presses less on the floor.

40 N30°0.20
work by pull—
work by friction—
net work = $\Delta K$—
final speed—

Work done by a variable force

A spring pulls harder the more it is stretched; the thrust of a rocket engine changes as it burns; the force of a bow on an arrow falls as the string straightens. For a force that changes with position we cannot just multiply $F$ by $d$. Instead, do what you did with the area under a $v$–$t$ graph in Chapter 2: cut the path into many small steps $\Delta x$. Over one small step the force is nearly constant, so the work is nearly $F(x)\,\Delta x$ — the area of a thin strip under the $F$–$x$ graph. Adding the strips and letting $\Delta x\to0$ gives the exact work.

📐 Result 5.3 — Work done by a variable force (one dimension)

If a force $F(x)$ acts along the $x$-axis while its point of application moves from $x_i$ to $x_f$, the work done is

$$W=\lim_{\Delta x\to0}\sum F(x)\,\Delta x=\int_{x_i}^{x_f}F(x)\,dx ,$$

which is the signed area under the $F$–$x$ graph between $x_i$ and $x_f$: area above the $x$-axis (force along $+x$) counts as positive work, area below it as negative work. For a constant force the area is a rectangle and $W=F\,(x_f-x_i)$, as before.

The work–energy theorem still holds. By Newton's second law, $F=m\dfrac{dv}{dt}$. Since $v$ depends on $x$ and $x$ on $t$, $\dfrac{dv}{dt}=\dfrac{dv}{dx}\cdot\dfrac{dx}{dt}=v\dfrac{dv}{dx}$ (this "chain rule" is proved in maths). So $F\,dx=mv\,dv$, and adding up both sides from start to finish,

$$\int_{x_i}^{x_f}F\,dx=\int_{u}^{v}mv\,dv=\tfrac12mv^2-\tfrac12mu^2 .$$

Graphically: the area under the $F$–$x$ graph is the kinetic energy gained. You can find that area by geometry (triangles, rectangles, trapezia) or, for a formula such as $F=kx$ or $F=ax^2$, by integration: $\displaystyle\int x\,dx=\tfrac12x^2+C$ and $\displaystyle\int x^2\,dx=\tfrac13x^3+C$ (you will meet these in maths).

✏️ Example 6 — Reading work from an $F$–$x$ graph
A $2.0\ \text{kg}$ block starts from rest at $x=0$ on a frictionless floor. A horizontal force acts on it along $+x$ as follows: it rises uniformly from $0$ to $10\ \text{N}$ between $x=0$ and $x=2\ \text{m}$, stays at $10\ \text{N}$ up to $x=6\ \text{m}$, then decreases uniformly to $-10\ \text{N}$ at $x=8\ \text{m}$. Find (a) the work done from $x=0$ to $x=8\ \text{m}$, (b) the speed at $x=8\ \text{m}$, (c) where the block is fastest, and that greatest speed.
Areas $0$–$2\ \text{m}$: triangle $\tfrac12(2)(10)=10\ \text{J}$. $\ 2$–$6\ \text{m}$: rectangle $4\times10=40\ \text{J}$. The last segment crosses zero at $x=7\ \text{m}$: $6$–$7\ \text{m}$ triangle $+\tfrac12(1)(10)=+5\ \text{J}$; $7$–$8\ \text{m}$ triangle $-5\ \text{J}$ (below the axis).
(a) $W=10+40+5-5=50\ \text{J}$.
(b) $\tfrac12(2.0)v^2=50$: $v=\sqrt{50}=7.1\ \text{m/s}$.
(c) The block speeds up while $F\gt0$ and slows down once $F\lt0$, so it is fastest at $x=7\ \text{m}$, where $W=55\ \text{J}$: $v_\text{max}=\sqrt{55}=7.4\ \text{m/s}$.
Check Choose "Example 6" in Interactive 2 and drag to $x=7$ and $x=8\ \text{m}$: the readouts show $55\ \text{J}$, $7.42\ \text{m/s}$ and $50\ \text{J}$, $7.07\ \text{m/s}$ ✓.
Interactive 2 — Work Is the Area under the F–x Graph

A $2.0\ \text{kg}$ block starts from rest on a frictionless floor and is pushed by a force that depends on its position $x$. Choose the force law and press Play (or drag the slider). The shaded area under the $F$–$x$ graph grows as the block moves — green where the force does positive work, red where it does negative work — and the block's kinetic energy always equals the net shaded area.

