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Work, Energy and Power – JEE Main Physics Formula Sheet & Class 11 Notes | W=Fs cosθ, KE, Work-Energy Theorem, Spring, Collision & All Formulas
This is the complete JEE Main Physics Formula Sheet and Class 11 Formula Sheet for Work, Energy and Power — Chapter 04 from the Aakash Rapid Revision & Formula Bank. This chapter covers: Work — W=F·s=Fs cosθ, W=∫F·ds (variable force), area under F-x graph; Kinetic Energy — KE=½mv²=p²/2m, work done by net force = ΔKE; Work-Energy Theorem — W_total=ΔKE, W_external+W_internal=ΔKE; Potential Energy — gravitational PE=mgh, elastic PE=½kx², relationship between F and U (F=–dU/dx); Conservation of Energy — total mechanical energy E=KE+PE=constant (for conservative forces); Power — P=W/t=F·v=Fv cosθ, average vs instantaneous power; Spring — work done W=½k(x₂²–x₁²), spring behaves as spring constant k=YA/l; Elastic Energy — U=½kx², energy density=½×stress×strain; Collision — coefficient of restitution e=relative velocity of separation/relative velocity of approach; Elastic Collision — e=1, KE conserved, velocity exchange formulas v₁=(m₁–m₂)u₁/(m₁+m₂)+2m₂u₂/(m₁+m₂), v₂=(m₂–m₁)u₂/(m₁+m₂)+2m₁u₁/(m₁+m₂); Inelastic Collision — e<1, KE lost = ½m₁m₂(u₁–u₂)²(1–e²)/(m₁+m₂); Perfectly Inelastic — e=0, bodies stick together; Oblique Elastic — both elastic and angle relations; Ball Bouncing — height after nth bounce hₙ=e²ⁿh₀, rebound speed after nth bounce=eⁿ√(2gh₀), total distance=h₀(1+e²)/(1–e²), time=√(2h₀/g)·(1+e)/(1–e). Work, Energy and Power contributes 3–4 questions in every JEE Main session. Download the Free PDF for all formulas in one JEE Main exam-ready reference.
Topics Covered in This Work, Energy and Power Formula Sheet
Work, Energy and Power JEE Main Formula Sheet PDF Preview
Scroll to explore all Work, Energy and Power formulas — JEE Main Physics Formula Sheet
Introduction: Why Work, Energy and Power Is a Guaranteed Scoring Chapter in JEE Main Physics
Work, Energy and Power bridges the gap between force-based mechanics (Laws of Motion) and energy-based mechanics — a paradigm shift that makes many complex problems solvable in seconds. Instead of applying F=ma step-by-step through a complicated path, the Work-Energy Theorem directly connects the net work done on a body to its change in kinetic energy: W_total = ΔKE. Conservation of energy makes it possible to find speeds at any point in a system — from springs to inclined planes to vertical circles — without ever computing force or acceleration explicitly.
For JEE Main physics, Work, Energy and Power contributes 3–4 questions per session. Questions test: work done by constant and variable forces, kinetic energy and the work-energy theorem, spring potential energy, elastic vs inelastic collision velocity formulas, coefficient of restitution and KE loss, and ball bouncing (hₙ = e²ⁿh₀, total distance, total time). The chapter also directly links to vertical circle from Laws of Motion, since energy conservation gives vH²=vL²–4gl.
Download the Free PDF for Work, Energy and Power to access all work formulas, kinetic energy, potential energy, power, spring energy, collision formulas (elastic, inelastic, oblique, coefficient of restitution), and ball bouncing results in one structured JEE Main physics revision reference.
Key Concepts and Formulas in Work, Energy and Power
Work — Definition, Formula, Sign Convention, Variable Force
Why W = F·s cosθ and the Area Under F-x Graph Are the Most Fundamental Work, Energy and Power JEE Main Formulas
Definition of Work (from Aakash PDF — Work, Energy and Power):
Work is done by a force on a body when the body undergoes displacement in the direction of the applied force. Work is a scalar quantity.
W = F·s = Fs cosθ
where F = magnitude of force, s = magnitude of displacement (s = s₂–s₁), θ = angle between the force vector F and displacement vector s.
Vector form: W = F·s (dot product of force and displacement vectors).
Sign Convention for Work (from Aakash PDF — Work, Energy and Power JEE Main):
Positive work: θ < 90° (force has component in direction of displacement). Example: applied force on a moving block.
Negative work: θ > 90° (force has component opposite to displacement). Example: friction on a moving block, gravity when a body moves upward.
Zero work: θ = 90° (force perpendicular to displacement). Example: normal force on a block moving horizontally, tension in circular motion, magnetic force on a moving charge.
