← Physics (312) · Class 12

L-6: Work, Energy and Power

Physics — Class 12 · NIOS Code 312 · Module 1 · Source: 312_Physics_Eng_Lesson6.pdf

From Force to Energy — Why This Lesson Matters

Newton's laws tell you how force changes motion. This lesson answers the next question: what does force accomplish over distance and time? That accomplishment is work. The capacity to do work is energy. The rate of doing work is power.

From muscular effort → animals → machines, civilization advanced by increasing power output. Modern society critically depends on usable energy — electrical, thermal, chemical, nuclear. Work and energy are therefore inseparable in physics and in daily life.

NIOS objectives: define and calculate work; state work–energy theorem; define power; calculate work by gravity; explain energy; derive gravitational and elastic potential energy; apply conservation of energy; apply momentum and energy laws in elastic collisions.

6.1 Work

In everyday language, "work" can mean mental effort. In physics, work has a precise definition: when a constant force causes displacement, work is the product of the force component along the displacement and the magnitude of displacement.

Fig 6.1 — Force at Angle θ displacement d → F θ W = Fd cos θ
Fig 6.1 — Work = force component along displacement × d

If force F acts at angle θ to displacement d:

W = Fd cos θ
W = Work (joule, J)
F = Magnitude of force (N)
d = Displacement (m)
θ = Angle between F and d
Vector form: W = F · d (dot product)

Key properties: If d = 0, W = 0 — pushing a wall does no work. Force and displacement are vectors, but work is a scalar (dot product).

1 J = 1 N·m
One joule = work by 1 N through 1 m.
Dimensional formula: [ML²T⁻²]
Same dimensions as energy.

Example: F = 6 N at 60° to horizontal, d = 2 m → W = 6×2×cos 60° = 6 J.

Example: Lifting 5 kg through 4 m: F = mg = 49 N → W = 49×4 = 196 J.

Fig 6.2 — Positive & Negative Work (a) Accelerator — W > 0 (b) Brakes — W < 0
Fig 6.2 — Same-direction force speeds up car; opposite force slows it

6.1.1 Positive and Negative Work

              WORK SIGN DECISION TREE
              =======================
                        |
                        v
              [Angle θ between F and d?]
                        |
         +--------------+--------------+
         |              |              |
    θ = 0°          θ = 90°        θ = 180°
    W = +Fd         W = 0          W = −Fd
    (accelerator)   (carry bag      (brakes)
                     horizontally)
         |              |              |
    Speed up        No energy        Speed down
    (positive)      transfer         (negative)
  • Positive work (0° ≤ θ < 90°): Force has component along displacement. Accelerator increases car speed.
  • Negative work (90° < θ ≤ 180°): Force opposes motion. Brakes reduce speed.
  • Zero work (θ = 90°): Force perpendicular to displacement — centripetal force in circular motion.
Fig 6.3 — Work by Gravity (a) Lifting — W_g = −mgh h↑ mg (b) Lowering — W_g = +mgh
Fig 6.3 — Gravity does negative work when mass is lifted, positive when lowered

6.1.2 Work Done by Gravity

Lifting mass m through height h: gravity acts down, displacement up (θ = 180°).

W_gravity (lift) = −mgh
Work by gravity when object rises: negative.
Work by person lifting: +mgh.

Lowering mass: gravity and displacement same direction (θ = 0°).

W_gravity (lower) = +mgh
Work by gravity when object falls: positive.
Work by person lowering: −mgh.
Assumed: motion without acceleration.

6.2 Work Done by a Variable Force

When force F(x) varies with position, divide displacement into small intervals Δx. Over each interval, treat F as constant:

ΔW = F(x) Δx
Small strip area under F–x curve.
Total work = sum of all strips.
Fig 6.4 & 6.6 — Variable Force & Spring x F shaded area = W ½kx² triangle
Fig 6.4/6.6 — Work = area under F–x curve; spring work = ½kx²

As Δx → 0, the sum becomes an integral — work equals the area under the F versus x graph between initial position xᵢ and final position x_f.

6.2.1 Work Done by a Spring (Hooke's Law)

For a spring obeying Hooke's law (small x):

F_s = −kx
k = Spring constant (N·m⁻¹)
x = Compression or extension from equilibrium
Restoring force opposite to displacement.

Compressing/extending: external force does positive work; spring force does negative work of equal magnitude.

W_ext = ½ kx²
Work by external agent to compress/ stretch spring by x.
Also = area of triangle under F–x graph (base x, height kx).
Stored as elastic potential energy.

Activity: Hang mass m, extension s at equilibrium → mg = ks → k = mg/s.

Example 6.4: m = 2 kg, k = 100 N·m⁻¹, x = 10 cm = 0.1 m → W = ½×100×(0.1)² = 0.5 J.

6.3 Power

Work alone does not tell how fast energy is transferred. Power is the rate of doing work.

P = ΔW / Δt
P = Average power (watt, W)
ΔW = Work done (J)
Δt = Time interval (s)
Instantaneous: P = dW/dt
1 W = 1 J/s
Power of 1 W = 1 joule of work per second.
1 kW = 10³ W; 1 MW = 10⁶ W
Dimensions: [ML²T⁻³]
1 hp = 746 W
Horsepower (British unit).
James Watt (1736–1819) improved steam engines.
1 kWh = 3.6 × 10⁶ J
Electrical energy unit.
1 kWh = 1 kW × 3600 s = 3.6 MJ
Domestic billing: 1 Unit = 1 kWh

Example: 100 kg lifted 8 m in 10 s → W = mgh = 7840 J → P = 784 W.

6.4 Work and Kinetic Energy

Objects in motion can do work before stopping — they possess kinetic energy (energy of motion). Consider mass m accelerated by constant F = ma from v₁ to v₂ over distance s:

K = ½ mv²
K = Kinetic energy (J)
m = Mass (kg)
v = Speed (m·s⁻¹)
Scalar; depends on v² — doubling speed quadruples K.

