Compiled from: NCERT Physics Part-1 (Work, Energy and Power) • H.C. Verma — Concepts of Physics Vol-1 (Ch. 8: Work and Energy) • D.C. Pandey — Understanding Physics: Mechanics Part-1 (Work, Energy & Power, Circular Motion) • Pradeep's Fundamental Physics XI • S.L. Arora • Cengage (B.M. Sharma) Mechanics-I • Errorless Physics. Target exams: Boards • JEE Main • JEE Advanced • NEET • NDA.
θ = angle between force and displacement. Work is a scalar (dot product!). SI unit: joule (J) = 1 N·m; dimension $[ML^2T^{-2}]$. Other units: 1 erg = 10⁻⁷ J; 1 eV = 1.6 × 10⁻¹⁹ J; 1 kWh = 3.6 × 10⁶ J; 1 calorie = 4.186 J.
| Sign of W | Condition | Classic Examples (Boards love these) |
|---|---|---|
| Positive (0° ≤ θ < 90°) | Force has a component along displacement | Gravity on a falling body; engine force on an accelerating car; stretching force on a spring (by the agent) |
| Negative (90° < θ ≤ 180°) | Force opposes the displacement | Friction on a sliding block; gravity on a rising ball; braking force; work by a spring on the stretching agent |
| Zero | θ = 90°, or s = 0, or F = 0 | Gravity on a body moved horizontally; centripetal force in circular motion; tension in a whirling string; a coolie walking with a load on his head (idealised); pushing a rigid wall (s = 0) |
Centripetal force does zero work because F ⊥ v at every instant — this is why speed stays constant in uniform circular motion. A man holding a heavy suitcase stationary does no work in physics (s = 0), though he tires physiologically. Both are repeat 1-markers in Boards/NEET.
$W = \vec{F}\cdot\vec{s} = F_xs_x + F_ys_y + F_zs_z$. Work depends on the frame of reference (displacement does!) — work by the same force can differ for observers in relative motion (an HCV subtlety asked in JEE Advanced).
W = Fs cos θ with the three sign cases
Same displacement, three different force directions — the sign of work is pure geometry.
Q. A body is displaced by $\vec{s} = (3\hat{i} + 4\hat{j})$ m under a force $\vec{F} = (4\hat{i} + \hat{j})$ N. Find the work done and the angle case (positive/negative/zero).
Area below the x-axis counts as negative work. This single idea handles springs, gravity over large heights, and every graph question of this chapter.
Q. A force $F = (3x^2 - 2x)$ N acts on a particle along x. Find the work done as it moves from x = 0 to x = 2 m.
Always positive, frame-dependent, scalar. Percentage tricks (NEET/NDA): if p increases by 100% (doubles), K becomes 4× (300% increase); if K increases by 300%, p doubles (100% increase); for small changes, %ΔK ≈ 2 × %Δp.
Statement: The net work done by all forces on a body equals the change in its kinetic energy: $W_{net} = K_f - K_i$.
Derivation (constant force): $v^2 = u^2 + 2as \Rightarrow \tfrac12mv^2 - \tfrac12mu^2 = (ma)s = Fs = W$ ✓
Derivation (variable force): $$W = \int F\,dx = \int m\frac{dv}{dt}\,dx = \int mv\,dv = \tfrac12mv^2 - \tfrac12mu^2$$ The theorem holds for all forces — conservative or not (friction included) and in any inertial frame. It is the integral form of Newton's second law.
Whenever a question gives forces/distances and asks for speed (or vice-versa) without asking for time, use the work–energy theorem instead of F = ma + kinematics — one line instead of three. Example: penetration of bullets into planks, blocks sliding on rough patches, pendulum speeds.