Work by Variable Force (from Aakash PDF — Work, Energy and Power JEE Main):
W = ∫F·ds (integrate force over the displacement path)
Graphically: W = area under the F-x graph (between initial and final position). Area above x-axis = positive work; area below = negative work.
For a spring (F = –kx): W_spring = –∫₀ˣ kx dx = –½kx² (work done BY spring on mass). If compressing or stretching from x₁ to x₂: W_spring = –½k(x₂²–x₁²).
Work done by various forces in common JEE Main Work, Energy and Power scenarios:
Work by gravity: W_g = –mgh (h = rise in height; negative when moving up). W_g = mgh (when moving down, h is positive descent).
Work by normal force: W_N = 0 (always perpendicular to motion on a surface).
Work by friction: W_f = –f_k × d = –μₖmg × d (negative, always opposing motion).
Work by tension in circular motion: W_T = 0 (tension always perpendicular to velocity).
Work by spring restoring force: W_spring = –½kx² (always negative when spring stretches/compresses from natural length).
Download the Free PDF for Work, Energy and Power for all work done examples for JEE Main.
Kinetic Energy, Potential Energy, Work-Energy Theorem, and Conservation of Energy
Why the Work-Energy Theorem and Conservation of Mechanical Energy Are the Most-Tested Work, Energy and Power Results in JEE Main
Kinetic Energy (from Aakash PDF — Work, Energy and Power JEE Main):
KE = ½mv²
In terms of momentum p = mv: KE = p²/2m
Useful relation: p = √(2m·KE). KE is always ≥ 0. KE = 0 only when v = 0 (body at rest).
Change in KE: ΔKE = ½mv₂² – ½mv₁² = ½m(v₂²–v₁²)
Work-Energy Theorem (from Aakash PDF — Work, Energy and Power JEE Main):
The total work done on a body by all forces = change in kinetic energy of the body.
W_total = ΔKE = ½mv² – ½mu²
This can be split into contributions: W_external + W_gravity + W_spring + W_friction + W_normal = ΔKE
Since W_normal = 0 (perpendicular) and W_gravity = mgh (when falling height h): effectively W_external + mgh – ½kx² – μₖmgd = ΔKE.
Gravitational Potential Energy (from Aakash PDF — Work, Energy and Power):
U_gravity = mgh (measured from a chosen reference level, h is height above reference).
Work by gravity = –ΔU_gravity = –mg(h₂–h₁) = mg(h₁–h₂). When body moves down (h decreases), gravity does positive work; PE decreases.
Elastic Potential Energy (from Aakash PDF — Work, Energy and Power):
U_spring = ½kx² (x = extension or compression from natural length)
Work by spring = –ΔU_spring = –[½kx₂² – ½kx₁²] = ½k(x₁²–x₂²)
Force from Potential Energy (from Aakash PDF — Work, Energy and Power JEE Main):
F = –dU/dx (the force is the negative gradient of potential energy)
For spring: F = –d(½kx²)/dx = –kx ✓ (restoring force). For gravity: F = –d(mgh)/dh = –mg ✓ (downward).
Equilibrium position: dU/dx = 0 (force = 0). Stable equilibrium: d²U/dx² > 0 (PE is minimum). Unstable equilibrium: d²U/dx² < 0 (PE is maximum). Neutral equilibrium: d²U/dx² = 0 (PE is constant).
Conservation of Mechanical Energy (from Aakash PDF — Work, Energy and Power JEE Main):
For a system with only conservative forces acting: total mechanical energy E = KE + PE = constant.
ΔKE + ΔPE = 0 → ΔKE = –ΔPE (gain in KE = loss in PE, and vice versa)
For non-conservative forces (like friction): E_final = E_initial – W_friction (work done against friction dissipates as heat)
KE₁ + PE₁ = KE₂ + PE₂ (conservative forces only)
½mv₁² + mgh₁ = ½mv₂² + mgh₂ (gravity only — most common JEE Main scenario)
½mv₁² + ½kx₁² = ½mv₂² + ½kx₂² (spring only — SHM energy conservation)
Download the Free PDF for Work, Energy and Power for all energy conservation examples for JEE Main.
Power — Definition, Formulas, Units, and Applications
Why P = Fv cosθ and Average Power Are Direct Work, Energy and Power JEE Main Questions
Definition of Power (from Aakash PDF — Work, Energy and Power JEE Main):
Power is the rate of doing work (or the rate of energy transfer).
Average power: P_avg = W_total / t_total
Instantaneous power: P = dW/dt = F·v = Fv cosθ
where θ = angle between force F and velocity v at that instant.
When force is parallel to velocity: P = Fv (maximum power for given F and v).
When force is perpendicular to velocity: P = 0 (e.g., centripetal force in uniform circular motion does zero work, hence zero power).
Units of Power (from Aakash PDF — Work, Energy and Power):
SI unit: Watt (W) = J/s = kg·m²/s³. Dimensional formula: [ML²T⁻³].