Using v₂² = v₁² + 2as and F = ma:

W = K₂ − K₁ = ΔK
Work–Energy Theorem:
Work done by the resultant of all forces = change in kinetic energy.
W_net = ½mv₂² − ½mv₁²
         WORK–ENERGY THEOREM — EXAM STEPS
         =================================
                    |
                    v
         [1] Find initial speed v₁, final v₂ (or use F=ma)
                    |
                    v
         [2] K₁ = ½mv₁² ,  K₂ = ½mv₂²
                    |
                    v
         [3] W_net = K₂ − K₁
                    |
                    v
         [4] Cross-check: W = F·s if force constant
             (Example 6.6: 30 N, 14 m → 420 J = 500−80 J)

Example 6.6: m = 10 kg, v₁ = 4 m/s, F = 30 N for 2 s → a = 3 m/s², v₂ = 10 m/s, s = 14 m, W = 420 J. K₁ = 80 J, K₂ = 500 J, ΔK = 420 J ✓

Kinetic energy cannot be negative (v² ≥ 0). If speed doubles, K becomes 4×; if mass halves, K halves.

6.5 Potential Energy

Energy due to position in a field is potential energy. Familiar case: gravitational potential energy in Earth's field.

6.5.1 Gravitational Potential Energy

Lifting mass from h₁ to h₂ (Δh = h₂ − h₁) against gravity:

U = mgh
U = Gravitational PE (J)
m = Mass (kg)
g ≈ 9.8 m·s⁻²
h = Vertical height above reference
Zero level is arbitrary — only Δh matters.

Example 6.7: Truck + load 10⁵ kg up 700 m in 1 h → W = 6.86×10⁸ J → P_avg ≈ 1.91×10⁵ W ≈ 256 hp.

Example 6.8 (Hydro): 10⁶ kg water falls 51 m in 1 s → W = 500 MJ → P = 500 MW (ideal, no friction loss).

6.5.2 Elastic Potential Energy

U_s = ½ kx²
Energy stored in compressed/stretched spring.
Released spring converts U_s → kinetic energy of attached mass.
Fig 6.10 — Energy Conservation (Free Fall) h — U=mgh K=0 P: K+U=mgh K=max
Fig 6.10 — Total mechanical energy constant during free fall

6.5.3 Conservation of Energy

Law of Conservation of Energy: Total energy of an isolated system remains constant. Energy transforms between forms but is neither created nor destroyed. The universe is an isolated system — total energy always constant.

Mechanical energy = K + U. For conservative forces only:

K₁ + U₁ = K₂ + U₂
Mechanical energy conserved when only conservative forces do work.
Free fall: mgh = ½mv² + mgh₂ at any point.

Free fall: Object at height h has U = mgh, K = 0. At point P after falling h₁: K = mgh₁, U = mg(h − h₁), total = mgh — unchanged.

Spring–mass oscillator: At x = 0: K = ½mv², U_s = 0. At max compression x_m: K = 0, U_s = ½kx_m². Equating: ½mv² = ½kx_m².

Example 6.9: m = 0.5 kg slides down 2.5 m → U_A = 12.25 J → at B (ground): ½mv² = 12.25 → v = 7 m/s.

6.5.4 Conservative and Non-Conservative Forces

    CONSERVATIVE vs NON-CONSERVATIVE FORCES
    =======================================
              CONSERVATIVE              NON-CONSERVATIVE
              (gravity, spring,         (friction, drag)
               electrostatic)
                    |                          |
         Work independent of path      Work depends on path
         Closed loop: W = 0            Closed loop: W ≠ 0
         Mechanical energy conserved   ME → heat (thermal)
         W = −ΔU                       Total ME not conserved

Conservative: Work between A and B depends only on vertical separation (gravity), not path. W_AB along path 1 = −W_BA along path 2 → round trip work = 0.

Non-conservative (friction): Block with speed v on rough surface stops at B — kinetic energy becomes thermal energy. Longer path → more energy dissipated. When non-conservative forces act, total mechanical energy is not conserved (though total energy including heat still is).

Inclined plane with friction: If KE at bottom is less than PE lost, difference = work done against friction.

Fig 6.19 — Head-on Elastic Collision Before A B After (identical balls) A Velocities exchange: v_Af = v_Bi , v_Bf = v_Ai
Fig 6.19 — Elastic head-on collision: identical balls swap velocities

6.6 Elastic and Inelastic Collisions

When two bodies interact in a closed system (no external force), it is a collision. Momentum is always conserved in all collisions. Kinetic energy may or may not be conserved.

  • Perfectly elastic: Conservative interaction forces → total KE before = total KE after.
  • Perfectly inelastic: Bodies stick together and move as one unit (bullet embedded in target).
         HEAD-ON ELASTIC COLLISION — SPECIAL CASES
         =========================================
    CASE I: m_A = m_B (identical balls)
         → velocities EXCHANGE: v_Af = v_Bi , v_Bf = v_Ai
         If B at rest: A stops, B moves with v_Ai
         (Neutron moderation in reactors)

    CASE II: m_B >> m_A , B at rest (ball hits wall)
         → v_Af ≈ −v_Ai (light ball rebounds)
         → v_Bf ≈ 0 (heavy object barely moves)
         (α-particle scattering from heavy nucleus)
m_A v_Ai + m_B v_Bi = m_A v_Af + m_B v_Bf
Momentum conservation (all collisions).
½m_A v_Ai² + ½m_B v_Bi² = ½m_A v_Af² + ½m_B v_Bf²
Kinetic energy conservation (elastic only).
Also: (v_Bf − v_Af) = −(v_Bi − v_Ai) for elastic head-on.

Inelastic example: Bullet (m, v) embeds in block (M, at rest). Common velocity v' = mv/(m+M). KE before > KE after — energy lost to deformation and heat.