8.00 m
force $F(x)$—
work so far $W$—
kinetic energy $K$—
speed $v$—
⚠️ Common mistakes

1. Writing $W=Fd$ when the force is at an angle. Only the component along the displacement counts: $W=Fd\cos\theta$. A force perpendicular to the motion does no work however large it is.

2. "Work by friction is always negative." Kinetic friction on a body sliding over a fixed floor is, but static friction on a box resting in an accelerating truck points forward and does positive work on the box. Always compare the force's direction with the body's displacement.

3. Using one force instead of the net work. The work–energy theorem needs the work done by all forces. The work done by your pull alone does not equal the change in kinetic energy when friction also acts (Interactive 1).

4. Effort is not work. Holding a heavy load still tires your muscles (they twitch and use chemical energy), but the force on the load does no work because the load does not move.

5. Vector kinetic energy. $K=\tfrac12mv^2$ is a scalar: never give it a direction, and never add the kinetic energies of components as vectors. $\tfrac12mv^2=\tfrac12mv_x^2+\tfrac12mv_y^2$ only because $v^2=v_x^2+v_y^2$.

Exercises

Take $g=9.8\ \text{m/s}^2$ unless told otherwise.

I
The scalar product.
  • 1 Find $\mathbf{A}\cdot\mathbf{B}$ for $\mathbf{A}=3\hat{\mathbf{i}}-2\hat{\mathbf{j}}+\hat{\mathbf{k}}$ and $\mathbf{B}=\hat{\mathbf{i}}+4\hat{\mathbf{j}}+5\hat{\mathbf{k}}$. What do you conclude?
  • 2 Find the angle between $\mathbf{P}=\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}$ and the $x$-axis.
  • 3 Two vectors have magnitudes $4$ and $5$ and scalar product $10$. Find the angle between them.
  • 4 Find $\lambda$ so that $2\hat{\mathbf{i}}+\lambda\hat{\mathbf{j}}-\hat{\mathbf{k}}$ is perpendicular to $3\hat{\mathbf{i}}-2\hat{\mathbf{j}}+4\hat{\mathbf{k}}$.
  • 5 Find the component of $\mathbf{F}=6\hat{\mathbf{i}}+2\hat{\mathbf{j}}-3\hat{\mathbf{k}}$ along the direction of $\mathbf{d}=2\hat{\mathbf{i}}-\hat{\mathbf{j}}+2\hat{\mathbf{k}}$.
  • 6 Show that $(\mathbf{A}+\mathbf{B})\cdot(\mathbf{A}-\mathbf{B})=A^2-B^2$. Deduce that if $|\mathbf{A}|=|\mathbf{B}|$, then $\mathbf{A}+\mathbf{B}$ is perpendicular to $\mathbf{A}-\mathbf{B}$.
  • 7 If $|\mathbf{A}+\mathbf{B}|=|\mathbf{A}-\mathbf{B}|$ for non-zero vectors, show that $\mathbf{A}$ and $\mathbf{B}$ are perpendicular.
1 $3-8+5=0$: the vectors are perpendicular.
2 $\cos\theta=\dfrac{\mathbf{P}\cdot\hat{\mathbf{i}}}{P}=\dfrac{1}{\sqrt3}$, so $\theta=54.7^\circ$.
3 $\cos\theta=\dfrac{10}{4\times5}=0.5$, $\theta=60^\circ$.
4 $6-2\lambda-4=0$, so $\lambda=1$.
5 $d=\sqrt{4+1+4}=3$; component $=\dfrac{\mathbf{F}\cdot\mathbf{d}}{d}=\dfrac{12-2-6}{3}=\dfrac43=1.33$.
6 Expanding (distributive law): $\mathbf{A}\cdot\mathbf{A}-\mathbf{A}\cdot\mathbf{B}+\mathbf{B}\cdot\mathbf{A}-\mathbf{B}\cdot\mathbf{B}=A^2-B^2$, since $\mathbf{A}\cdot\mathbf{B}=\mathbf{B}\cdot\mathbf{A}$. If $A=B$ the product is zero, so the two vectors (the diagonals of a rhombus) are perpendicular.
7 Square both sides: $A^2+2\mathbf{A}\cdot\mathbf{B}+B^2=A^2-2\mathbf{A}\cdot\mathbf{B}+B^2$, so $4\,\mathbf{A}\cdot\mathbf{B}=0$ and $\theta=90^\circ$.
II
Work done by constant forces.
