Chapter 6: Work, Energy and Power

Newton's laws describe motion in the language of forces; energy offers a complementary, often more powerful, viewpoint. Once we recognise certain scalar quantities that change in predictable ways, problems that would be intractable by force alone — collisions, oscillations, orbits — become straightforward. The bedrock identity is the work–energy theorem.

Concept Map

• 6.1 — Scalar (dot) product
• 6.2 — Notion of work for constant and variable force
• 6.3 — Work–energy theorem
• 6.4 — Kinetic energy
• 6.5 — Work done by variable forces (spring, gravity)
• 6.6 — Potential energy; conservative vs non-conservative forces
• 6.7 — PE of spring and gravitational PE near Earth
• 6.8 — Conservation of mechanical energy
• 6.9 — Power
• 6.10 — Collisions in one and two dimensions

6.1 The Scalar (Dot) Product

Definition

For vectors $\vec{A}$ and $\vec{B}$ with angle $\theta$ between them,

$\vec{A}\cdot\vec{B} = AB\cos\theta$

Equivalently, in Cartesian components,

$\vec{A}\cdot\vec{B} = A_xB_x + A_yB_y + A_zB_z$

Key Properties

• Commutative: $\vec{A}\cdot\vec{B} = \vec{B}\cdot\vec{A}$.
• Distributive: $\vec{A}\cdot(\vec{B}+\vec{C}) = \vec{A}\cdot\vec{B} + \vec{A}\cdot\vec{C}$.
• $\vec{A}\cdot\vec{A} = A^2$.
• Two vectors are perpendicular iff their dot product is zero.
• Unit-vector identities: $\hat{i}\cdot\hat{i}=1$, $\hat{i}\cdot\hat{j}=0$, etc.

Worked Example

Find the angle between $\vec{A} = 2\hat{i}+3\hat{j}$ and $\vec{B} = 4\hat{i}-\hat{j}$.

$\vec{A}\cdot\vec{B} = 8 - 3 = 5, \quad A = \sqrt{13}, \quad B = \sqrt{17}$

$\cos\theta = \frac{5}{\sqrt{13\cdot 17}} \approx 0.336 \Rightarrow \theta \approx 70.4^\circ$

Common Mistakes

• Treating the dot product as a vector — the result is a scalar.
• Forgetting the $\cos\theta$ when force and displacement are perpendicular.

6.2 Notion of Work

Definition (Constant Force)

The work done by a constant force $\vec{F}$ on a particle that undergoes displacement $\vec{d}$ is

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

SI unit: joule (J) $= 1\,\text{N·m}$.

Variable Force — Line Integral

If the force varies along the path, we break the path into infinitesimal pieces $d\vec{r}$ and sum:

$W = \int_A^B \vec{F}\cdot d\vec{r}$

For motion along a single axis $x$ with force $F(x)$,

$W = \int_{x_1}^{x_2} F(x)\,dx$

This is the area under the $F$–$x$ curve.

Sign Conventions

• $0 \le \theta < 90^\circ$: $W > 0$ (force aids motion).
• $\theta = 90^\circ$: $W = 0$ (force perpendicular, e.g., centripetal force).
• $90^\circ < \theta \le 180^\circ$: $W < 0$ (force opposes motion, e.g., friction during sliding).

Derivation (work by gravity for projectile)

A projectile starts at height $0$ and ends at height $h$. Gravity acts downward $-mg\hat{j}$, displacement is $\Delta y\,\hat{j} = h\hat{j}$.

$W_g = -mg\cdot h = -mgh$

The work depends only on the vertical displacement, not the path.

Special Cases

• A man holding a heavy bag stationary does no mechanical work on it.
• Normal force on a block sliding down an incline does zero work (perpendicular to motion).
• Friction force on the rolling contact of a wheel does zero work (the contact point is instantaneously at rest).

Worked Example

A horizontal force of $20\,\text{N}$ pulls a block over a distance of $5\,\text{m}$ on a rough floor with friction $8\,\text{N}$. Find net work.

Work by pull: $W_1 = 20\cdot 5 = 100\,\text{J}$.

Work by friction: $W_2 = -8\cdot 5 = -40\,\text{J}$.

Net work: $W = 60\,\text{J}$.

Common Mistakes

• Forgetting the sign on friction or negative work.
• Forgetting that work is path-dependent in general; it equals area under $F$–$x$ for 1D forces only.

6.3 Work–Energy Theorem

Statement

The net work done on a particle equals the change in its kinetic energy.

