NCERT Plus One (Class 11) Physics • Chapter 5

Work, Energy and Power

Q: Express 1 kilowatt-hour (kWh) in Joules. Which physical quantity does kWh represent?

Key (1 Mark): 1 kWh represents Energy (not Power). 1 kWh = 1000 W × 3600 s = 3.6 × 10⁶ J .

Q: Define work done by a constant force. Give one physical example each for positive, zero, and negative work.

Work is the scalar dot product of force and displacement: W = F · d = F d cos θ . • Positive: Work by gravity on a falling coconut ( θ = 0° ). • Zero: Work by centripetal force in uniform circular motion, or coolie walking level against gravity ( θ = 90° ). • Negative: Work done by friction on a sliding box ( θ = 180° ).

Q: State and prove the Work-Energy Theorem for a constant force.

Statement: The change in kinetic energy of a particle equals the net work done on it: W net = K f − K i . Proof: Using v² − u² = 2as , multiply both sides by ½m : ½mv² − ½mu² = m a s = F s = W (since F = ma along displacement s ). Hence ΔK = W .

Q: Derive an expression for the potential energy of an elastic spring stretched by displacement x, and sketch the variation of V(x) with x.

By Hooke's law, restoring force is F s = −kx . External force to slowly stretch is F ext = +kx . W = ∫ 0 x kx dx = [½kx²] 0 x = ½kx² . This stored work equals spring potential energy V(x) = ½kx² — an upward-opening parabola symmetric for both compression ( x 0 ).

Q: Prove that the total mechanical energy of a freely falling body of mass m remains constant throughout its fall.

Let body fall from height H from rest: 1. At top (height H): v = 0 ⇒ K = 0 , V = mgH ⇒ E = mgH . 2. At distance x below top (height H − x): v² = 2gx ⇒ K = ½m(2gx) = mgx ; V = mg(H − x) . Total E = mgx + mgH − mgx = mgH . 3. At ground (height 0): V = 0 , v² = 2gH ⇒ K = mgH ; Total E = mgH . Hence K + V = mgH stays conserved at every point.

Q: An elevator of total mass 1800 kg moves upward at a uniform speed of 2 m/s while a constant frictional force of 4000 N opposes motion. Calculate the minimum power delivered by the motor in Watts and Horsepower (g = 10 m/s²).

Upward force required: F = mg + f = (1800 × 10) + 4000 = 22000 N . Power P = F v = 22000 × 2 = 44000 W (44 kW) . In horsepower: 44000 / 746 ≈ 59 hp .

Q: A light body and a heavy body have the same linear momentum. Which one has greater kinetic energy? Justify.

Using K = p² / (2m) , since momentum p is identical, K ∝ 1 / m . Therefore, the light body (smaller mass) has greater kinetic energy !

Q: Show that in a perfectly inelastic one-dimensional collision where a mass m₁ moving with speed u₁ sticks to a stationary mass m₂, some kinetic energy is always lost. Find the expression for this loss.

By momentum conservation: m₁u₁ = (m₁ + m₂)v ⇒ v = m₁u₁ / (m₁ + m₂) . Initial KE: K i = ½ m₁ u₁² . Final KE: K f = ½ (m₁ + m₂) v² = ½ m₁² u₁² / (m₁ + m₂) . Loss ΔK = K i − K f = ½ m₁ u₁² [1 − m₁/(m₁ + m₂)] = ½ · [m₁m₂ / (m₁ + m₂)] · u₁² > 0 .

Dot product work, kinetic energy theorem, conservation of mechanical energy, spring energy, power & elastic/inelastic collisions.

📄 Official NCERT Textbook PDF

Topic 5.1 – 5.3: Scalar Dot Product & Work by Constant / Variable Force

Picture carrying a heavy rice bag from your scooter into your Kochi home. Walking flat across the courtyard makes your arms ache, yet climbing the staircase leaves you panting. Weirdly, Physics says walking level at steady speed does zero work against gravity, while climbing stairs does real work! Why? Because Physics only counts force pulling along the direction you actually move!