  • 1 A horizontal push of $50\ \text{N}$ moves a box $6.0\ \text{m}$ across a floor. How much work does the push do?
  • 2 A child pulls a sledge $20\ \text{m}$ over level snow with a rope at $37^\circ$ above the horizontal; the tension is $60\ \text{N}$. Find the work done by the tension.
  • 3 A weightlifter holds a $120\ \text{kg}$ barbell above his head for $5.0\ \text{s}$. How much work does he do on it? How much did he do lifting it through $2.0\ \text{m}$ at steady speed?
  • 4 A $2.0\ \text{kg}$ ball is thrown straight up and rises $5.0\ \text{m}$. Find the work done by gravity on the way up, on the way back down to the hand, and for the round trip.
  • 5 A constant force $\mathbf{F}=(4\hat{\mathbf{i}}-3\hat{\mathbf{j}}+2\hat{\mathbf{k}})\ \text{N}$ produces a displacement $\mathbf{d}=(3\hat{\mathbf{i}}+2\hat{\mathbf{j}}+4\hat{\mathbf{k}})\ \text{m}$. Find the work done.
  • 6 A $5.0\ \text{kg}$ box slides $3.0\ \text{m}$ down a $30^\circ$ incline with $\mu_k=0.25$. Find the work done on it by gravity, by the normal force, by friction, and the net work.
  • 7 A stone tied to a string is whirled in a horizontal circle at constant speed. How much work does the tension do in one revolution? Explain.
  • 8 A $60\ \text{kg}$ shopper stands still on an escalator that carries her $8.0\ \text{m}$ higher at constant speed. Find the work done on her by the escalator and by gravity, and the net work.
1 $W=50\times6.0=3.0\times10^2\ \text{J}$.
2 $W=60\times20\times\cos37^\circ=958\ \text{J}\approx9.6\times10^2\ \text{J}$.
3 Holding still: $d=0$, so $W=0$. Lifting: the upward force equals the weight, $W=mgh=120\times9.8\times2.0=2.4\times10^3\ \text{J}$.
4 Up: $W_g=-mgh=-2.0\times9.8\times5.0=-98\ \text{J}$; down: $+98\ \text{J}$; round trip: $0$.
5 $W=12-6+8=14\ \text{J}$.
6 Gravity: $mgd\sin30^\circ=5.0\times9.8\times3.0\times0.5=73.5\ \text{J}\approx74\ \text{J}$. Normal force: $0$ (perpendicular). Friction: $-\mu_kmg\cos30^\circ\,d=-0.25\times5.0\times9.8\times0.866\times3.0=-32\ \text{J}$. Net: $73.5-31.8=42\ \text{J}$.
7 Zero: the tension points to the centre, perpendicular to the velocity at every instant, so each small displacement contributes $F\,\Delta s\cos90^\circ=0$.
8 Escalator: $+mgh=60\times9.8\times8.0=4.7\times10^3\ \text{J}$; gravity: $-4.7\times10^3\ \text{J}$; net: $0$ — her speed does not change.
III
Kinetic energy and the work–energy theorem.
  • 1 Find the kinetic energy of a $0.16\ \text{kg}$ cricket ball bowled at $40\ \text{m/s}$.
  • 2 Find the kinetic energy of a $1500\ \text{kg}$ car at $72\ \text{km/h}$.
  • 3 By what percentage does a body's kinetic energy increase if its speed increases by $10\%$?
  • 4 A $2.0\ \text{kg}$ body has momentum $10\ \text{kg m/s}$. Find its kinetic energy.
  • 5 A $5.0\ \text{kg}$ trolley moving at $2.0\ \text{m/s}$ is pushed with a net force of $25\ \text{N}$ in its direction of motion for $8.0\ \text{m}$. Find its final speed using the work–energy theorem, and check by kinematics.
  • 6 A $0.50\ \text{kg}$ ice-hockey puck sliding at $12\ \text{m/s}$ comes to rest after $36\ \text{m}$. Find the friction force (assumed constant) and the coefficient of kinetic friction.
  • 7 Bodies of mass $1.0\ \text{kg}$ and $4.0\ \text{kg}$ have equal kinetic energies. Find the ratio of their momenta.