1 kilowatt (kW) = 1000 W. 1 megawatt (MW) = 10⁶ W.
1 Horsepower (HP) = 746 W ≈ 750 W (British unit, used in engine ratings).
1 kWh = 3.6×10⁶ J (commercial unit of energy = energy consumed at 1 kW power for 1 hour).
Power in common Work, Energy and Power JEE Main scenarios:
Vehicle on level road at constant speed v with resistance f: P = fv (engine power = resistive force × speed).
Vehicle climbing incline at constant speed v: P = (mg sinθ + f)v where f is frictional resistance.
At maximum speed on level road: engine power P = f × v_max → v_max = P/f (if f is constant resistance).
Water pump: P = ρAv³/2 + ρAgh·v (kinetic power + potential power for lifting water).
Pumping water from depth h at flow rate Q = Av: P_minimum = ρgQh (to lift only; ignoring KE).
Download the Free PDF for Work, Energy and Power for all power examples for JEE Main.
Spring — Work Done, Elastic Energy, Wire as Spring, Energy Density
Why Spring Potential Energy and Wire Elastic Energy Are Essential Work, Energy and Power JEE Main Formulas
Spring Force and Work (from Aakash PDF — Work, Energy and Power JEE Main):
Spring restoring force: F = –kx (Hooke's Law; k = spring constant). A wire behaves like a spring with constant:
k = YA/l where Y = Young's modulus, A = cross-sectional area, l = natural length.
Work Done in Stretching a Spring (from Aakash PDF — Work, Energy and Power JEE Main):
When stretched from natural length to extension x:
(a) Work done by external force: W_ext = +½kx² (positive — external force does positive work)
(b) Work done by restoring force (spring on mass): W_spring = –½kx² (negative — restoring force opposes displacement)
(c) Elastic potential energy stored: U = ½kx²
When stretched from x₁ to x₂: W_ext = ½k(x₂²–x₁²); W_spring = –½k(x₂²–x₁²)
(d) Heat produced when spring is stretched by external force F (and held) equals zero IF spring is ideal (no friction). When spring is released and comes to natural length: KE gained = ½kx² (from stored PE).
Elastic Energy and Energy Density (from Aakash PDF — Work, Energy and Power JEE Main):
For a wire/rod: elastic PE stored = ½ × F × Δl = ½ × (stress × A) × (strain × l) × (1/Al) × Al = ½ × stress × strain × volume
Energy density (energy per unit volume) = ½ × stress × strain = stress²/(2Y) = Y × strain²/2
With k = YA/l and extension Δl: U = ½kΔl² = ½(YA/l)Δl² = ½·Y·A·Δl²/l
Energy density = U/(Volume) = ½·Y·Δl²/l² = ½·Y·(strain)² = ½·stress·strain
Rod hanging from support (from Aakash PDF — Work, Energy and Power JEE Main):
For a rod of mass M, length l, cross-section A, Young's modulus Y, hanging from ceiling:
Extension due to its own weight: Δl = Mgl/(2AY) [factor 2 because average tensile stress = Mg/2A]
Elastic PE stored = ½ × (average stress) × strain × volume = M²g²l/(6AY) [using U=½kΔl² approach]
Heat produced (from Aakash PDF): when a wire is stretched by external force F to extension Δl and then released: W_ext = ½FΔl = U_stored. If inelastic deformation occurs, some energy converts to heat. Download the Free PDF for Work, Energy and Power for all spring and elastic energy examples for JEE Main.
Collisions — Coefficient of Restitution, Elastic, Inelastic, Perfectly Inelastic
Why Collision Velocity Formulas and KE Loss Formula Are the Most Calculation-Heavy Work, Energy and Power JEE Main Questions
Coefficient of Restitution (from Aakash PDF — Work, Energy and Power JEE Main):
For a collision along a straight line between objects A and B:
e = (velocity of separation along line of impact) / (velocity of approach along line of impact)
e = (v₂–v₁)/(u₁–u₂) [where u₁>u₂ means A approaches B from left]
Range: 0 ≤ e ≤ 1. e=1 (perfectly elastic); e=0 (perfectly inelastic); 0
KE Lost in Collision (from Aakash PDF — Work, Energy and Power JEE Main):
ΔKE = ½ × m₁m₂/(m₁+m₂) × (u₁–u₂)² × (1–e²)
For e=1 (elastic): ΔKE=0. For e=0 (perfectly inelastic): ΔKE = ½m₁m₂(u₁–u₂)²/(m₁+m₂) [maximum KE loss].