Quick Revision — What You Have Learnt

  • Work: W = F·d = Fd cos θ; scalar; joule; area under F–x graph for variable F.
  • Spring: W_ext = ½kx²; k in N·m⁻¹; restoring work = −½kx².
  • Power: P = W/t; watt; 1 hp = 746 W; 1 kWh = 3.6×10⁶ J.
  • KE: K = ½mv²; work–energy theorem W = ΔK.
  • PE: U_g = mgh; U_s = ½kx²; zero reference arbitrary.
  • Conservation: Isolated system → total energy constant; ME conserved if only conservative forces.
  • Collisions: Momentum always conserved; KE conserved only in elastic collisions.
20 cards · click any card to flip
Work (physics)
W = F·d = Fd cos θ. Product of force component along displacement and displacement. Scalar. SI unit: joule (J).
Zero work
No displacement (d = 0), OR force ⊥ displacement (θ = 90°). Example: pushing a wall, centripetal force in uniform circular motion.
Positive work
Force has component along displacement (0° ≤ θ < 90°). W = +Fd when θ = 0°. Speed often increases.
Negative work
Force opposes displacement (90° < θ ≤ 180°). W = −Fd when θ = 180°. Brakes do negative work on a car.
Joule
1 J = 1 N·m. Work done by 1 N force through 1 m displacement. Dimensional formula [ML²T⁻²].
kWh
1 kWh = 1 kW × 3600 s = 3.6 × 10⁶ J. Used for electrical energy billing (1 Unit = 1 kWh).
Variable force work
W = area under F–x graph. W = Σ F(x)Δx → ∫ F(x) dx from xᵢ to x_f.
Hooke's law
|F_s| = kx (small displacements). Restoring force F_s = −kx. Spring constant k in N·m⁻¹.
Work on a spring
External force: W_ext = +½kx². Spring restoring force: W_spring = −½kx².
Power
Rate of doing work: P = ΔW/Δt. Instantaneous P = dW/dt. SI unit: watt (W) = J/s.
Horsepower
1 hp = 746 W (British). James Watt introduced hp; SI unit watt named after him.
Kinetic energy
K = ½mv². Energy of motion. Scalar. Depends on mass and speed squared.
Work–energy theorem
Net work by all forces = change in kinetic energy: W = K_f − K_i = ΔK.
Gravitational PE
U = mgh (near Earth, g constant). Stored by position. Zero level is arbitrary; only Δh matters.
Elastic PE
U_s = ½kx² stored in compressed/stretched spring. Converts to KE when spring released.
Conservation of energy
Total energy of an isolated system remains constant. Energy transforms but is neither created nor destroyed.
Conservative force
Work independent of path; closed-loop work = 0. Examples: gravity, elastic force, electrostatic force.
Non-conservative force
Work depends on path. Friction converts mechanical energy to heat. Total mechanical energy not conserved.
Elastic collision
Momentum AND kinetic energy both conserved. Conservative interaction forces. Head-on: velocities exchange if m_A = m_B.
Inelastic collision
Momentum conserved; KE not conserved. Perfectly inelastic: bodies stick and move together after collision.

Q1. The SI unit of work is:

Q2. A force acts on a body but the body does not move. The work done is:

Q3. Work done by a force is zero when the angle between force and displacement is:

Q4. According to the work–energy theorem, work done by net force equals:

Q5. Kinetic energy of a body of mass m moving with speed v is:

Q6. The SI unit of power is:

Q7. Gravitational potential energy near Earth's surface is:

Q8. Energy stored in a spring compressed by distance x is:

Q9. Which is a conservative force?

Q10. In a perfectly elastic collision:

W = Fd cos θ
1 J = 1 N·m
W_gravity (lift) = −mgh
W_gravity (lower) = +mgh
ΔW = F(x) Δx
F_s = −kx
W_ext = ½ kx²
P = ΔW / Δt
1 W = 1 J/s
1 hp = 746 W
1 kWh = 3.6 × 10⁶ J
K = ½ mv²
W = K₂ − K₁ = ΔK
U = mgh
U_s = ½ kx²
K₁ + U₁ = K₂ + U₂
m_A v_Ai + m_B v_Bi = m_A v_Af + m_B v_Bf
½m_A v_Ai² + ½m_B v_Bi² = ½m_A v_Af² + ½m_B v_Bf²

1. Formulas & Definitions

Full Ch 6 study guide — work, energy, power, conservation, and collisions.

W = Fd cos θ

Definition: Work by constant force equals force component along displacement times distance.

Derivation

W = F·d (dot product). Only the parallel component F cos θ does work.

Variables

W (J) · F (N) · d (m) · θ = angle between F and d

Why it works

Pushing at an angle — only the along-path part transfers energy. Perpendicular force (θ=90°) does zero work.

Historical context

Classical mechanics formalised work as ∫F·dr; scalar product links vectors to energy transfer.

Deep understanding

Work is a scalar. d=0 means W=0 even with huge force (pushing a wall).

2. Diagrams & Visuals

W = Fd cos θ d → F θ between F and d

Color-coded visual · step-by-step breakdown below

  1. Draw F and displacement d
  2. Find angle θ between them
  3. Compute F cos θ (component along d)
  4. W = (F cos θ) × d

3. Solved Examples

Basic

Q: 10 N along motion, d=3 m.

Solution: θ=0, cosθ=1

Answer: W = 30 J

Intermediate

Q: F=6 N at 60°, d=2 m.

Solution: W=6×2×cos60°

Answer: W = 6 J (Fig 6.1)

Advanced

Q: Force perpendicular to 5 m displacement.

Solution: θ=90°, cosθ=0

Answer: W = 0

Exam

Q: MCQ Q3: zero work angle?

Solution: θ = 90°

Answer: Option C

1 J = 1 N·m

Definition: One joule is work done by 1 newton through 1 metre.

Derivation

SI definition linking force and energy dimensions [ML²T⁻²].

Variables

J = joule · N = newton · m = metre

Why it works

Same unit for work AND energy — they are the same physical quantity in different contexts.

Historical context

Named after James Prescott Joule who demonstrated mechanical equivalent of heat.

Deep understanding

1 kJ = 10³ J. Energy billing uses kWh (different unit for electrical energy).

2. Diagrams & Visuals

1 J = 1 N × 1 m

Color-coded visual · step-by-step breakdown below

  1. Identify force in N
  2. Identify displacement in m
  3. Multiply for work in joules

3. Solved Examples

Basic

Q: 1 N pushes box 1 m.

Solution: W=1×1

Answer: 1 J

Intermediate

Q: 49 N lifts 5 kg 4 m.

Solution: W=49×4

Answer: 196 J

Advanced

Q: Convert 500 J to N·m.

Solution: Same dimension

Answer: 500 N·m

Exam

Q: MCQ Q1: SI unit of work?

Solution: Joule

Answer: Option B

W_gravity (lift) = −mgh

Definition: Work by gravity when mass is lifted upward is negative.