  • 8 A car stops in $8.0\ \text{m}$ from $36\ \text{km/h}$. With the same braking force, how far does it need from $108\ \text{km/h}$?
1 $K=\tfrac12(0.16)(40)^2=128\ \text{J}\approx1.3\times10^2\ \text{J}$.
2 $72\ \text{km/h}=20\ \text{m/s}$: $K=\tfrac12(1500)(20)^2=3.0\times10^5\ \text{J}$.
3 $K\propto v^2$: $1.10^2=1.21$, an increase of $21\%$.
4 $K=\dfrac{p^2}{2m}=\dfrac{100}{4.0}=25\ \text{J}$.
5 $K_i=\tfrac12(5.0)(2.0)^2=10\ \text{J}$, $W=25\times8.0=200\ \text{J}$, $K_f=210\ \text{J}$, $v=\sqrt{\tfrac{2\times210}{5.0}}=9.2\ \text{m/s}$. Kinematics: $a=5.0\ \text{m/s}^2$, $v^2=4.0+2(5.0)(8.0)=84$, $v=9.2\ \text{m/s}$ ✓.
6 $K_i=\tfrac12(0.50)(12)^2=36\ \text{J}$; $f\times36=36$, so $f=1.0\ \text{N}$; $\mu_k=\dfrac{f}{mg}=\dfrac{1.0}{4.9}=0.20$.
7 $p=\sqrt{2mK}$, so $\dfrac{p_1}{p_2}=\sqrt{\dfrac{1.0}{4.0}}=\dfrac12$: the heavier body has twice the momentum.
8 $s\propto u^2$ and the speed is tripled ($10\to30\ \text{m/s}$): $s=9\times8.0=72\ \text{m}$.
IV
Variable forces and applications.
  • 1 A force along $+x$ increases uniformly from $0$ to $20\ \text{N}$ as $x$ goes from $0$ to $4.0\ \text{m}$. Find the work done.
  • 2 Find the work done by $F=(5+2x)\ \text{N}$ ($x$ in metres) from $x=0$ to $x=3\ \text{m}$, (a) by integration, (b) as the area of a trapezium.
  • 3 Find the work done by $F=6x^2\ \text{N}$ from $x=0$ to $x=2\ \text{m}$.
  • 4 The force of part 2 acts on a $4.0\ \text{kg}$ block, initially at rest at $x=0$ on a frictionless floor. Find its speed at $x=3\ \text{m}$.
  • 5 A force is $12\ \text{N}$ from $x=0$ to $3\ \text{m}$, then decreases uniformly to zero at $x=7\ \text{m}$, then is $-6\ \text{N}$ from $x=7$ to $9\ \text{m}$. Sketch the $F$–$x$ graph and find the work done from $x=0$ to $9\ \text{m}$.
  • 6 A $1.5\ \text{kg}$ ball dropped from rest at a height of $10\ \text{m}$ hits the ground at $12\ \text{m/s}$. Find the work done by air resistance, and the average air-resistance force.
  • 7 (Modelling) A $4.0\times10^5\ \text{kg}$ train runs at $25\ \text{m/s}$; its brakes give a constant retarding force of $1.0\times10^5\ \text{N}$. Find its braking distance. Why must railway warning signals be placed well over a kilometre before a danger point?
1 Triangle: $W=\tfrac12(4.0)(20)=40\ \text{J}$.
2 (a) $\displaystyle\int_0^3(5+2x)\,dx=\left[5x+x^2\right]_0^3=15+9=24\ \text{J}$. (b) $F(0)=5$, $F(3)=11$: $\tfrac12(5+11)(3)=24\ \text{J}$ ✓.
3 $\displaystyle\int_0^2 6x^2\,dx=\left[2x^3\right]_0^2=16\ \text{J}$.
4 $\tfrac12(4.0)v^2=24$: $v=\sqrt{12}=3.5\ \text{m/s}$.
5 Rectangle $12\times3=36\ \text{J}$, triangle $\tfrac12(4)(12)=24\ \text{J}$, rectangle below the axis $-6\times2=-12\ \text{J}$: $W=48\ \text{J}$.
6 $W_g=mgh=147\ \text{J}$, $K_f=\tfrac12(1.5)(12)^2=108\ \text{J}$. $W_g+W_\text{air}=K_f$, so $W_\text{air}=-39\ \text{J}$; average force $=\tfrac{39}{10}=3.9\ \text{N}$ upward.
7 $K=\tfrac12(4.0\times10^5)(25)^2=1.25\times10^8\ \text{J}$; $s=\dfrac{K}{F}=1.25\times10^3\ \text{m}=1.25\ \text{km}$. A driver must see the signal at least this far ahead (plus reaction distance, and more on wet rails or downhill), so signals are placed more than $1.5\ \text{km}$ before the danger.
🎯 Exam corner BoardJEE MainJEE AdvancedNEET