Elastic Collision (e=1) — Velocity Formulas (from Aakash PDF — Work, Energy and Power JEE Main):
From momentum conservation and e=1 (v₂–v₁=u₁–u₂):
v₁ = (m₁–m₂)u₁/(m₁+m₂) + 2m₂u₂/(m₁+m₂)
v₂ = 2m₁u₁/(m₁+m₂) + (m₂–m₁)u₂/(m₁+m₂)
Special cases (u₂=0, m₂ initially at rest):
(i) m₁=m₂: v₁=0, v₂=u₁ — complete velocity exchange. (Ball A stops, ball B moves with initial velocity of A.)
(ii) m₁ >> m₂ (heavy hits light): v₁ ≈ u₁ (nearly unchanged); v₂ ≈ 2u₁ (light ball moves at twice the heavy ball speed).
(iii) m₁ << m₂ (light hits heavy): v₁ ≈ –u₁ (light ball bounces back); v₂ ≈ 0 (heavy ball barely moves).
Inelastic Collision (0 < e < 1) (from Aakash PDF — Work, Energy and Power):
From momentum conservation: m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂ …(i)
From restitution: v₂–v₁ = e(u₁–u₂) …(ii)
Solving (i) and (ii): v₁ = (m₁u₁+m₂u₂)/(m₁+m₂) – m₂e(u₁–u₂)/(m₁+m₂)
v₂ = (m₁u₁+m₂u₂)/(m₁+m₂) + m₁e(u₁–u₂)/(m₁+m₂)
Perfectly Inelastic Collision (e=0) (from Aakash PDF — Work, Energy and Power JEE Main):
Bodies stick together after collision. Common velocity: v = (m₁u₁ + m₂u₂)/(m₁+m₂) [from momentum conservation only].
KE lost = ½m₁m₂(u₁–u₂)²/(m₁+m₂) [maximum possible KE loss for given initial conditions].
Download the Free PDF for Work, Energy and Power for all collision examples for JEE Main.
Oblique Elastic Collision and Ball Bouncing — All Formulas
Why Oblique Collision Angle Result and Ball Bouncing Series Formulas Are Key Work, Energy and Power JEE Main Results
Oblique Elastic Collision (from Aakash PDF — Work, Energy and Power JEE Main):
A body of mass m moving with velocity u collides obliquely with a stationary body of the same mass m. After collision, the two bodies move at angles φ and ψ to the original direction.
From momentum conservation and KE conservation (elastic, e=1):
(a) Conservation of momentum along original direction: u = v₁ cosφ + v₂ cosψ
(b) Conservation of momentum perpendicular: 0 = v₁ sinφ – v₂ sinψ
(c) Conservation of KE: u² = v₁² + v₂²
From (a), (b), (c): squaring and using (c):
φ + ψ = 90° (for oblique elastic collision between EQUAL masses)
The two bodies move at right angles to each other after an oblique elastic collision of equal masses.
Also: v₁cosφ+v₂cosψ=u; v₁²+v₂²=u².
Special case of head-on elastic (ψ=0°, φ=180°): v₁=0, v₂=u (velocity exchange) ✓.
Coefficient of Restitution for Oblique Impact (from Aakash PDF — Work, Energy and Power JEE Main):
For oblique impact, e applies only along the line of impact (normal to surfaces at contact), not along the tangential direction:
Normal direction: v₂_n – v₁_n = e(u₁_n – u₂_n)
Tangential direction: v₁_t = u₁_t and v₂_t = u₂_t (tangential components unchanged in smooth collision)
v = √(v₁_t² + v₁_n²) [speed of each ball after oblique collision]
tanψ (new angle with original direction) from new velocity components.
Ball Bouncing — All Formulas (from Aakash PDF — Work, Energy and Power JEE Main):
Ball of mass m dropped from height h₀ on an inelastic floor. Coefficient of restitution = e (between ball and floor).
(a) Maximum height after nth bounce: hₙ = e²ⁿh₀
(b) Speed of rebound after nth bounce: vₙ = eⁿ√(2gh₀)
(c) Total distance travelled before ball comes to rest: S_total = h₀(1+e²)/(1–e²)
Derivation: S = h₀ + 2h₁ + 2h₂ + … = h₀ + 2e²h₀ + 2e⁴h₀ + … = h₀ + 2e²h₀/(1–e²) = h₀(1–e²+2e²)/(1–e²) = h₀(1+e²)/(1–e²).
(d) Total time before ball comes to rest: t_total = √(2h₀/g) × (1+e)/(1–e)
Derivation: t = √(2h₀/g) + 2√(2h₁/g) + … = √(2h₀/g)[1 + 2e + 2e² + …] = √(2h₀/g)[(1+e)/(1–e)] since 1+2e+2e²+… = 1+2e/(1–e) = (1+e)/(1–e).
(e) Average force exerted on ground = mg (by impulse-momentum theorem applied to entire motion).
(f) Final displacement of ball when it stops = h₀ (ball stops at the floor, same level as where it was dropped).