Derivation

F_g = mg down, displacement up → θ=180°, cosθ=−1, W = −mg×h.

Variables

m (kg) · g ≈ 9.8 m/s² · h (m)

Why it works

Gravity removes energy from the lifted object-Earth system; person lifting does +mgh.

Historical context

Gravitational work independence from path led to potential energy concept.

Deep understanding

Person does +mgh; gravity does −mgh; net on system depends on who you track.

2. Diagrams & Visuals

mg↓ h↑ W_g = −mgh

Color-coded visual · step-by-step breakdown below

  1. Mass m lifted through height h
  2. Gravity force mg opposite displacement
  3. W_g = −mgh
  4. Person/agent does +mgh

3. Solved Examples

Basic

Q: 2 kg lifted 1 m.

Solution: W_g=−2×9.8×1

Answer: −19.6 J

Intermediate

Q: 10 kg to shelf 2 m high.

Solution: W_g=−196 J

Answer: Person does +196 J

Advanced

Q: Why negative?

Solution: Force opposite displacement

Answer: Energy removed from system

Exam

Q: Lowering vs lifting: sign of W_g?

Solution: Lift: negative; lower: +mgh

Answer: Fig 6.3

W_gravity (lower) = +mgh

Definition: Work by gravity when mass is lowered is positive.

Derivation

Displacement and mg same direction → θ=0°, W = +mgh.

Variables

m · g · h as above

Why it works

Gravity gains energy from falling — used in hydro power (water falling height h).

Historical context

Water wheels and modern hydroelectric plants exploit positive gravitational work.

Deep understanding

Person lowering gently does −mgh (negative work) to control descent.

2. Diagrams & Visuals

W_g = +mgh

Color-coded visual · step-by-step breakdown below

  1. Object moves down distance h
  2. mg and displacement same way
  3. W_g = +mgh

3. Solved Examples

Basic

Q: 3 kg falls 2 m (no friction).

Solution: W_g=3×9.8×2

Answer: +58.8 J

Intermediate

Q: Ex 6.8: 10⁶ kg water falls 51 m in 1 s.

Solution: W=mgΔh

Answer: ≈ 500 MJ

Advanced

Q: Compare lift vs lower W_g.

Solution: Sign flips

Answer: + vs −

Exam

Q: Hydro power uses which?

Solution: Positive W_g as water falls

Answer: Ex 6.8

ΔW = F(x) Δx

Definition: Small work strip when force varies with position.

Derivation

Sum all strips; as Δx→0, W = ∫F(x)dx = area under F–x graph.

Variables

F(x) = force at x · Δx = small displacement

Why it works

Real forces (springs, engines) change with position — cannot use constant F formula alone.

Historical context

Integral calculus (Newton, Leibniz) made variable-force work precise.

Deep understanding

Total work = shaded area between x_i and x_f on F vs x plot (Fig 6.4).

2. Diagrams & Visuals

area = W

Color-coded visual · step-by-step breakdown below

  1. Plot F vs x
  2. Divide into strips Δx
  3. Sum F(x)Δx
  4. Limit → integral/area

3. Solved Examples

Basic

Q: Constant F=5 N, Δx=2 m strip.

Solution: ΔW=10 J

Answer: 10 J

Intermediate

Q: Spring: F grows with x.

Solution: Use strips then integral

Answer: → ½kx²

Advanced

Q: Why area under graph?

Solution: Each strip is work

Answer: Calculus limit

Exam

Q: Variable force work equals?

Solution: Area under F–x curve

Answer: Sec 6.2

F_s = −kx

Definition: Spring restoring force (Hooke's law) opposite to displacement.

Derivation

Empirical for small x; k = spring constant from F vs x slope.

Variables

F_s (N) · k (N/m) · x = extension/compression (m)

Why it works

Minus sign: spring always pulls back toward equilibrium — restoring force.

Historical context

Robert Hooke (1678): ut tensio sic vis — extension proportional to force.

Deep understanding

At equilibrium hanging mass: mg = ks gives k = mg/s (activity in notes).

2. Diagrams & Visuals

F=-kx

Color-coded visual · step-by-step breakdown below

  1. Measure extension x from natural length
  2. Identify k
  3. F_s = −kx (direction toward equilibrium)

3. Solved Examples

Basic

Q: k=200 N/m, x=0.05 m.

Solution: F=−200×0.05

Answer: −10 N

Intermediate

Q: 2 kg mass, extension 0.02 m at equilibrium.

Solution: k=mg/s=9.8/0.02

Answer: k = 490 N/m

Advanced

Q: Why negative sign?

Solution: Restoring — opposite x

Answer: Always toward x=0

Exam

Q: Unit of k?

Solution: N·m⁻¹

Answer: Spring constant

W_ext = ½ kx²

Definition: Work by external agent to stretch/compress spring by x.

Derivation

Area of triangle under F=kx graph: ½ × base × height = ½x·kx = ½kx².

Variables

W_ext (J) · k (N/m) · x (m)

Why it works

Energy stored as elastic PE; spring force does −½kx² (equal magnitude, opposite sign).

Historical context

Elastic potential energy explains clocks, toys, vehicle suspension, trampolines.

Deep understanding

Stored as U_s = ½kx² — converts to kinetic when released.

2. Diagrams & Visuals

½kx² triangle

Color-coded visual · step-by-step breakdown below

  1. Find k and compression x
  2. W_ext = ½kx²
  3. Equals elastic PE stored

3. Solved Examples

Basic

Q: k=100, x=0.1 m.

Solution: W=½×100×0.01

Answer: 0.5 J (Ex 6.4)

Intermediate

Q: k=50 N/m, x=0.2 m.

Solution: W=½×50×0.04

Answer: 1 J

Advanced

Q: Double x — how does W change?

Solution: W ∝ x²

Answer: 4× work

Exam

Q: MCQ Q8: spring energy?

Solution: ½kx²

Answer: Option C

P = ΔW / Δt

Definition: Average power is rate of doing work.

Derivation

P = dW/dt for instantaneous power.

Variables

P (W) · ΔW (J) · Δt (s)

Why it works

Same work done faster needs higher power — why motors are rated in kW/hp.

Historical context

James Watt coined horsepower comparing steam engines to draft horses.