$W_{\text{net}} = \Delta K = K_f - K_i$

Derivation (1D, constant force)

Start with $F = ma$. Use $v^2 = u^2 + 2as$ from kinematics:

$v^2 - u^2 = 2as = \frac{2Fs}{m} \quad \Rightarrow \quad Fs = \tfrac{1}{2}mv^2 - \tfrac{1}{2}mu^2$

That is, $W = \Delta K$.

Derivation (1D, variable force)

$W = \int_{x_1}^{x_2} F\,dx = \int m\frac{dv}{dt}\,dx = \int m\,dv\cdot\frac{dx}{dt} = \int_{v_1}^{v_2} mv\,dv = \tfrac{1}{2}mv_2^2 - \tfrac{1}{2}mv_1^2$

Derivation (3D, variable force)

$W = \int \vec{F}\cdot d\vec{r} = \int m\frac{d\vec{v}}{dt}\cdot d\vec{r} = \int m\,d\vec{v}\cdot\vec{v} = \tfrac{1}{2}m\Delta(v\cdot v) = \Delta\!\left(\tfrac{1}{2}mv^2\right)$

Worked Example

A $0.5\,\text{kg}$ ball moving at $4\,\text{m/s}$ is brought to rest in $2\,\text{m}$ by friction. Find the friction force.

$\Delta K = 0 - \tfrac{1}{2}(0.5)(16) = -4\,\text{J}$

$-f\cdot 2 = -4 \Rightarrow f = 2\,\text{N}$

Common Mistakes

• Applying the theorem to individual forces — it is the net work that equals $\Delta K$.
• Forgetting that $\Delta K$ can be negative.

6.4 Kinetic Energy

Definition

$K = \tfrac{1}{2}mv^2$

Relation to Momentum

$K = \frac{p^2}{2m}$

This is useful in collision problems: equal momentum does not mean equal kinetic energy; the lighter body has more KE for the same $p$.

Worked Example

A $1000\,\text{kg}$ car at $20\,\text{m/s}$ and a $1\,\text{kg}$ bullet at $400\,\text{m/s}$. Compare KEs and momenta.

Car: $p = 20000\,\text{kg m/s}$, $K = 2\times 10^5\,\text{J}$.

Bullet: $p = 400\,\text{kg m/s}$, $K = 8\times 10^4\,\text{J}$.

The bullet has much smaller momentum but the car has more KE.

Common Mistakes

• Treating KE as a vector.
• Forgetting that KE scales with $v^2$, so doubling the speed quadruples the KE.

6.5 Work Done by Variable Force — Spring and Gravity

Spring (Hooke's Law)

A spring stretched by $x$ exerts a restoring force $F = -kx$. Work done by the spring as the body moves from $x_1$ to $x_2$:

$W_s = \int_{x_1}^{x_2} (-kx)\,dx = -\tfrac{1}{2}k(x_2^2 - x_1^2)$

If we stretch from natural length ($x_1 = 0$) to $x_2 = x$, the spring does $-\tfrac{1}{2}kx^2$ of work; the external agent does $+\tfrac{1}{2}kx^2$.

Gravity (Near the Surface)

Work done by gravity when a body of mass $m$ falls from height $h_1$ to $h_2$:

$W_g = -mg(h_2 - h_1) = mg(h_1 - h_2)$

Path-independent.

Worked Example

A spring of $k = 200\,\text{N/m}$ is compressed by $0.1\,\text{m}$. Find work stored.

$W_{\text{stored}} = \tfrac{1}{2}kx^2 = \tfrac{1}{2}(200)(0.01) = 1\,\text{J}$

Common Mistakes

• Using $W = Fx$ for a spring (force varies — must integrate).
• Forgetting that the spring's work on the body is negative when storing energy.

6.6 Potential Energy — Conservative vs Non-conservative Forces

Definition

A force is conservative if the work it does between two points is independent of the path taken. Equivalently, the work on any closed loop is zero. For such a force we can define a potential energy $U(\vec{r})$ such that

$\vec{F} = -\vec{\nabla} U$

In 1D: $F(x) = -\dfrac{dU}{dx}$.

Examples

• Conservative: gravity, spring, electrostatic force.
• Non-conservative: friction, viscous drag, air resistance.

Test for Conservation (1D)

If $F(x)$ depends only on $x$ (not on velocity or time), it is conservative.

Worked Example

For $U(x) = \tfrac{1}{2}kx^2$, find $F(x)$.