Download the Free PDF for Work, Energy and Power for all collision and bouncing examples for JEE Main.
Download Free PDF — Work, Energy and Power JEE Main Physics Formula Sheet
All Work, Energy and Power formulas from the Aakash Rapid Revision PDF: W=Fs cosθ=F·s (dot product), variable force W=∫F·ds = area under F-x graph, positive/negative/zero work sign rules, KE=½mv²=p²/2m, p=√(2m·KE), Work-Energy Theorem W_total=ΔKE=½m(v²–u²), work by gravity (mgh down/–mgh up), work by normal and tension (zero), work by friction (–μₖmgd), U_gravity=mgh, U_spring=½kx², F=–dU/dx, equilibrium conditions (dU/dx=0; stable d²U/dx²>0; unstable d²U/dx²<0), conservation of mechanical energy (KE+PE=constant), energy lost to friction = W_friction, instantaneous power P=dW/dt=F·v=Fv cosθ, average power P=W/t, 1HP=746W, 1kWh=3.6×10⁶J, spring k=YA/l, W_ext=+½kx² (spring), W_spring=–½kx², U=½kx², energy density=½×stress×strain=σ²/2Y, rod hanging Δl=Mgl/2AY, coefficient of restitution e=(v₂–v₁)/(u₁–u₂), 0≤e≤1, ΔKE=½m₁m₂(u₁–u₂)²(1–e²)/(m₁+m₂), elastic (e=1): v₁=(m₁–m₂)u₁+2m₂u₂)/(m₁+m₂) and v₂=(2m₁u₁+(m₂–m₁)u₂)/(m₁+m₂), special cases (equal mass velocity exchange; heavy-light 2u₁; light-heavy –u₁), perfectly inelastic (e=0): v=(m₁u₁+m₂u₂)/(m₁+m₂), oblique elastic equal mass (φ+ψ=90°, v₁²+v₂²=u²), ball bouncing (hₙ=e²ⁿh₀; vₙ=eⁿ√(2gh₀); S=h₀(1+e²)/(1–e²); t=√(2h₀/g)×(1+e)/(1–e); avg force=mg; final displacement=h₀).
Why Work, Energy and Power Is a Cross-Chapter Tool in JEE Main Physics
The Work-Energy Theorem W_total=ΔKE is the fastest method for almost every mechanics problem in JEE Main. When a problem asks for the final speed of an object after complex motion — sliding down a curved surface, released from a spring, swinging on a pendulum — the Work-Energy Theorem bypasses all intermediate force calculations. Just compute total work done by all forces and set it equal to ΔKE. If friction is absent, use KE+PE=constant (even simpler). The key skill: identifying which forces do work (those not perpendicular to motion) and computing their work correctly.
KE = p²/2m is the critical formula when JEE Main Work, Energy and Power questions link momentum and kinetic energy. If two bodies have the same momentum, the lighter one has greater KE. If they have the same KE, the heavier one has greater momentum. These comparative statements are tested as 1-mark conceptual questions. Also: when a body splits into pieces (explosion), total KE increases (internal energy converts) while total momentum is conserved.
Elastic collision velocity formulas v₁ and v₂ are the most formula-intensive Work, Energy and Power results. The equal-mass result (velocity exchange: v₁=0, v₂=u₁) is the most frequently tested in JEE Main. The heavy-hitting-light result (v₂≈2u₁) is conceptually important for understanding neutron moderation and similar problems. Memorising these three special cases of elastic collision eliminates the need to derive from scratch during the exam.
Ball bouncing formulas (hₙ=e²ⁿh₀, S=h₀(1+e²)/(1–e²)) are geometric series results that appear repeatedly in JEE Main Work, Energy and Power questions. They derive directly from e = (velocity of separation)/(velocity of approach) and geometric series sums. Once the coefficient of restitution e is known, the height after any number of bounces, total distance, and total time are all computable in one step. Download the Free PDF for Work, Energy and Power to have all formulas ready.
Who Should Use This Work, Energy and Power Formula Sheet?
Learning Outcomes After Completing Work, Energy and Power
After working through Work, Energy and Power using this formula sheet, a student should confidently accomplish the following for JEE Main physics. On work: compute W=Fs cosθ for any constant force; determine sign of work (positive/negative/zero); compute W=∫F·ds for variable forces; read work from area under F-x graph; compute work by gravity, normal, friction, tension, spring in any scenario.
On energy: compute KE=½mv² and KE=p²/2m; convert between p and KE; apply Work-Energy Theorem W_total=ΔKE; compute PE (gravitational=mgh, spring=½kx²); apply F=–dU/dx; identify equilibrium type from U-x graph; apply conservation of mechanical energy (KE+PE=constant); account for energy lost to friction.
On power: compute average power P=W/t; compute instantaneous power P=F·v=Fv cosθ; identify when P=0 (force perpendicular to velocity); convert between watts, kW, HP; solve vehicle engine power problems (constant speed, maximum speed).