DOT PRODUCT & WORKW = F · d = F d cos θ (Scalar, SI unit: Joule)

SIGN OF WORKθ < 90°: Positive • θ = 90°: Zero • θ > 90°: Negative

VARIABLE FORCEW = ∫ F(x) dx = Area under F–x graph

🧠 Memory Trick: Think "Same-Side Push = Work!" At 90° (like a suitcase carried horizontally), force and motion are strangers, so cos 90° = 0!

💡 Simple Kid-Check: Two arrows multiplied by a dot product (A · B = AB cos θ) give a plain number (scalar) — how much one arrow helps the other!

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Scalar Dot Product Rule: For vectors A = A_xî + A_yĵ + A_zk̂ and B, A · B = AB cos θ = A_xB_x + A_yB_y + A_zB_z. Perpendicular vectors give 0!
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Constant Force Work: W = (F cos θ) d. Pushing forward gives positive work; friction pulling backward (θ = 180°) gives negative work (−Fd); centripetal force or gravity on level ground (θ = 90°) does zero work!
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Variable Force Work: When force stretches or shrinks along x, chop the path into tiny strips and sum them: W = ∫_xᵢ^x_f F(x) dx = shaded area under the F vs x curve.

Units & Conversions: 1 Joule (J) = 1 N·m = 10⁷ erg | Dimensions: [ML²T⁻²]

🧒 Super-Easy Example: A boy pulls a toy cart 5 m across a floor with 20 N rope tilted at 60°.
W = F d cos 60° = 20 × 5 × ½ = 50 J!

Watch this classic Kerala HSE 1-mark trap: A porter walks 50 m on a flat railway platform carrying a 30 kg trunk on his head. Work done against gravity is ZERO (θ = 90°, cos 90° = 0), NOT mgh or mgd!

Topic 5.4 – 5.6: Kinetic Energy & the Work-Energy Theorem

Picture cruising on a scooter along Thrissur MG Road at 30 km/h versus 60 km/h. If a stray dog dashes across and you jam the brakes with the exact same braking force, how much farther will the scooter skid at double speed? Not twice as far — FOUR times as far! That quadratic jump comes straight from Kinetic Energy!

KINETIC ENERGYK = ½mv² = p² / (2m) (Scalar, always ≥ 0)

WORK-ENERGY THEOREMW_net = ΔK = ½mv² − ½mu²

STOPPING DISTANCEF_brake · d = ½mu² ⇒ d ∝ u²!

🧠 Memory Trick: Net Work = Cash Deposit in your Speed Bank! Positive net work boosts K; negative braking work drains K!

💡 Simple Kid-Check: Anything moving holds punch-power called Kinetic Energy — heavier things and faster things pack a bigger punch (with speed squared)!

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Formula & Momentum Link: K = ½mv². Since momentum p = mv, we get the super-useful shortcut K = p² / (2m) or p = √(2mK).
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Work-Energy Theorem Proof Idea: From v² − u² = 2as, multiply by ½m: ½mv² − ½mu² = (ma)s = F_net s = W_net! Even for curved paths and changing forces, W_net = K_f − K_i holds universally!
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Meaning of Signs: If W_net > 0, body speeds up; if W_net < 0 (friction/brakes), body slows down; if W_net = 0, speed stays steady.

🧒 Super-Easy Example: A 2 kg football kicked from rest speeds up to 10 m/s. Net work done by the kick = K_f − 0 = ½ × 2 × 10² = 100 J!

Big exam favourite: If linear momentum p increases by 100% (doubles), what happens to Kinetic Energy? Since K = p²/(2m), K becomes 4 times (a 300% increase), NOT double! Also remember W_net sums every force acting (gravity + friction + push).

Topic 5.7 – 5.8: Potential Energy & Conservation of Mechanical Energy

Look up at a ripe coconut dangling 10 m high on a Palakkad palm tree. Even though it sits completely still, it holds silent trapped energy! The second the stem snaps, gravity swaps every metre of height into pure falling speed — and just before hitting the ground, all stored energy becomes kinetic energy!