$F(x) = -\frac{dU}{dx} = -kx$

This recovers Hooke's law.

Common Mistakes

• Defining $U$ for non-conservative forces (impossible).
• Forgetting the minus sign in $F = -dU/dx$.

6.7 PE of Spring and Gravitational PE Near Earth

Spring PE

Taking $U(0) = 0$ at natural length,

$U_{\text{spring}}(x) = \tfrac{1}{2}kx^2$

Independent of whether compression or stretch.

Gravitational PE (Near Earth)

Taking $U = 0$ at ground level,

$U_g = mgh$

This is valid only near the Earth's surface where $g$ is roughly constant.

General Gravitational PE (Chapter 8 preview)

For an inverse-square gravitational force,

$U = -\frac{GMm}{r}$

with $U(\infty) = 0$.

Common Mistakes

• Treating $U = mgh$ as universally valid (it isn't — only near surface).
• Forgetting that the choice of $U = 0$ is a convention — only differences in $U$ are physical.

6.8 Conservation of Mechanical Energy

Statement

If only conservative forces act on a system, its total mechanical energy $E = K + U$ is conserved.

Derivation

For a conservative force, work–energy theorem gives $W = \Delta K$, and also $W = -\Delta U$ (by definition of $U$). Hence

$\Delta K + \Delta U = 0 \quad \Rightarrow \quad K + U = \text{constant}$

Worked Example (Free fall)

A body of mass $m$ falls from height $h$. At the top, $K = 0$, $U = mgh$. At the bottom, $K = \tfrac{1}{2}mv^2$, $U = 0$. Equating,

$v = \sqrt{2gh}$

which agrees with kinematics.

Worked Example (Spring + Block)

A block of mass $m$ moving at $v$ hits a spring of constant $k$ on a smooth floor. Find maximum compression.

At max compression, $v = 0$. Conservation:

$\tfrac{1}{2}mv^2 = \tfrac{1}{2}kx_{\max}^2 \quad \Rightarrow \quad x_{\max} = v\sqrt{\frac{m}{k}}$

Worked Example (Pendulum)

A simple pendulum of length $L$ is pulled aside by angle $\theta_0$ and released. Find the speed at the bottom.

Height drop: $h = L(1 - \cos\theta_0)$.

$v = \sqrt{2gL(1-\cos\theta_0)}$

When Non-conservative Forces Act

$W_{\text{nc}} = \Delta E = (K_f + U_f) - (K_i + U_i)$

Energy is "lost" to heat, sound, etc., but total energy (including these forms) is still conserved.

Common Mistakes

• Applying mechanical-energy conservation when friction is present.
• Choosing inconsistent reference levels for $U$ at start and end.

6.9 Power

Definition

Power is the rate at which work is done:

$P_{\text{avg}} = \frac{W}{t}, \quad P_{\text{inst}} = \frac{dW}{dt} = \vec{F}\cdot\vec{v}$

SI unit: watt (W) $= 1\,\text{J/s}$.

Other units: $1\,\text{HP} = 746\,\text{W}$.

Derivation

Starting from $dW = \vec{F}\cdot d\vec{r}$ and dividing by $dt$,

$P = \frac{dW}{dt} = \vec{F}\cdot\frac{d\vec{r}}{dt} = \vec{F}\cdot\vec{v}$

Worked Example

A pump lifts $200\,\text{kg}$ of water per second through a height of $5\,\text{m}$. Find the power. ($g = 10\,\text{m/s}^2$)

$P = \frac{dW}{dt} = \frac{m g h}{t} = 200\cdot 10\cdot 5 = 10000\,\text{W} = 10\,\text{kW}$

Worked Example (Constant Power Engine)

A car of mass $m$ accelerates from rest with constant engine power $P$. Find $v(t)$.

$P = Fv = mv\frac{dv}{dt}$

Separating variables,

$\int_0^v mv\,dv = \int_0^t P\,dt$

$\tfrac{1}{2}mv^2 = Pt \quad \Rightarrow \quad v = \sqrt{\frac{2Pt}{m}}$

Common Mistakes

• Confusing average and instantaneous power.
• Forgetting that $P = Fv$ uses instantaneous values.

6.10 Collisions

Classification

• Elastic: both momentum and kinetic energy conserved.
• Inelastic: momentum conserved; KE not conserved (some lost to heat, deformation).
• Perfectly inelastic: bodies stick together after collision.

1D Elastic Collision — Derivation of Final Velocities

Two masses $m_1, m_2$ with initial velocities $u_1, u_2$. Final velocities $v_1, v_2$.