E1. Board · 3 marks State the work–energy theorem. Prove it for a constant force acting on a body moving in a straight line.

Statement The work done by the net force on a body equals the change in its kinetic energy: $W_\text{net}=K_f-K_i$. [1]
Proof A constant net force $F$ gives a constant acceleration $a=F/m$. Over a displacement $d$ along the force, $v^2=u^2+2ad$. [1]
Finish Multiply by $\tfrac12m$: $\tfrac12mv^2-\tfrac12mu^2=mad=Fd=W_\text{net}$. [1]
Exam tip Say "net" force and "all" forces — the theorem is false for the work of a single force when others also act.

E2. NEET The work done by a friction force on a body (a) is always negative (b) is always zero (c) can be positive, negative or zero (d) is always positive

Idea Compare the friction force with the body's displacement. Kinetic friction on a block sliding on a fixed floor: negative. Static friction on a crate in an accelerating truck: forward, so positive. Static friction on a crate resting on a stationary floor while you push it gently: zero (no displacement).
Answer (c).
Exam tip "Always negative" is the most common trap; friction opposes relative sliding, not motion as seen from the ground.

E3. JEE Main (Numerical value) A particle moves along the $x$-axis under a force $F=(3x^2-2x)\ \text{N}$, $x$ in metres. It is at rest at $x=1\ \text{m}$. No other force does work. Find its kinetic energy (in J) when it reaches $x=3\ \text{m}$.

Work $W=\displaystyle\int_1^3(3x^2-2x)\,dx=\left[x^3-x^2\right]_1^3=(27-9)-(1-1)=18\ \text{J}$.
Answer $K=0+18=18\ \text{J}$.
Exam tip Watch the lower limit: integrating from $0$ instead of $1$ also gives $18$ here only by coincidence ($1^3-1^2=0$) — always substitute both limits.

E4. JEE Main NEET The kinetic energy of a body increases by $44\%$. Its momentum increases by (a) $20\%$ (b) $22\%$ (c) $44\%$ (d) $12\%$

Relation $p=\sqrt{2mK}$, so $\dfrac{p_f}{p_i}=\sqrt{1.44}=1.20$.
Answer (a) $20\%$.
Exam tip (b) halves the percentage — valid only for small changes ($\Delta p/p\approx\tfrac12\Delta K/K$); for $44\%$ use exact square roots.

E5. JEE Advanced (One or more correct) A block of mass $m$ is pulled a distance $d$ up a rough incline of angle $\theta$ (coefficient of kinetic friction $\mu_k$) at constant speed by a force parallel to the incline. Which statements are correct? (A) The net work done on the block is zero. (B) The work done by the pulling force is $mgd(\sin\theta+\mu_k\cos\theta)$. (C) The work done by gravity is $+mgd\sin\theta$. (D) The work done by the normal force is $\mu_kmgd\cos\theta$.