CONSERVATIVE FORCEWork is path-independent & zero in closed loops (e.g. Gravity, Spring)

GRAVITY POTENTIAL ENERGYV(h) = mgh | General link: F = −dV/dx

TOTAL ENERGY LAWE = K + V = constant (when only conservative forces act!)

🧠 Memory Trick: Think "Two Energy Cups" — pouring water from the Potential cup V into the Kinetic cup K keeps the total water E = K + V unchanged!

💡 Simple Kid-Check: Sliding down a straight slide vs a curvy spiral slide of the same height h gives the exact same speed v = √(2gh) at the bottom (ignoring friction) — because gravity only cares about height dropped!

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Conservative vs Non-Conservative Forces: Gravity and spring force are conservative (work depends only on start & end heights; closed-trip work = 0). Friction is non-conservative (drains energy into heat!).
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Potential Energy Formula: Stored energy due to position: V(h) = mgh (strictly, change ΔV = −W_conservative).
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Conservation of Mechanical Energy: When only conservative forces do work, K_i + V_i = K_f + V_f. Dropping from rest at height H gives speed at height h: v = √[2g(H − h)], and at bottom v = √(2gH)!

🧒 Super-Easy Example: A 1 kg coconut falls from 20 m (take g = 10 m/s²). Initial E = mgh = 200 J. Halfway down (10 m), V = 100 J so K = 100 J; at ground, V = 0 so K = 200 J and v = √(2×10×20) = 20 m/s!

Potential energy is ONLY defined for conservative forces — there is NO such thing as "potential energy of friction"! Also remember: shifting the zero-height floor changes the value of V, never ΔV or speeds.

Topic 5.9: Potential Energy of a Spring (Hooke's Force)

Click a spring ballpoint pen or bounce on a spring sofa: push the spring inward and it pushes right back outward; pull it outward and it tugs right back inward! Either way, you are loading elastic energy into the coils — just like winding up an old mechanical clock!

HOOKE'S RESTORING FORCEF_s = −k x (always opposes displacement x!)

SPRING ENERGYV(x) = ½ k x² (Symmetric U-parabola, always ≥ 0!)

MAX SPEED AT MEAN½ m v_max² = ½ k x_m² ⇒ v_max = x_m √(k/m)

🧠 Memory Trick: Area of Triangle = ½ × base (x) × height (kx) = ½kx²! Works identically for BOTH stretching & squashing!

💡 Simple Kid-Check: A stiff shock-absorber spring on a bike has a huge spring constant k (measured in N/m), while a soft pen spring has a tiny k!

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Hooke's Law: Spring restoring force is proportional and opposite to stretch/compression x from natural length: F_s = −kx.
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Integrating / Triangle Area: Since force grows linearly from 0 to kx, work stored by stretching (or compressing!) by x is the triangle area: V(x) = ½kx².
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Oscillation Energy Trade: At extremes x = ±x_m, K = 0 and V = ½kx_m². Passing equilibrium x = 0, V = 0 and K = ½mv_max² = ½kx_m²!

🧒 Super-Easy Example: A spring with k = 200 N/m is compressed by 0.1 m. Stored potential energy V = ½ × 200 × (0.1)² = 1.0 J. If it launches a 0.5 kg block, speed v = √(2×1.0/0.5) = 2 m/s!

Watch two classic mistakes: (1) Compressing a spring (x = −0.1 m) STILL stores POSITIVE potential energy +½kx² because (−x)² = +x²! (2) Work to stretch from x₁ to x₂ is ½k(x₂² − x₁²), NEVER ½k(x₂ − x₁)²!

Topic 5.10: Power & Commercial Unit of Energy (kWh)

Suppose two electric motors pump water to fill the exact same rooftop tank in Alappuzha. Both lift the same weight up the same height, so both do identical work! Yet a 1 hp motor finishes in 10 minutes while a 0.5 hp motor takes 20 minutes — because the faster motor delivers double the Power!