Conservation of momentum:

$m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2 \quad (\star)$

Conservation of KE:

$\tfrac{1}{2}m_1 u_1^2 + \tfrac{1}{2}m_2 u_2^2 = \tfrac{1}{2}m_1 v_1^2 + \tfrac{1}{2}m_2 v_2^2 \quad (\star\star)$

Rewriting $(\star)$ and $(\star\star)$:

$m_1(u_1 - v_1) = m_2(v_2 - u_2)$ $m_1(u_1^2 - v_1^2) = m_2(v_2^2 - u_2^2)$

Dividing the second by the first:

$u_1 + v_1 = u_2 + v_2 \quad \Rightarrow \quad u_1 - u_2 = -(v_1 - v_2)$

This is the famous result: in 1D elastic collision, relative velocity of approach equals relative velocity of separation.

Solving with $(\star)$:

$v_1 = \frac{(m_1 - m_2)u_1 + 2m_2 u_2}{m_1 + m_2}$ $v_2 = \frac{(m_2 - m_1)u_2 + 2m_1 u_1}{m_1 + m_2}$

Special Cases

1. Equal masses ($m_1 = m_2$): $v_1 = u_2$, $v_2 = u_1$ — velocities are exchanged.
2. Light hitting stationary heavy ($m_2 \gg m_1$, $u_2 = 0$): $v_1 \approx -u_1$, $v_2 \approx 0$ — light bounces back.
3. Heavy hitting stationary light ($m_1 \gg m_2$, $u_2 = 0$): $v_1 \approx u_1$, $v_2 \approx 2u_1$.

Perfectly Inelastic Collision

Both bodies stick: $v_1 = v_2 = V$.

$V = \frac{m_1 u_1 + m_2 u_2}{m_1 + m_2}$

Energy lost:

$\Delta K = \frac{1}{2}\frac{m_1 m_2}{m_1 + m_2}(u_1 - u_2)^2$

Coefficient of Restitution

$e = \frac{|v_2 - v_1|}{|u_1 - u_2|} = \frac{\text{relative velocity of separation}}{\text{relative velocity of approach}}$

• $e = 1$: perfectly elastic.
• $0 < e < 1$: partially elastic.
• $e = 0$: perfectly inelastic.

For ball bouncing on floor, $e = v_{\text{after}}/v_{\text{before}}$. Height ratio: $h_2/h_1 = e^2$.

2D Elastic Collision

A particle of mass $m_1$ with velocity $u$ hits a stationary particle of mass $m_2$. After collision, $m_1$ moves at angle $\theta_1$, $m_2$ at angle $\theta_2$ (on opposite sides of original direction).

• $x$-momentum: $m_1 u = m_1 v_1\cos\theta_1 + m_2 v_2\cos\theta_2$
• $y$-momentum: $0 = m_1 v_1\sin\theta_1 - m_2 v_2\sin\theta_2$
• KE: $\tfrac{1}{2}m_1 u^2 = \tfrac{1}{2}m_1 v_1^2 + \tfrac{1}{2}m_2 v_2^2$

For equal masses ($m_1 = m_2$) the angle between $v_1$ and $v_2$ after collision is $90^\circ$ — a famous result useful in billiards and Compton scattering.

Worked Example (1D, elastic, equal masses)

A $2\,\text{kg}$ ball at $5\,\text{m/s}$ hits an identical stationary ball. Find final velocities.

By symmetry: $v_1 = 0$, $v_2 = 5\,\text{m/s}$.

Worked Example (Perfectly Inelastic)

$0.02\,\text{kg}$ bullet at $300\,\text{m/s}$ embeds in $1.98\,\text{kg}$ block (smooth floor). Final common velocity?

$V = \frac{0.02\cdot 300}{2.0} = 3\,\text{m/s}$

Worked Example (Ball Drop)

A ball dropped from $1\,\text{m}$ rebounds to $0.64\,\text{m}$. Find $e$.

$e^2 = \frac{h_2}{h_1} = 0.64 \quad \Rightarrow \quad e = 0.8$

Common Mistakes

• Applying KE conservation to inelastic collisions.
• Forgetting that 2D collision needs three equations (two momentum + one KE) for elastic.
• Forgetting the sign of relative velocity in $e$.

Solved Problems

Problem 1

A block of $4\,\text{kg}$ is pulled by a $50\,\text{N}$ force at $60^\circ$ above horizontal over $10\,\text{m}$ on a rough floor with $\mu_k = 0.1$, $g = 10\,\text{m/s}^2$. Find the work–energy budget.