(A) Constant speed means $\Delta K=0$, so $W_\text{net}=0$ ✓.
(B) Constant speed: $F=mg\sin\theta+\mu_kmg\cos\theta$, along the displacement, so $W_F=mgd(\sin\theta+\mu_k\cos\theta)$ ✓.
(C) The block rises $d\sin\theta$ while gravity points down: $W_g=-mgd\sin\theta$ ✗.
(D) The normal force is perpendicular to the incline and the motion: $W_N=0$ ✗ ($\mu_kN$ is the friction, not the normal force).
Answer (A) and (B). Check: $W_F+W_g+W_f+W_N=mgd(\sin\theta+\mu_k\cos\theta)-mgd\sin\theta-\mu_kmgd\cos\theta+0=0$ ✓.

E6. NEET Two bodies of masses $1\ \text{g}$ and $4\ \text{g}$ have equal kinetic energies. The ratio of the magnitudes of their momenta is (a) $4:1$ (b) $\sqrt2:1$ (c) $1:2$ (d) $1:16$

Idea $p=\sqrt{2mK}$ with equal $K$: $\dfrac{p_1}{p_2}=\sqrt{\dfrac{m_1}{m_2}}=\sqrt{\dfrac14}=\dfrac12$.
Answer (c) $1:2$.
Exam tip (d) squares instead of taking the root; (a) inverts the ratio. With equal kinetic energy the heavier body has the larger momentum.
🔬 Research–Teaching NexusResearch-tutored · optional, about 40 minutes

Research question. How do crumple zones and airbags reduce the force on a car passenger in a crash, and by how much?

Why it matters. Crash-test programmes in Europe, the USA, India and elsewhere rate cars by the forces measured on instrumented dummies. The physics is the work–energy theorem: the passenger's kinetic energy must be removed by the work of the restraining forces, $\bar F\,s=\tfrac12mv^2$. The same energy removed over a longer stopping distance $s$ means a smaller average force. Engineers design the front of the car to crush in a controlled way, and the seat belt and airbag to stretch and deflate, precisely to lengthen $s$.

What to do.

  1. For a $70\ \text{kg}$ passenger at $50\ \text{km/h}$, compute the kinetic energy that must be removed.
  2. Find the average stopping force (in newtons and as a multiple of the passenger's weight) if the passenger stops in (i) $0.10\ \text{m}$ (hitting a rigid dashboard), (ii) $0.30\ \text{m}$, (iii) $0.60\ \text{m}$ (crumple zone plus belt and airbag).
  3. Plot average force against stopping distance. What shape is the graph?
  4. Repeat for $64\ \text{km/h}$ (a common test speed). By what factor does the force grow for the same stopping distance?
  5. Look up the head-injury and chest-acceleration limits used in crash testing and compare with your results.

Tools: Spreadsheet; published crash-test reports (they give speeds, crush depths and dummy accelerations).

Hand in: A table and graph of force against stopping distance, and a half-page note explaining to a non-scientist why "a car that crumples is a safer car".

Going further. The force in a real crash is not constant. Sketch a realistic force–distance curve with the same area as your constant-force rectangle; why do engineers try to make it as flat as possible?

Instructor note: $50\ \text{km/h}=13.9\ \text{m/s}$, $K=6.75\times10^3\ \text{J}$. Average forces: $6.75\times10^4\ \text{N}$ ($\approx98$ times body weight) at $0.10\ \text{m}$, $2.25\times10^4\ \text{N}$ ($\approx33\,mg$) at $0.30\ \text{m}$, $1.13\times10^4\ \text{N}$ ($\approx16\,mg$) at $0.60\ \text{m}$; $\bar F\propto1/s$ (a hyperbola). At $64\ \text{km/h}$ the force is $(64/50)^2=1.64$ times larger. A flat force curve gives the smallest peak force for a given area (energy).

🧪 Self-check quiz

0/5 correct

Q1. A stone moves in a horizontal circle at constant speed on a string. In one full revolution the work done by the tension is

Q2. For $\mathbf{A}=2\hat{\mathbf{i}}+3\hat{\mathbf{j}}$ and $\mathbf{B}=3\hat{\mathbf{i}}-2\hat{\mathbf{j}}$, the scalar product $\mathbf{A}\cdot\mathbf{B}$ is

Q3. One joule is equal to

Q4. If the speed of a car is tripled, its kinetic energy becomes

Q5. The area under a force–position ($F$–$x$) graph between two positions represents

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