AVERAGE & INST. POWERP_av = W / t | P = dW/dt = F · v = Fv cos θ

SI & ENGINE UNITS1 Watt = 1 J/s | 1 hp = 746 W

ELECTRICITY BILL UNIT1 kWh = 1000 W × 3600 s = 3.6 × 10⁶ J (ENERGY!)

🧠 Memory Trick: Work = Total Homework Done; Power = Homework Speed per Hour! And 1 kWh = 3.6 MJ is your KSEB electricity "1 unit"!

💡 Simple Kid-Check: Running up stairs and walking slowly up the same stairs do the exact same Work (mgh), but running takes less time so your Power is much higher!

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Definition: Power is the time rate of doing work: P_av = W / t. Instantaneous power uses the dot product: P = F · v = Fv cos θ. Like work and energy, Power is a scalar!
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Units: SI unit is Watt (W) = 1 J/s (dimensions [ML²T⁻³]). Mechanical motors use Horsepower: 1 hp = 746 W.
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Commercial Electrical Energy (kWh): Running a 1000 W (1 kW) iron box for 1 hour (3600 s) consumes 1 kilowatt-hour (1 kWh) = 3.6 × 10⁶ J!

🧒 Super-Easy Example: An elevator lifts a 1800 kg load at constant 2 m/s against 4000 N friction (g = 10 m/s²).
Motor force F = mg + f = 18000 + 4000 = 22000 N → Power P = Fv = 22000 × 2 = 44000 W = 44 kW!

Guaranteed 1-mark DHSE trap: kilowatt-hour (kWh) is a unit of ENERGY, NOT POWER! Why? Because kW (power) multiplied by hour (time) equals Energy (3.6 × 10⁶ J). Don't fall for "kW means power so kWh means power"!

Topic 5.11: Elastic & Inelastic Collisions in 1D and 2D

Set up a carrom board on a rainy afternoon in Kottayam! Flick your striker head-on into a single red queen coin sitting still: if it were an ideal elastic collision between equal masses, your striker would freeze dead on the spot while the coin rockets off at the striker's exact speed!

ALWAYS CONSERVEDLinear Momentum (p_i = p_f) & Total Energy in ALL collisions!

ELASTIC vs INELASTICElastic (e = 1): K conserved! Inelastic (e < 1): K lost to heat! Sticky (e = 0): max loss!

2D EQUAL-MASS TRICKGlancing elastic hit (m₁=m₂, u₂=0) → fly apart at 90° (θ₁ + θ₂ = 90°)!

🧠 Memory Trick: Equal Masses Swap Velocities! In 1D elastic collisions, if m₁ = m₂, ball 1 takes ball 2's speed and ball 2 takes ball 1's speed!

💡 Simple Kid-Check: Drop a superball (nearly elastic, bounces high, keeps kinetic energy) vs a lump of wet clay (perfectly inelastic, splats and sticks on impact, loses kinetic energy)!

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1D Elastic Formulas (u₂ = 0): Both p and K stay conserved, so relative speed of separation equals speed of approach (v₂ − v₁ = u₁ − u₂):
v₁ = m₁ − m₂m₁ + m₂ u₁ | v₂ = 2m₁m₁ + m₂ u₁.
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3 Special 1D Elastic Cases: (a) Equal mass (m₁ = m₂) → v₁ = 0, v₂ = u₁ (velocities swap!). (b) Heavy truck hits tiny marble (m₁ ≫ m₂) → truck keeps going (v₁ ≈ u₁), marble rockets at 2u₁! (c) Light ball hits massive wall (m₁ ≪ m₂) → ball bounces back at −u₁!
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Perfectly Inelastic (Sticky, e = 0): Combined speed v = m₁u₁ + m₂u₂m₁ + m₂. Loss in kinetic energy (when u₂ = 0): ΔK = 12 m₁m₂m₁ + m₂ u₁²!