Deep understanding

Instantaneous: P = F·v when force parallel to velocity (constant F).

2. Diagrams & Visuals

P = ΔW / Δt more work per second → more power

Color-coded visual · step-by-step breakdown below

  1. Calculate total work ΔW
  2. Measure time Δt
  3. P = ΔW/Δt

3. Solved Examples

Basic

Q: 100 J in 5 s.

Solution: P=20

Answer: 20 W

Intermediate

Q: 7840 J in 10 s (100 kg, 8 m).

Solution: P=784

Answer: 784 W

Advanced

Q: Truck 10⁵ kg up 700 m in 1 h.

Solution: W=6.86×10⁸ J

Answer: P ≈ 191 kW (Ex 6.7)

Exam

Q: MCQ Q6: SI unit of power?

Solution: Watt

Answer: Option C

1 W = 1 J/s

Definition: One watt equals one joule of work per second.

Derivation

From P = ΔW/Δt definition with SI base units.

Variables

W (watt) · J/s

Why it works

Links energy transfer rate to everyday appliance ratings (60 W bulb, 2000 W heater).

Historical context

Watt unit named after James Watt; standardised in SI.

Deep understanding

Dimensions [ML²T⁻³]. 1 kW = 10³ W; 1 MW = 10⁶ W.

2. Diagrams & Visuals

1 W = 1 J / 1 s

Color-coded visual · step-by-step breakdown below

  1. Express work in joules
  2. Divide by time in seconds
  3. Result in watts

3. Solved Examples

Basic

Q: 1 J each second.

Solution: P=1 J/s

Answer: 1 W

Intermediate

Q: 3600 J in 1 hour.

Solution: P=1 J/s

Answer: 1 W average

Advanced

Q: 2 kW device for 30 min. Energy?

Solution: W=P×t

Answer: 3600 kJ = 1 kWh

Exam

Q: Difference J vs W?

Solution: J=energy, W=power

Answer: Don't confuse units

1 hp = 746 W

Definition: Horsepower — British power unit.

Derivation

Historical comparison: horse lifting 550 lb·ft/s ≈ 746 W.

Variables

hp · W

Why it works

Still used for car engines; convert to watts for SI calculations.

Historical context

James Watt (1736–1819) marketed steam engines using horsepower.

Deep understanding

Ex 6.7: 1.91×10⁵ W ≈ 256 hp for truck climbing.

2. Diagrams & Visuals

1 hp ≈ 746 W

Color-coded visual · step-by-step breakdown below

  1. Given power in hp
  2. Multiply by 746 for watts
  3. Or divide watts by 746 for hp

3. Solved Examples

Basic

Q: 1 hp in watts?

Solution: ×746

Answer: 746 W

Intermediate

Q: 1500 W in hp?

Solution: 1500/746

Answer: ≈ 2 hp

Advanced

Q: 256 hp to kW?

Solution: 256×746/1000

Answer: ≈ 191 kW

Exam

Q: Who is watt/hp named after?

Solution: James Watt

Answer: Steam engine pioneer

1 kWh = 3.6 × 10⁶ J

Definition: Kilowatt-hour — electrical energy unit on bills.

Derivation

1 kWh = 1 kW × 3600 s = 1000 × 3600 J = 3.6 MJ.

Variables

kWh · J

Why it works

Domestic meter reads kWh (1 Unit = 1 kWh on Indian bills).

Historical context

Utility companies use kWh because power×time is natural for billing.

Deep understanding

Not a power unit — it is energy. 100 W bulb for 10 h = 1 kWh.

2. Diagrams & Visuals

1 kWh = 1 kW × 3600 s = 3.6 MJ

Color-coded visual · step-by-step breakdown below

  1. Power in kW × time in hours
  2. Gives kWh
  3. ×3.6×10⁶ for joules

3. Solved Examples

Basic

Q: 1 kW for 1 h.

Solution: 1 kWh

Answer: 3.6×10⁶ J

Intermediate

Q: 2 kW heater, 3 h.

Solution: 6 kWh

Answer: 21.6 MJ

Advanced

Q: Convert 5 kWh to J.

Solution: 5×3.6×10⁶

Answer: 18×10⁶ J

Exam

Q: Is kWh power or energy?

Solution: Energy

Answer: Common trap

K = ½ mv²

Definition: Kinetic energy — energy of motion.

Derivation

From W = Fs = mas and v²=v₁²+2as → W = ½mv²−½mv₁².

Variables

K (J) · m (kg) · v (m/s)

Why it works

Moving objects can do work before stopping — faster motion stores more energy (v² dependence).

Historical context

Vis viva debates in 17th–18th century; ½mv² won as kinetic energy.

Deep understanding

Double speed → 4× K. K ≥ 0 always (scalar from v²).

2. Diagrams & Visuals

K = ½mv² 2× speed → 4× K

Color-coded visual · step-by-step breakdown below

  1. Mass m in kg
  2. Speed v (not velocity for K magnitude)
  3. K = ½mv²

3. Solved Examples

Basic

Q: 2 kg at 3 m/s.

Solution: K=½×2×9

Answer: 9 J

Intermediate

Q: 10 kg: 4 m/s → 10 m/s.

Solution: K₁=80, K₂=500

Answer: ΔK=420 J

Advanced

Q: Speed doubles. K becomes?

Solution: K ∝ v²

Answer: 4× original

Exam

Q: MCQ Q5: kinetic energy?

Solution: ½mv²

Answer: Option B

W = K₂ − K₁ = ΔK

Definition: Work–energy theorem: net work equals change in kinetic energy.

Derivation

W_net = ∫F·ds = ½mv₂² − ½mv₁² using F=ma and v²=v₁²+2as.

Variables

W_net (J) · K₁, K₂ (J)

Why it works

Powerful shortcut — find speed change without time if you know net work.

Historical context

Central bridge between Newton's force view and energy methods in mechanics.

Deep understanding

Use ALL forces' net work (or net force). Conservative + non-conservative included.

2. Diagrams & Visuals

W_net = ΔK K₂ - K₁ = work by resultant force

Color-coded visual · step-by-step breakdown below

  1. Find v₁, v₂
  2. K₁=½mv₁², K₂=½mv₂²
  3. W_net = K₂−K₁
  4. Cross-check with F·s if constant F

3. Solved Examples

Basic

Q: K changes 20 J to 50 J.