Q. A bullet loses 1/20 of its velocity passing through one plank. How many such planks will just stop it?
Q. A 2 kg block moving at 10 m/s slides onto a rough patch with μ = 0.5 (g = 10 m/s²). How far does it travel before stopping?
| Property | Conservative (gravity, spring, electrostatic) | Non-conservative (friction, viscous drag) |
|---|---|---|
| Work over a path | Depends only on end points | Depends on the path (longer path → more loss) |
| Work over a closed loop | Zero | Non-zero (negative) |
| Potential energy | Can be defined: $\Delta U = -W_{cons}$ | Cannot be defined |
| Mechanical energy | Conserved | Dissipated as heat/sound |
Only differences in U are physical — the zero level is our choice. Force points from higher U toward lower U ("downhill" on the U–x curve).
| Type | Condition | U–x shape | Behaviour on displacement |
|---|---|---|---|
| Stable | $\frac{dU}{dx}=0$, $\frac{d^2U}{dx^2} > 0$ | Minimum (valley) | Restoring force pushes it back (oscillates) |
| Unstable | $\frac{dU}{dx}=0$, $\frac{d^2U}{dx^2} < 0$ | Maximum (hilltop) | Force pushes it further away |
| Neutral | $\frac{dU}{dx}=0$, $\frac{d^2U}{dx^2} = 0$ | Flat region | Stays wherever displaced |
Potential-energy curve with all three equilibria, allowed regions and turning points
The particle oscillates in the well, easing to rest exactly at the two turning points where E = U — trapped because K = E − U can never go negative.
Q. The potential energy of a particle is $U = (x^2 - 4x + 3)$ J. Find the equilibrium position and its type, and the force at x = 0.
Hooke's law: $F_{spring} = -kx$ (restoring, opposite to stretch/compression x; k = spring constant, N/m; stiffer spring → larger k).
Derivation of spring PE. Work done by the agent in stretching slowly from 0 to x:
$$W = \int_0^x kx'\,dx' = \tfrac12kx^2 \;\Rightarrow\; U_{spring} = \tfrac12kx^2$$Work done by the spring in the same stretch = −½kx². From x₁ to x₂: $W_{spring} = \tfrac12k(x_1^2 - x_2^2)$ — depends only on end deformations (conservative). U is the area of the triangle under the F–x line.
Spring F–x graph, the energy triangle, and the block–spring collision
The block decelerates into the spring, pauses at maximum compression, and is thrown back at the same speed — kinetic energy parked briefly as ½kx² and returned in full.
Q. A spring (k = 200 N/m) is stretched by 5 cm. Find the work needed to stretch it by a further 5 cm.
Statement: If only conservative forces do work, the total mechanical energy E = K + U of a system remains constant.
Proof for a freely falling body (dropped from height H; take ground as U = 0):
E is the same everywhere; K and U merely interconvert. With friction: $\Delta E = -W_{friction}$ (energy lost = heat) — use the modified theorem $K_f + U_f = K_i + U_i - |W_{fric}|$.
Free-fall energy conservation with live K/U bars and three snapshots
The bars are driven by the same acceleration curve as the ball, so U + K stays pinned to the bracket at every instant — conservation you can watch.
Q. A ball dropped from 80 m: at what height is its KE equal to its PE? What is its speed there? (g = 10 m/s²)
Setup: a bob on a string of length L whirled in a vertical circle. At the TOP, string tension and gravity both point toward the centre: $T_{top} + mg = \dfrac{mv_{top}^2}{L}$. The critical condition is T_top = 0:
$$v_{top(min)} = \sqrt{gL}$$Energy conservation top → bottom (height difference 2L):
$$\tfrac12mv_{bottom}^2 = \tfrac12mv_{top}^2 + mg(2L)$$At the bottom: $T_{bottom} = \dfrac{mv_b^2}{L} + mg$ (minimum 6mg for a full loop). If $v_{bottom} < \sqrt{2gL}$: oscillates like a pendulum (never reaches horizontal); between $\sqrt{2gL}$ and $\sqrt{5gL}$: leaves the circle somewhere above the horizontal and becomes a projectile. For a bead on a rigid rod/track (can push), the top condition relaxes to v_top ≥ 0 ⟹ v_bottom ≥ $\sqrt{4gL} = 2\sqrt{gL}$.
Vertical circle: forces at top, bottom and side, with the three speed regimes
The bob's speed is computed from energy conservation (v² = gL(3 + 2cos θ), the critical case): watch it crawl over the top and whip through the bottom — the string rotates rigidly with it.
Q. A 0.5 kg stone on a 2 m string must just complete a vertical circle (g = 10 m/s²). Find the minimum speeds at the top and bottom and the tension at the bottom in that case.
Scalar. SI unit watt (W) = J/s; dimension $[ML^2T^{-3}]$. 1 hp = 746 W. kWh is a unit of ENERGY, not power: 1 kWh = 3.6 × 10⁶ J (the "unit" on an electricity bill).
