Work, Energy & Power

The work–energy theorem is Advanced's favourite shortcut: it turns messy variable-force problems into a single scalar equation. The depth comes from spring systems, vertical-circle motion, potential-energy curves (stable/unstable equilibrium), and conservative-force fields where F = −dU/dx.

Key results

• Work–energy theorem: W(net) = ΔKE = ½mv² − ½mu², valid even for variable forces and curved paths.
• Conservative force: F(x) = −dU/dx. Equilibrium where dU/dx = 0; stable if d²U/dx² > 0 (U minimum), unstable if < 0.
• Spring PE: U = ½kx². Work by spring is path-independent.
• Vertical circle (string/track): minimum speed at top = √(gR); at bottom = √(5gR). Tension at bottom − top = 6mg.
• Power: P = F·v (instantaneous). For a vehicle at constant power, a = P/(mv).

Quantity	Formula
Work by variable force	∫F·dx (area under F–x)
Spring energy	½kx²
Top-of-loop min speed	√(gR)
Bottom-of-loop speed (min case)	√(5gR)

Exam Tricks & Tips

• 🎯 Use W–E theorem for variable forces (springs, drag) instead of integrating equations of motion — one scalar line replaces vectors.
• 🎯 Read equilibrium off the U(x) graph: minima are stable, maxima unstable; the force points toward decreasing U.
• 🎯 Vertical circle: the critical condition is that tension (or normal force) ≥ 0 at the top, giving v_top ≥ √(gR).
• 🎯 At constant power, acceleration falls as speed rises (a = P/mv), so top speed is where a → 0 (F = drag).
• 🎯 Work by static friction can be zero or positive (e.g. friction propelling a block on an accelerating truck) — do not assume friction always removes energy.
• ❌ Common mistake: using ½kx² for the work done to stretch from x1 to x2 — it is ½k(x2² − x1²), not ½k(x2 − x1)².

Expected exam pattern

1–2 questions, often a spring-block-on-incline energy problem or a vertical-circle/loop question. Potential-energy-curve conceptual questions appear as multiple-correct.

Quick recap

Reach for the work–energy theorem whenever force varies. Use F = −dU/dx and the sign of d²U/dx² for equilibrium. Memorise the vertical-circle speeds (√gR top, √5gR bottom) and P = Fv.

Cover the answer, recall, then check. 12 cards on work, energy and power for JEE Advanced.

Q1. State the work–energy theorem and its validity.
A1. W(net) = ΔKE; valid for variable forces and curved paths.

Q2. Relation between a conservative force and its potential energy.
A2. F(x) = −dU/dx.

Q3. How to identify stable equilibrium from U(x)?
A3. dU/dx = 0 and d²U/dx² > 0 (U is a minimum).

Q4. Minimum speed at the top of a vertical circle (string).
A4. √(gR), where tension just reaches zero.

Q5. Minimum speed at the bottom to complete a vertical loop.
A5. √(5gR).

Q6. Work to stretch a spring from x1 to x2.
A6. ½k(x2² − x1²), not ½k(x2 − x1)².

Q7. Instantaneous power in terms of force and velocity.
A7. P = F·v (dot product).

Q8. A car engine delivers constant power P. How does acceleration vary with speed?
A8. a = P/(mv), so acceleration decreases as speed increases.

Q9. Can friction do positive work?
A9. Yes — e.g. static friction accelerating a block riding on an accelerating surface.

Q10. Tension difference between bottom and top of a vertical circle (min case).
A10. T_bottom − T_top = 6mg.

Q11. Work done by a centripetal force in uniform circular motion?
A11. Zero — force is perpendicular to velocity.

Q12. Potential energy stored in a compressed spring of compression x.
A12. ½kx², regardless of the sign of x.

Work–energy converts hard force problems into simple scalar bookkeeping — when a problem asks "speed after moving through this messy path", energy usually beats Newton's laws. JEE Advanced tests whether you know exactly which energy theorem applies and when energy is not conserved.