Solution: W=30 J

Answer: 30 J

Intermediate

Q: Ex 6.6: 10 kg, 4→10 m/s.

Solution: ΔK=500−80

Answer: 420 J = F×14 m

Advanced

Q: Net work negative means?

Solution: K decreases

Answer: Braking

Exam

Q: MCQ Q4: work–energy theorem?

Solution: ΔK

Answer: Option B

U = mgh

Definition: Gravitational potential energy near Earth's surface.

Derivation

Work to lift: W_agent = mgh = −W_gravity; store as U.

Variables

U (J) · m · g · h (m above reference)

Why it works

Position energy — can convert to K when object falls. Reference level arbitrary; only ΔU matters.

Historical context

Potential energy formalises energy stored in gravitational fields.

Deep understanding

Conservative: W_gravity = −ΔU. Round trip lift+lower: net W_g = 0.

2. Diagrams & Visuals

h U=mgh

Color-coded visual · step-by-step breakdown below

  1. Choose reference zero level
  2. Height h above reference
  3. U = mgh

3. Solved Examples

Basic

Q: 1 kg at 10 m.

Solution: U=98 J

Answer: 98 J

Intermediate

Q: 0.5 kg, 2.5 m height (Ex 6.9).

Solution: U=12.25 J

Answer: → v=7 m/s at ground

Advanced

Q: ΔU only matters — prove.

Solution: Move reference

Answer: Same physics

Exam

Q: MCQ Q7: gravitational PE?

Solution: mgh

Answer: Option B

U_s = ½ kx²

Definition: Elastic potential energy in deformed spring.

Derivation

Equals work W_ext = ½kx² to compress/stretch.

Variables

U_s (J) · k · x

Why it works

Spring stores energy when deformed; releases to K when let go.

Historical context

Mechanical watches, vehicle springs, and oscillators use elastic PE ↔ K exchange.

Deep understanding

Oscillator: ½mv² + ½kx² = constant (no friction).

2. Diagrams & Visuals

U_s=½kx²

Color-coded visual · step-by-step breakdown below

  1. Measure x from equilibrium
  2. U_s = ½kx²
  3. Conservation: U_s ↔ K

3. Solved Examples

Basic

Q: k=100, x=0.1.

Solution: U=0.5 J

Answer: 0.5 J

Intermediate

Q: Max compression x_m: all K→U_s.

Solution: ½mv²=½kx_m²

Answer: Solve x_m

Advanced

Q: x doubles — U_s?

Solution: ×4

Answer: Quadratic in x

Exam

Q: Same as W_ext to compress?

Solution: Yes, ½kx²

Answer: MCQ Q8

K₁ + U₁ = K₂ + U₂

Definition: Mechanical energy conservation (conservative forces only).

Derivation

W_cons = −ΔU and W_net = ΔK → ΔK+ΔU=0 when only conservative forces.

Variables

K + U = mechanical energy E

Why it works

Track energy instead of forces — powerful for free fall, springs, smooth tracks.

Historical context

Conservation of mechanical energy follows from work done by conservative fields.

Deep understanding

With friction: K₁+U₁ = K₂+U₂ + heat lost. Total energy still conserved.

2. Diagrams & Visuals

top: K=0, U=mgh bottom: K=max, U=0 K+U = constant (Fig 6.10)

Color-coded visual · step-by-step breakdown below

  1. Identify initial and final points
  2. Write K₁+U₁ and K₂+U₂
  3. Set equal (no friction)
  4. Solve unknown

3. Solved Examples

Basic

Q: Ball at height h, falls to ground.

Solution: mgh = ½mv²

Answer: v = √(2gh)

Intermediate

Q: Ex 6.9: 0.5 kg, 2.5 m.

Solution: 12.25=½×0.5×v²

Answer: v = 7 m/s

Advanced

Q: Incline with friction — ME conserved?

Solution: No — heat loss

Answer: Use energy balance + friction work

Exam

Q: When does ME conserve?

Solution: Only conservative forces

Answer: MCQ Q9: gravity

m_A v_Ai + m_B v_Bi = m_A v_Af + m_B v_Bf

Definition: Momentum conservation in all collisions (isolated system).

Derivation

From Newton's laws — no external impulse.

Variables

m_A, m_B · velocities before (i) and after (f)

Why it works

Always true in closed collisions even when KE is not conserved.

Historical context

Links Lesson 3 momentum to energy chapter collision analysis.

Deep understanding

Use with energy equation only for elastic collisions.

2. Diagrams & Visuals

before → after Σp = const

Color-coded visual · step-by-step breakdown below

  1. Draw before/after diagram
  2. Write total p before
  3. Write total p after
  4. Equate and solve

3. Solved Examples

Basic

Q: 2 kg at 3 m/s hits 2 kg at rest, stick together.

Solution: 6=4v′

Answer: v′=1.5 m/s

Intermediate

Q: Identical elastic head-on: velocities swap.

Solution: v_Af=v_Bi

Answer: Fig 6.19

Advanced

Q: Heavy wall (m_B>>m_A): light ball rebounds.

Solution: v_Af≈−v_Ai

Answer: Case II in notes

Exam

Q: MCQ Q10: elastic collision conserves?

Solution: p AND K

Answer: Option C

½m_A v_Ai² + ½m_B v_Bi² = ½m_A v_Af² + ½m_B v_Bf²

Definition: Kinetic energy conservation — elastic collisions only.

Derivation

Conservative internal forces during elastic impact.

Variables

KE before = KE after (elastic)

Why it works

Distinguishes elastic vs inelastic — billiard balls nearly elastic; mud ball inelastic.

Historical context

Coefficient of restitution quantifies how elastic a collision is.

Deep understanding

Elastic head-on: v_Bf−v_Af = −(v_Bi−v_Ai). Inelastic: bodies may stick (v′ = mv/(m+M)).

2. Diagrams & Visuals

Elastic: KE before = KE after Inelastic: KE lost to heat/deformation

Color-coded visual · step-by-step breakdown below

  1. Confirm collision is elastic
  2. Sum ½mv² before
  3. Sum ½mv² after
  4. Equate with momentum eqn

3. Solved Examples

Basic

Q: Elastic: same speeds exchanged (equal masses).