Classic exam trick: In an inelastic collision, Total Energy IS conserved (converted to heat/sound/deformation), and Momentum IS conserved — ONLY Kinetic Energy is NOT conserved! Also remember coefficient of restitution e = (v₂ − v₁)/(u₁ − u₂) is 1 for elastic and 0 for sticky collisions.

Solved Previous Year Questions (Kerala DHSE & NCERT)

Q1 (Kerala HSE March • 1 Marks): Express 1 kilowatt-hour (kWh) in Joules. Which physical quantity does kWh represent?

Key (1 Mark): 1 kWh represents Energy (not Power).
1 kWh = 1000 W × 3600 s = 3.6 × 10⁶ J.

Q2 (Kerala HSE Say • 2 Marks): Define work done by a constant force. Give one physical example each for positive, zero, and negative work.

Work is the scalar dot product of force and displacement: W = F · d = F d cos θ.
• Positive: Work by gravity on a falling coconut (θ = 0°).
• Zero: Work by centripetal force in uniform circular motion, or coolie walking level against gravity (θ = 90°).
• Negative: Work done by friction on a sliding box (θ = 180°).

Q3 (Kerala HSE March • 3 Marks): State and prove the Work-Energy Theorem for a constant force.

Statement: The change in kinetic energy of a particle equals the net work done on it: W_net = K_f − K_i.
Proof: Using v² − u² = 2as, multiply both sides by ½m:
½mv² − ½mu² = m a s = F s = W (since F = ma along displacement s). Hence ΔK = W.

Q4 (Kerala HSE Model • 3 Marks): Derive an expression for the potential energy of an elastic spring stretched by displacement x, and sketch the variation of V(x) with x.

By Hooke's law, restoring force is F_s = −kx. External force to slowly stretch is F_ext = +kx.
W = ∫₀^x kx dx = [½kx²]₀^x = ½kx².
This stored work equals spring potential energy V(x) = ½kx² — an upward-opening parabola symmetric for both compression (x < 0) and extension (x > 0).

Q5 (Kerala HSE March • 4 Marks): Prove that the total mechanical energy of a freely falling body of mass m remains constant throughout its fall.

Let body fall from height H from rest:
1. At top (height H): v = 0 ⇒ K = 0, V = mgH ⇒ E = mgH.
2. At distance x below top (height H − x): v² = 2gx ⇒ K = ½m(2gx) = mgx; V = mg(H − x). Total E = mgx + mgH − mgx = mgH.
3. At ground (height 0): V = 0, v² = 2gH ⇒ K = mgH; Total E = mgH.
Hence K + V = mgH stays conserved at every point.

Q6 (Kerala HSE Say • 2 Marks): An elevator of total mass 1800 kg moves upward at a uniform speed of 2 m/s while a constant frictional force of 4000 N opposes motion. Calculate the minimum power delivered by the motor in Watts and Horsepower (g = 10 m/s²).

Upward force required: F = mg + f = (1800 × 10) + 4000 = 22000 N.
Power P = F v = 22000 × 2 = 44000 W (44 kW).
In horsepower: 44000 / 746 ≈ 59 hp.

Q7 (Kerala HSE March • 2 Marks): A light body and a heavy body have the same linear momentum. Which one has greater kinetic energy? Justify.

Using K = p² / (2m), since momentum p is identical, K ∝ 1 / m.
Therefore, the light body (smaller mass) has greater kinetic energy!

Q8 (NCERT / HSE Model • 4 Marks): Show that in a perfectly inelastic one-dimensional collision where a mass m₁ moving with speed u₁ sticks to a stationary mass m₂, some kinetic energy is always lost. Find the expression for this loss.

By momentum conservation: m₁u₁ = (m₁ + m₂)v ⇒ v = m₁u₁ / (m₁ + m₂).
Initial KE: K_i = ½ m₁ u₁².
Final KE: K_f = ½ (m₁ + m₂) v² = ½ m₁² u₁² / (m₁ + m₂).
Loss ΔK = K_i − K_f = ½ m₁ u₁² [1 − m₁/(m₁ + m₂)] = ½ · [m₁m₂ / (m₁ + m₂)] · u₁² > 0.