On springs: use k=YA/l for wire as spring; compute W_ext=+½kx² and W_spring=–½kx²; compute elastic PE stored U=½kx²; compute energy density=½×stress×strain.
On collisions: compute e=(v₂–v₁)/(u₁–u₂); compute ΔKE from e; apply elastic collision velocity formulas for v₁ and v₂; use all three special cases (equal mass, heavy-light, light-heavy); compute v for perfectly inelastic from momentum conservation; apply oblique equal-mass elastic (φ+ψ=90°); compute hₙ=e²ⁿh₀, vₙ=eⁿ√(2gh₀), total distance h₀(1+e²)/(1–e²), total time √(2h₀/g)(1+e)/(1–e). Download the Free PDF for Work, Energy and Power to test all outcomes before your JEE Main exam.
Get the Free PDF for Work, Energy and Power — JEE Main Quick Revision
The Aakash Rapid Revision & Formula Bank PDF for Work, Energy and Power contains all work formulas, kinetic and potential energy, Work-Energy Theorem, power, spring elastic energy, all collision formulas (elastic, inelastic, oblique), coefficient of restitution, and ball bouncing series results in one structured JEE Main physics reference.
Conclusion — Work, Energy and Power: The Energy Language of JEE Main Physics
Work, Energy and Power provides the energy perspective on mechanics — a parallel framework to Newton's laws that is often faster and more elegant for solving complex problems. While Newton's laws track forces and accelerations at every instant, the energy approach tracks only initial and final states: what was the total energy at start, what is it at end, and where did the difference go (spring, height, friction heat).
The two most-used results in Work, Energy and Power for JEE Main: (1) Work-Energy Theorem W=ΔKE and conservation of mechanical energy KE₁+PE₁=KE₂+PE₂ — these together solve almost every projectile, pendulum, spring, and inclined plane energy question; (2) Elastic collision velocity exchange and ball bouncing height hₙ=e²ⁿh₀ — these two collision results cover all collision and restitution JEE Main questions. Use this page and the Free PDF Download for Work, Energy and Power as your complete JEE Main revision foundation.
Frequently Asked Questions — Work, Energy and Power Formulas
What is the Work-Energy Theorem in Work, Energy and Power and how is it applied in JEE Main?
In Work, Energy and Power, the Work-Energy Theorem states: the net work done by ALL forces on a body equals the change in kinetic energy. W_total = ΔKE = ½mv²–½mu². Application in JEE Main: Step 1 — identify all forces on the body (gravity, normal, friction, applied force, spring, tension, etc.). Step 2 — compute work done by each force: W_gravity=mgh (body falls h) or –mgh (rises h); W_normal=0; W_tension=0 (in circular motion); W_friction=–μₖmgd; W_spring=–½kx²; W_applied=F×d×cosθ. Step 3 — sum all works. Step 4 — set equal to ΔKE=½mv²–½mu². Example: block slides down rough incline of length l, angle θ: W_gravity=mgl sinθ; W_friction=–μₖmgl cosθ; W_normal=0; total W=(mgl sinθ–μₖmgl cosθ)=ΔKE=½mv². So v=√(2gl(sinθ–μₖcosθ)). This bypasses finding acceleration and using kinematic equations — one energy equation gives v directly. These Work Energy Power applications appear in every JEE Main session.
What is the difference between elastic and inelastic collision in Work, Energy and Power?
In Work, Energy and Power, collisions are classified by the coefficient of restitution e=(relative speed of separation)/(relative speed of approach): Elastic collision (e=1): both momentum AND kinetic energy are conserved. Coefficient of restitution = 1. Velocity formulas: v₁=(m₁–m₂)u₁+2m₂u₂)/(m₁+m₂); v₂=(2m₁u₁+(m₂–m₁)u₂)/(m₁+m₂). KE lost = 0. Inelastic collision (0
What is the coefficient of restitution in Work, Energy and Power and how does it relate to KE loss?
In Work, Energy and Power, coefficient of restitution e = (velocity of separation along impact line)/(velocity of approach along impact line) = (v₂–v₁)/(u₁–u₂) [assuming u₁>u₂]. Range: 0≤e≤1. e=1: elastic (no KE loss). e=0: perfectly inelastic (maximum KE loss). KE lost: ΔKE=½m₁m₂(u₁–u₂)²(1–e²)/(m₁+m₂). Note: the factor (1–e²) shows that for e=1: ΔKE=0; for e=0: ΔKE is maximum. For e between 0 and 1: ΔKE increases as e decreases. Derivation: from momentum conservation (m₁u₁+m₂u₂=m₁v₁+m₂v₂) and restitution (v₂–v₁=e(u₁–u₂)), solve for v₁ and v₂, then compute KE_initial–KE_final. Result: ΔKE=½μ(u₁–u₂)²(1–e²) where μ=m₁m₂/(m₁+m₂) is the reduced mass. Ball on floor: e=(rebound speed)/(impact speed)=√(h₁/h₀) where h₁ is first rebound height. After nth bounce: hₙ=e²ⁿh₀. These Work Energy Power coefficient of restitution results are directly tested in JEE Main.