Solution: KE total same

Answer: Velocities swap

Intermediate

Q: Prove energy lost in inelastic stick.

Solution: KE_after < KE_before

Answer: Difference = heat

Advanced

Q: Bullet embeds: v′=mv/(m+M).

Solution: Use momentum only

Answer: KE not conserved

Exam

Q: MCQ Q10: perfectly elastic?

Solution: p and KE both

Answer: Option C

5. Special Features & Extras

Complete study guide for Work, Energy and Power.

Exam Tips & Tricks

  • Work is a scalar — use W = Fd cos θ; θ=90° → zero work (MCQ Q3).
  • Work–energy theorem: W_net = ΔK — not ΔU (MCQ Q4).
  • Spring energy and work: ½kx² — not kx or ½kx (MCQ Q8).
  • Conservative force: gravity, spring — not friction (MCQ Q9).
  • Elastic collision: both momentum and KE conserved (MCQ Q10).
  • Variable force: work = area under F–x graph.

Common Student Mistakes

  • Confusing watt (power) with joule (energy) or kWh (energy)
  • Using K = mv instead of ½mv²
  • Forgetting cos θ in W = Fd cos θ
  • Applying KE conservation in inelastic collisions
  • Choosing wrong reference height for mgh (only ΔU matters)
  • Counting work by force that is perpendicular to motion

Memory Aids & Mnemonics

WORK: When Only Radial part of force × distance — cos θ matters
KE: "Half em vee squared" — K = ½mv²
Spring: "Half k x squared" for both W_ext and U_s
Power: P = Work/time — "how fast you earn joules"

Which Formula When?

  • Constant force at angle? → W = Fd cos θ
  • Force varies with x? → Area under F–x or ½kx² for spring
  • Find speed from forces? → W_net = ΔK
  • Height / gravity? → U = mgh; free fall: mgh = ½mv²
  • Spring stored energy? → U_s = ½kx²
  • How fast energy used? → P = ΔW/Δt
  • Collision? → Momentum always; add KE eqn if elastic

QUICK REFERENCE — Ch 6 Work, Energy & Power

W = Fd cos θ1 J = 1 N·mW_gravity (lift) = −mghW_gravity (lower) = +mghΔW = F(x) ΔxF_s = −kxW_ext = ½ kx²P = ΔW / Δt1 W = 1 J/s1 hp = 746 W1 kWh = 3.6 × 10⁶ JK = ½ mv²W = K₂ − K₁ = ΔKU = mghU_s = ½ kx²K₁ + U₁ = K₂ + U₂m_A v_Ai + m_B v_Bi = m_A v_Af + m_B v_Bf½m_A v_Ai² + ½m_B v_Bi² = ½m_A v_Af² + ½m_B v_Bf²

Units: J = N·m · W = J/s · hp = 746 W · 1 kWh = 3.6×10⁶ J

Tip: Conservative forces only → K₁+U₁ = K₂+U₂. With friction, energy goes to heat.

PYQ — Previous Year Questions

Extracted from NIOS Physics (312) board exam papers in your PDF. Chapter L6 — Work, Energy and Power only. Use Model Answer for marking points; Explanation for concept clarity.

Chapter 6 — Work, Energy and Power (L6)

19 questions · Section A (MCQ & objective) + Section B (short/long) · Sources: 312/TUS/104A, 312/MAY/204A–C, 68/ESS/1-312-A, Marking Scheme

Section A — Multiple Choice (1 mark)

PYQ1. In which of the following situations is work done against the gravity of Earth? — (A) A monkey climbs up a tree   (B) A car moves on a horizontal road   (C) A person tries to lift a load but fails   (D) A person lifts a load on a natural satellite

1 mark · Section A Q1 · 68/ESS/1-312-A

Model Answer

Answer: (A) A monkey climbs up a tree

Monkey gains height → displacement opposite to weight → external agent does positive work against gravity.

Explanation

Work against gravity: W = mgh when height increases. Horizontal motion (B) has no vertical displacement. Failed lift (C) has zero displacement. On satellite (D), effective g ≈ 0.

PYQ2. Through wave motion — (A) only energy is transmitted   (B) only particles are transmitted   (C) energy and particles both   (D) neither

1 mark · Section A Q2 · 68/ESS/1-312-A

Model Answer

Answer: (A) only energy is transmitted

Explanation

In wave propagation, medium particles oscillate about equilibrium — they are not transported. Energy and momentum propagate with the wave (L6: energy transfer without bulk matter transfer).

PYQ3. A body of mass m is thrown vertically upward with initial velocity v. Its kinetic energy at height h will be — (A) equal to ½mv²   (B) more than ½mv²   (C) less than ½mv²   (D) mgh − ½mv²

1 mark · Section A Q4 · 312/TUS/104A

Model Answer

Answer: (C) less than ½mv²

At height h, speed < v → K = ½mv_h² < ½mv². Energy converts to gravitational PE.

Explanation

Conservation: ½mv² = ½mv_h² + mgh. Hence K at height h = ½mv² − mgh < initial K. Option (D) is not a kinetic energy expression.

PYQ4. A particle is projected at 60° to the horizontal with kinetic energy E. Its kinetic energy at the highest point will be — (A) 0   (B) E/2   (C) E/4   (D) E

1 mark · Section A Q5 · 68/ESS/1-312-A

Model Answer

Answer: (C) E/4

At top: v_y = 0, v_x = v cos 60° = v/2. K_top = ½m(v/2)² = ¼(½mv²) = E/4.

Explanation

Horizontal component unchanged; vertical component zero at apex. cos 60° = ½ → speed at top is half → KE is one-quarter of launch KE.

PYQ5. A 2 kg body moves at constant velocity v = 5 m·s⁻¹ under a constant force of 3 N. The power loss due to friction is — (A) Zero   (B) 15 W   (C) −15 W   (D) 30 W

1 mark · Section A Q13 · 68/ESS/1-312-A

Model Answer

Answer: (B) 15 W

Uniform velocity ⇒ net force = 0 ⇒ friction = 3 N. P = F·v = 3 × 5 = 15 W.

Explanation

Applied force balances friction. Power dissipated against friction P = f_k v (L6: P = ΔW/Δt = Fv when F ∥ v).