Q. A pump lifts 600 kg of water per minute from a 10 m deep well and ejects it at 5 m/s. Find the power of the pump if its efficiency is 80%. (g = 10 m/s²)
Q. A car of mass 1000 kg moves up at a constant 15 m/s against total resistance 500 N on a level road. Find the engine power.
In every collision (no external force during the brief impact) momentum is conserved. Kinetic energy may or may not be conserved — that defines the type.
| Type | Momentum | Kinetic Energy | Coefficient of restitution e |
|---|---|---|---|
| Elastic | Conserved | Conserved | e = 1 |
| Inelastic | Conserved | Partly lost (heat, sound, deformation) | 0 < e < 1 |
| Perfectly inelastic | Conserved | Maximum possible loss; bodies stick together | e = 0 |
Ball dropped from height h rebounds to $h' = e^2h$; after n bounces $h_n = e^{2n}h$ (a NEET regular).
Given: m₁ at u₁ hits m₂ at u₂ (u₁ > u₂), head-on, elastic.
Conservation equations:
$$m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2$$ $$\tfrac12m_1u_1^2 + \tfrac12m_2u_2^2 = \tfrac12m_1v_1^2 + \tfrac12m_2v_2^2$$Rearrange each and divide (the classic trick): $u_1 + v_1 = u_2 + v_2$, i.e. relative velocity reverses: $u_1 - u_2 = -(v_1 - v_2)$ (this IS e = 1). Solving the linear pair:
$$v_1 = \frac{(m_1-m_2)u_1 + 2m_2u_2}{m_1+m_2} \qquad v_2 = \frac{(m_2-m_1)u_2 + 2m_1u_1}{m_1+m_2}$$| Case (elastic, u₂ = 0 unless stated) | Result |
|---|---|
| Equal masses (m₁ = m₂) | They exchange velocities: v₁ = u₂, v₂ = u₁ (Newton's cradle!) |
| Heavy hits light (m₁ ≫ m₂) | v₁ ≈ u₁ (barely slows); v₂ ≈ 2u₁ (light one shoots off at double speed) |
| Light hits heavy (m₁ ≪ m₂) | v₁ ≈ −u₁ (bounces back with same speed); v₂ ≈ 0 (ball off a wall) |
| Energy transfer | Fraction transferred to m₂: $\dfrac{4m_1m_2}{(m_1+m_2)^2}$ — maximum (100%) when m₁ = m₂ (why moderators in reactors use light nuclei!) |
Sticking together: common velocity $v = \dfrac{m_1u_1 + m_2u_2}{m_1+m_2}$.
KE loss (u₂ = 0): $\Delta K = \dfrac{m_1m_2}{2(m_1+m_2)}u_1^2 = \dfrac{1}{2}\mu_{red}\,u_{rel}^2$ where $\mu_{red} = \dfrac{m_1m_2}{m_1+m_2}$ (reduced mass form — elegant and exam-usable).
Ballistic pendulum (bullet embeds in a hanging block): first momentum conservation for the impact, then energy conservation for the swing: $h = \dfrac{v_{common}^2}{2g}$. Never apply energy conservation ACROSS the impact! In 2-D elastic collision of equal masses (one at rest), the two final velocities are perpendicular (θ₁ + θ₂ = 90°) — billiards physics, a JEE favourite.
Collision gallery: elastic special cases and the perfectly inelastic stick-together
Both animations obey momentum exactly: in the exchange, B departs at A's full speed; in the sticky collision, the joined pair moves at precisely half speed (segment timings encode the velocities).
Q. A 2 kg body moving at 6 m/s collides with a stationary 4 kg body and sticks to it. Find the common velocity and the kinetic energy lost.
Q. A 10 g bullet at 400 m/s embeds in a 1.99 kg block hanging from long strings. To what height does the block rise? (g = 10 m/s²)
One worked model of each distinct question type asked from this chapter, tagged by exam. Attempt each yourself first, then check.
Q. State and prove the work–energy theorem for a variable force (5 marks). / Prove conservation of mechanical energy for a freely falling body (5 marks). / Derive the velocities after a 1-D elastic collision and discuss the equal-mass case (5 marks).
Q. The momentum of a body increases by 50%. By what percentage does its kinetic energy increase?
Q. A force on a particle varies as: F = 10 N (constant) from x = 0 to 2 m, then decreases linearly to 0 at x = 4 m, then is −5 N (constant) from 4 to 6 m. Find the total work.