Core concept: the work–energy theorem W_net = ΔKE always holds; mechanical energy is conserved only when non-conservative forces (friction, drag, applied) do no work.

Core ideas

Beginner — work and the theorem

Work W = ∫F·dr = ∫F cosθ ds. It is a scalar, can be negative (friction, gravity going up). The work–energy theorem W_net = ½mv² − ½mu² is universal — it holds even with friction, variable forces, and curved paths. That universality is why it rescues problems where forces vary.

Intermediate — conservative forces and potential energy

A force is conservative if its work is path-independent (gravity, spring, electrostatic). For these, define potential energy U with F = −dU/dx. Spring PE = ½kx²; gravitational PE (near Earth) = mgh. Mechanical energy E = KE + U is conserved when only conservative forces act. Equilibrium points are where dU/dx = 0; stable if d²U/dx² > 0 (potential well), unstable if < 0.

Advanced — power, variable forces and the potential curve

Instantaneous power P = F·v; average power = total work / time. A common JEE setup: a vehicle at constant power has a = P/(mv), so acceleration falls as speed rises, and v(t) comes from ∫mv dv = ∫P dt giving v = √(2Pt/m) from rest. Reading a U(x) graph is a favourite: turning points are where E = U(x); the particle is confined to regions where U ≤ E; force is −slope, pointing "downhill". A particle released in a well oscillates between the two turning points.

Worked example

A block of mass m is pushed against a spring (constant k), compressing it by x, then released on a rough horizontal surface (coefficient μ). How far does it travel after leaving the spring? Energy stored = ½kx². Friction dissipates μmg·d over distance d. If it leaves the spring at the natural length and then all KE is eaten by friction over further distance d: ½kx² = μmg(x + d) if friction acts during release too — read the problem carefully. Taking friction only after release: ½kx² = μmg·d ⇒ d = kx²/(2μmg). The trap is deciding over what distance friction acts.

How JEE Advanced tests this

Potential-energy-graph questions (identify stable/unstable equilibria, turning points, speed at a point); constant-power kinematics; spring–block–friction combinations; and vertical-circle problems solved by energy plus the critical-speed condition.

Exam tricks & shortcuts

• If forces vary or the path curves, reach for W_net = ΔKE before Newton's laws.
• On a U(x) graph: slope = −force, wells = stable, hills = unstable.
• Mnemonic: "Conservative? Then energy's safe." Any friction/drag/applied work breaks conservation.

Key formulas

• Work: W = F⃗·d⃗ = Fd cosθ; variable force W = ∫F dx.
• Kinetic energy: KE = ½mv² = p²/2m; work–energy theorem W_net = ΔKE.
• Potential energy: U_grav = mgh; U_spring = ½kx²; F = −dU/dx.
• Conservation: KE + U = constant (conservative forces).
• Power: P = W/t = F⃗·v⃗; 1 hp = 746 W.

Use energy conservation for problems with variable forces; power is the rate of energy transfer.

Worked Example

Problem: A block of mass 2 kg moving at 4 m/s on a rough horizontal floor (μ = 0.2) runs into and compresses a spring of stiffness k = 800 N/m fixed to a wall. Find the maximum compression of the spring. (g = 10 m/s².)

Solution:
At maximum compression the block is momentarily at rest, so all its initial kinetic energy has gone into (i) elastic potential energy of the spring and (ii) work done against friction over the compression distance x.

Energy balance:
½ m v² = ½ k x² + μ m g x

Substitute numbers:
½ (2)(4²) = ½ (800) x² + (0.2)(2)(10) x
16 = 400 x² + 4 x

Rearrange into a quadratic:
400 x² + 4 x − 16 = 0, i.e. 100 x² + x − 4 = 0.

Solve using the quadratic formula:
x = [−1 + √(1 + 1600)] / 200 = (−1 + √1601)/200 = (−1 + 40.01)/200 ≈ 0.195 m.

Answer: Maximum compression ≈ 0.195 m (about 19.5 cm).