What are all the ball bouncing formulas in Work, Energy and Power?
In Work, Energy and Power, ball dropped from height h₀ with coefficient of restitution e between ball and floor: (a) Height after nth bounce: hₙ=e²ⁿh₀. Proof: speed just before first impact v₀=√(2gh₀); after bounce v₁=ev₀; height h₁=v₁²/2g=e²h₀. After nth bounce: vₙ=eⁿv₀; hₙ=vₙ²/2g=e²ⁿh₀. (b) Speed at nth rebound: vₙ=eⁿ√(2gh₀). (c) Total distance before stopping: S=h₀+2h₁+2h₂+…=h₀+2e²h₀+2e⁴h₀+…=h₀+2e²h₀/(1–e²)=h₀(1–e²+2e²)/(1–e²)=h₀(1+e²)/(1–e²). (d) Total time: t₀=√(2h₀/g) (first fall); t₁=2√(2h₁/g)=2e√(2h₀/g) (first bounce); t₂=2e²√(2h₀/g); total t=t₀+t₁+t₂+…=√(2h₀/g)[1+2e+2e²+…]=√(2h₀/g)×(1+e)/(1–e). (e) Average force on ground=mg (since by impulse-momentum over entire time, total impulse=m×v₀–(–m×0)=mv₀ upward? Actually: Favg×t_total=change in momentum=0 (starts and ends at rest) → Favg=mg for the overall process.) (f) Final displacement=h₀ (ball is at floor level). These Work Energy Power bouncing formulas are tested directly in JEE Main.
What is the elastic potential energy of a spring in Work, Energy and Power?
In Work, Energy and Power, spring elastic potential energy U=½kx² where x is the extension or compression from the natural length and k is the spring constant. For a wire behaving as spring: k=YA/l (Y=Young's modulus, A=cross-section area, l=length). Work by external force stretching spring from 0 to x: W_ext=+½kx² (positive, external force in same direction as displacement). Work by spring restoring force: W_spring=–½kx² (negative, restoring force opposes displacement). From x₁ to x₂: W_ext=½k(x₂²–x₁²); W_spring=–½k(x₂²–x₁²). Energy density (energy per unit volume) in stretched wire: u=½×stress×strain=σ²/2Y=Yε²/2 where σ=F/A (stress), ε=Δl/l (strain). Total PE in wire = u × Volume = (½×stress×strain)×(A×l). Equilibrium extension under weight mg: kx₀=mg → x₀=mg/k. At displacement x from equilibrium in SHM: KE=½k(A²–x²); PE=½kx² (measured from natural length); total E=½kA² (constant). These Work Energy Power spring formulas appear in JEE Main as direct substitution questions on spring compression, spring release, and spring combinations.
What is the formula for power in Work, Energy and Power and what are its applications?
In Work, Energy and Power, instantaneous power P=F·v=Fv cosθ where θ=angle between force and velocity. Average power=W/t. Units: Watt (W=J/s), 1HP=746W, 1kWh=3.6×10⁶J. When P=0: force perpendicular to velocity (centripetal force, normal force, magnetic force on charge). When P=Fv maximum: force parallel to velocity. JEE Main Work Energy Power power applications: (1) Vehicle at constant speed on level road: engine force = friction f; P=fv. (2) Maximum speed on level: at terminal speed, P=f×v_max → v_max=P/f. (3) Vehicle climbing incline at constant speed v: P=(mg sinθ+f)v. (4) Water pump lifting water from depth h at rate Q (volume/second): minimum power P=ρgQh (just to lift, ignoring KE). If also giving KE: P=ρgQh+½ρQv²=ρQ(gh+v²/2). (5) Force by water jet: F=ρAv²; power to create jet P=Fv=ρAv³. (6) Efficiency: η=P_output/P_input×100%; for vehicle: useful power=mg sinθ×v; efficiency=mg sinθ×v/P_engine. These Work Energy Power power formulas are tested in JEE Main as 4-mark direct questions.
What is potential energy and how is equilibrium defined using F=–dU/dx in Work, Energy and Power?