Q. A 2 kg block slides from rest down a rough incline of height 5 m and reaches the bottom at 8 m/s (g = 10 m/s²). Find the energy lost to friction.
Q. A uniform chain of mass M and length L lies on a table with one-third of its length hanging over the edge. Find the work required to pull the hanging part back onto the table.
Q. A 1 kg block at 4 m/s hits a spring (k = 100 N/m) on a floor with μ = 0.2 (g = 10). Find the maximum compression. (Quadratic expected!)
Q. A ball dropped from 20 m rebounds to 5 m. Find e and the height after the second bounce.
Q. A 2 kg body starts from rest and is driven by a machine delivering a constant power of 16 W. Find its speed and displacement after 4 s.
Q. A neutron (mass m) collides elastically head-on with a stationary carbon nucleus (mass 12m). What fraction of its kinetic energy does it transfer?
Q. A particle slides from rest from the top of a smooth hemisphere of radius R. At what height (from the ground) does it leave the surface?
Read this the night before the exam:
| Concept | Formula | Condition / Note |
|---|---|---|
| Work (constant F) | $W = Fs\cos\theta = \vec{F}\cdot\vec{s}$ | Zero at θ = 90°; 1 kWh = 3.6×10⁶ J |
| Work (variable F) | $W = \int F\,dx$ | = signed area under F–x graph |
| Kinetic energy | $K = \tfrac12mv^2 = p^2/2m$ | $p = \sqrt{2mK}$; %ΔK ≈ 2%Δp (small) |
| Work–energy theorem | $W_{net} = \Delta K$ | All forces, any frame (inertial) |
| PE ↔ force | $F = -dU/dx$ | Stable eq.: U min; unstable: U max |
| Gravitational PE | $U = mgh$ | Near surface; zero level arbitrary |
| Spring PE | $U = \tfrac12kx^2$ | $W_{x_1\to x_2} = \tfrac12k(x_2^2-x_1^2)$ |
| Springs combined | Series: $\tfrac1{k} = \tfrac1{k_1}+\tfrac1{k_2}$ | Parallel: $k = k_1+k_2$; n cuts → each nk |
| Energy conservation | $K + U = $ constant | Conservative forces only; else ΔE = −|W_fric| |
| Vertical circle | $v_{top}=\sqrt{gL},\ v_{bot}=\sqrt{5gL}$ | $T_{bot}-T_{top} = 6mg$; rod: $v_{bot} = 2\sqrt{gL}$ |
| Leaving a smooth sphere | $\cos\theta = 2/3$ | Leaves at height 2R/3 from ground |
| Power | $P = W/t = \vec{F}\cdot\vec{v}$ | 1 hp = 746 W; pump: P = (mgh+½mv²)/t ÷ η |
| Constant power motion | $v = \sqrt{2Pt/m}$ | $x \propto t^{3/2}$ |
| Elastic collision (1-D) | $v_1 = \frac{(m_1-m_2)u_1+2m_2u_2}{m_1+m_2}$ | Swap indices for v₂; equal masses exchange v |
| Energy transfer fraction | $\frac{4m_1m_2}{(m_1+m_2)^2}$ | Max (=1) when m₁ = m₂ |
| Perfectly inelastic | $v = \frac{m_1u_1+m_2u_2}{m_1+m_2}$ | KE lost $= \frac{m_1m_2}{2(m_1+m_2)}u_{rel}^2$ |
| Restitution | $e = \frac{v_2-v_1}{u_1-u_2}$ | Bounce: $h' = e^2h$; n-th: $e^{2n}h$ |
| Chain pull-up | $W = \frac{MgL}{2n^2}$ | Fraction 1/n hanging over the edge |
How to use these notes: Day 1: Sections 1–3 (work, W–E theorem) + Fig 1; redo both theorem derivations. Day 2: Sections 4–6 (PE, springs, conservation) + Figs 2–4; sketch the U–x curve analysis from memory. Day 3: Sections 7–8 (vertical circle, power) + Fig 5; derive √(5gL) unaided. Day 4: Section 9 (collisions) + Fig 6, then all ten Types of Section 10 without looking, followed by the mistakes checklist. Finish every session by writing the formula sheet from memory.
VARDAAN COMET