Normal: $N = mg - F\sin 60^\circ = 40 - 50(0.866) = -3.3\,\text{N}$ — negative! The body lifts off; let's take instead a less-steep angle.

Recompute assuming the body remains on the floor: $N = mg - F\sin 60^\circ$ would be negative, so the block leaves the floor and slides essentially without friction (treat $f = 0$).

Then $W_{\text{applied}} = F\cos 60^\circ\cdot d = 25\cdot 10 = 250\,\text{J}$. $W_g = 0$. $\Delta K = 250\,\text{J}$, so $v = \sqrt{2\cdot 250/4} = \sqrt{125}\approx 11.2\,\text{m/s}$.

Problem 2

A spring of natural length $0.5\,\text{m}$ and $k = 100\,\text{N/m}$ is stretched to $0.7\,\text{m}$. Find the work done by the external agent.

$x = 0.2\,\text{m}$. $W = \tfrac{1}{2}k x^2 = \tfrac{1}{2}(100)(0.04) = 2\,\text{J}$.

Problem 3

A ball of $0.2\,\text{kg}$ is dropped from $20\,\text{m}$. Just before hitting the ground, find KE. ($g = 10$)

$K = mgh = 0.2\cdot 10\cdot 20 = 40\,\text{J}$.

Problem 4

A $0.5\,\text{kg}$ ball collides elastically with a $2\,\text{kg}$ ball at rest. The first ball has initial speed $6\,\text{m/s}$. Find final velocities.

$v_1 = \frac{(0.5-2)\cdot 6}{2.5} = -3.6\,\text{m/s}, \quad v_2 = \frac{2\cdot 0.5\cdot 6}{2.5} = 2.4\,\text{m/s}$

The first ball rebounds; second moves forward.

Problem 5

A pump fills a tank of volume $20\,\text{m}^3$ at height $30\,\text{m}$ in $1000\,\text{s}$. Density of water $\rho = 1000\,\text{kg/m}^3$. Find power. ($g = 10$)

Mass lifted: $M = 20000\,\text{kg}$. Total work: $Mgh = 6\times 10^6\,\text{J}$. Power: $P = 6000\,\text{W} = 6\,\text{kW}$.

Problem 6

A car of mass $1000\,\text{kg}$ moves at constant $20\,\text{m/s}$ on a level road against friction $200\,\text{N}$. Engine power?

$P = Fv = 200\cdot 20 = 4000\,\text{W} = 4\,\text{kW}$.

Problem 7

A bullet of $10\,\text{g}$ embeds in a $1.99\,\text{kg}$ pendulum bob hanging by a string $1\,\text{m}$ long. The pendulum rises by $0.2\,\text{m}$. Find bullet's initial speed. ($g = 10$)

Common velocity just after impact: $V = \sqrt{2gh} = \sqrt{4} = 2\,\text{m/s}$.

Momentum conservation: $0.01\cdot v = 2.0\cdot 2$, $v = 400\,\text{m/s}$.

JEE/NEET Edge Cases

1. Variable mass with constant power: Use $P = Fv$ and integrate carefully.
2. Spring with pulley problems require identifying displacements of both ends correctly.
3. Inelastic collision in CM frame: In the CM frame, perfectly inelastic collision converts all CM-frame KE into heat.
4. Bouncing on inclined surface: velocity components parallel and normal behave differently — the parallel component is conserved (smooth incline), normal reverses with factor $e$.
5. Work by static friction can be non-zero when the contact point moves (rolling friction does no work, but static friction on a wheel that drives a car does — over a non-zero displacement).
6. Energy method vs Newton's laws: energy gives speed at endpoints quickly; forces are needed for time/acceleration.

Quick Recap

• Work: $W = \int \vec{F}\cdot d\vec{r}$, scalar.
• Work–energy theorem: $W_{\text{net}} = \Delta K$.
• Conservative force has path-independent work; $\vec{F} = -\vec{\nabla} U$.
• Mechanical energy $K + U$ is conserved if only conservative forces act.
• Power: $P = \vec{F}\cdot\vec{v}$.
• Elastic 1D collision: $v_1, v_2 =$ standard formula; relative velocities of approach and separation are equal.
• Perfectly inelastic: bodies stick; energy lost is $\tfrac{1}{2}\mu(u_1-u_2)^2$ with $\mu = m_1 m_2/(m_1+m_2)$.

Formula Sheet