In Work, Energy and Power, potential energy is the energy stored by virtue of configuration or position. F=–dU/dx (the conservative force is the negative gradient of potential energy in 1D). In 3D: F=–∇U. Equilibrium: dU/dx=0 → force=0. Three types: Stable equilibrium: d²U/dx²>0 (U is at minimum; displaced → restoring force pulls back). Example: ball at bottom of bowl. Unstable equilibrium: d²U/dx²<0 (U is at maximum; displaced → force pushes further away). Example: ball on top of hill. Neutral equilibrium: d²U/dx²=0 (U is constant in the region; displaced → no force). Example: ball on a flat horizontal surface. For spring: U=½kx²; F=–dU/dx=–kx ✓ (restoring). For gravity: U=mgh; F=–dU/dh=–mg ✓ (downward). For U=ax²–bx (two-term potential): equilibrium at dU/dx=0 → 2ax–b=0 → x=b/2a. d²U/dx²=2a>0 (if a>0) → stable. Conservation (only conservative forces): ΔKE+ΔU=0 → KE+U=constant. These Work Energy Power potential energy results are tested in JEE Main as U-x graph analysis questions.
What is the result of oblique elastic collision between equal masses in Work, Energy and Power?
In Work, Energy and Power, for oblique elastic collision between equal masses (m₁=m₂=m), one initially at rest: from momentum conservation along x: u=v₁cosφ+v₂cosψ …(i). Momentum along y: 0=v₁sinφ–v₂sinψ …(ii). KE conservation: u²=v₁²+v₂² …(iii). Squaring (i): u²=v₁²cos²φ+v₂²cos²ψ+2v₁v₂cosφcosψ. Squaring (ii): 0=v₁²sin²φ+v₂²sin²ψ–2v₁v₂sinφsinψ. Adding: u²=v₁²+v₂²+2v₁v₂cos(φ+ψ). From (iii): u²=v₁²+v₂². So 2v₁v₂cos(φ+ψ)=0. Since v₁≠0 and v₂≠0 (both balls move after collision), cos(φ+ψ)=0 → φ+ψ=90°. Result: the two equal masses always move at right angles after an oblique elastic collision. For head-on case: φ=180°, ψ=0° → φ+ψ=180° ≠ 90°? Actually for head-on with u₂=0: v₁=0, v₂=u (velocity exchange), and ball 1 stops → φ is undefined (it doesn't move). So the 90° result applies only when both balls move after collision (oblique case, not head-on). In JEE Main Work Energy Power: oblique equal-mass elastic collision → the two velocities are perpendicular: φ+ψ=90°.
What is the relationship between KE and momentum in Work, Energy and Power?
In Work, Energy and Power, KE=½mv²=p²/2m where p=mv is linear momentum. Key relations: p=√(2m×KE); KE=p²/2m. Important comparative results tested in JEE Main Work Energy Power: (1) If two bodies have same momentum (p₁=p₂=p): KE₁/KE₂=p²/2m₁ ÷ p²/2m₂ = m₂/m₁. Lighter body has more KE for same momentum. (2) If two bodies have same KE: p₁²/2m₁=p₂²/2m₂ → p₁/p₂=√(m₁/m₂). Heavier body has more momentum for same KE. (3) Work-Energy Theorem: W=ΔKE=Δ(p²/2m)=p Δp/m (if m constant). (4) Kinetic energy in terms of impulse J=Δp: if body starts from rest, J=mv → KE=J²/2m. (5) If two bodies are acted on by equal forces for equal times: they receive equal impulse → equal change in momentum → if starting from rest, equal momentum → lighter body has more KE (KE=p²/2m). (6) If two bodies are acted on by equal forces through equal distances: equal work done → equal KE gained → same KE → heavier body has more momentum. These Work Energy Power KE-momentum relations are tested in JEE Main as concept-MCQs.
What is the connection between Work, Energy and Power and the vertical circle in Laws of Motion?
In Work, Energy and Power, energy conservation directly connects to the vertical circle results from Laws of Motion Chapter 03. For a mass m on string of length l completing vertical circle: at bottom (speed vL) and top (speed vH), using conservation of mechanical energy: ½mvL²=½mvH²+mg(2l) → vH²=vL²–4gl. This is the energy conservation link. Applying Newton's 2nd law at top: TH+mg=mvH²/l → TH=mvH²/l–mg. At bottom: TL–mg=mvL²/l → TL=mvL²/l+mg. Difference: TL–TH=(mvL²/l+mg)–(mvH²/l–mg)=m(vL²–vH²)/l+2mg=m(4gl)/l+2mg=6mg. So TL–TH=6mg comes from combining Work Energy Power (energy conservation: vL²–vH²=4gl) with Newton's 2nd law. Three conditions: vL=√(5gl) is the minimum for completing the circle → from vH=√(gl) (minimum at top, TH=0) and vH²=vL²–4gl → vL=√(gl+4gl)=√(5gl). This integration of Work Energy Power with Laws of Motion is a common JEE Main multi-step question.
Related Formula Sheets — JEE Main Physics
Work, Energy and Power – JEE Main Physics Formula Sheet
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