PYQ6. A 100 W bulb is connected to a 220 V supply. The current through the bulb is — (A) 5/11 A   (B) 10/11 A   (C) 11/5 A   (D) 11/10 A

1 mark · Section A Q14 · 68/ESS/1-312-A

Model Answer

Answer: (A) 5/11 A

I = P/V = 100/220 = 5/11 A ≈ 0.45 A.

Explanation

Electrical power P = VI. Rated power 100 W at 220 V gives operating current. Power rating is energy per unit time (watt = joule/second).

PYQ7. Fill in the blanks (any two): (a) SI unit of energy is _____ (b) Joule per second is called _____ (c) 1 kWh is the unit of _____ (d) 1 horsepower = _____ watt.

2 marks (1×2) · Section A Q20 · 312/TUS/104A

Model Answer

  • (a) joule (J)
  • (b) watt (W)
  • (c) energy (electrical energy consumed)
  • (d) 746 watt

Explanation

1 J = 1 N·m. Power = energy/time → 1 W = 1 J/s. kWh is energy (power × time), not power. 1 hp = 746 W (L6 formula sheet).

PYQ8. Match Column I with II (any two): (a) Conservation of energy → ?   (b) Non-attainability of 100% efficiency → ?   Options: (i) Zeroth law   (ii) Kelvin-Planck   (iii) First law   (iv) Clausius

2 marks (1×2) · Section A Q22 · 312/TUS/104A

Model Answer

(a) Conservation of energy ↔ (iii) First law of thermodynamics

(c) 100% efficiency impossible ↔ (ii) Kelvin-Planck statement

Explanation

First law: ΔU = Q − W (energy conservation). Kelvin-Planck: no engine converts all heat to work. Links to L6 universal conservation of energy.

PYQ9. Passage — Conservation of energy: Apparent breach in beta-decay led to discovery of — ? Law applicable to — ? (A) electron / mechanical only   (B) proton / chemical only   (C) neutron / nuclear only   (D) neutrino / every system of the universe

2 marks (1×2) · Section A Q17 · 68/ESS/1-312-A

Model Answer

(i) (D) Neutrino

(ii) (D) every system of the universe

Explanation

Missing energy in beta-decay was accounted by Pauli’s neutrino hypothesis. L6: conservation of energy is universal — mechanical, thermal, nuclear, chemical systems.

PYQ10. Which form of energy is most closely associated with heat? — (A) Potential   (B) Magnetic   (C) Sound   (D) Kinetic

1 mark · Passage Q(b) · 312/MAY/204A–C

Model Answer

Answer: (D) Kinetic energy

Heat is random thermal motion of molecules — microscopic kinetic energy.

Explanation

Temperature measures average molecular KE. First law equates heat with energy transfer; internally heat manifests as disordered kinetic energy of particles.

PYQ11. Draw a restoring force vs displacement graph for a helical spring. Write an expression for energy stored at maximum displacement.

2 marks · Section B Q32 · Marking Scheme

Model Answer

Graph: straight line through origin, slope = −k (F = −kx).

Energy stored = work done stretching = area under F–x curve = U_s = ½kx²max (or ½ × base × height = ½ × x_m × kx_m).

Explanation

Elastic PE equals work done against spring force. Linear restoring force gives triangular area → ½kx². L6: U_s = ½kx², W_ext = ½kx².

PYQ12. Show that 1 kWh of energy is equal to 3.6 × 10⁶ J.

2 marks · Section B Q33 · Marking Scheme

Model Answer

1 kWh = 1 kW × 1 h = 1000 W × 3600 s

= 1000 J/s × 3600 s = 3.6 × 10⁶ J

Explanation

kWh is energy (power × time), not power. One "unit" on electricity bills = 1 kWh = 3.6 MJ. Essential L6 conversion.

PYQ13. Convert: (a) 7460 watt into hp   (b) 360 kJ into kWh.

3 marks · Section B Q41 · 312/TUS/104A

Model Answer

(a) hp = 7460/746 = 10 hp

(b) 360 kJ = 360/(3.6×10⁶) kWh = 360/3600 = 0.1 kWh

Explanation

Use 1 hp = 746 W and 1 kWh = 3.6×10⁶ J = 3600 kJ. Unit conversions test power vs energy distinction.

PYQ14. Two particles of different masses have equal kinetic energies. Find the ratio of their linear momenta and velocities.

3 marks · Section B Q39 · 312/MAY/204A/B/C

Model Answer

½m₁v₁² = ½m₂v₂² ⇒ v₁/v₂ = √(m₂/m₁)

p = mv ⇒ p₁/p₂ = (m₁v₁)/(m₂v₂) = √(m₁/m₂)

Velocity ratio: √(m₂/m₁)  ·  Momentum ratio: √(m₁/m₂)

Explanation

From K = p²/2m: p = √(2mK). Equal K ⇒ p ∝ √m. Heavier particle has larger momentum but smaller speed.

PYQ15. A particle undergoes displacement d = (3î + 4ĵ) m under force F = (5î + 3ĵ) N. Calculate the work done.

3 marks · Section B Q39 (OR) · 312/MAY/204A/B/C

Model Answer

W = F · d = F_x d_x + F_y d_y

= 5×3 + 3×4 = 15 + 12 = 27 J

Explanation

Work is dot product of force and displacement. Only the component of force along displacement contributes. L6: W = Fd cos θ generalises to W = F⃗ · d⃗.

PYQ16. A raindrop of mass 1 g falls from 1 km height and hits the ground at 50 m·s⁻¹. Calculate (I) loss of P.E. (II) gain in K.E. (g = 10 m·s⁻²).

3 marks · Section B Q39 · Marking Scheme

Model Answer

m = 10⁻³ kg, h = 1000 m, v = 50 m·s⁻¹

(I) Loss in PE = mgh = 10⁻³ × 10 × 1000 = 10 J

(II) Gain in KE = ½mv² = ½ × 10⁻³ × 50² = 1.25 J

Remaining ~8.75 J dissipated (air resistance, sound, etc.).

Explanation

Not all PE converts to KE — non-conservative air drag removes energy. Illustrates conservation with energy "lost" to heat.

PYQ17. A body of mass 0.5 kg moves in a straight line with v = ax^(3/2), where a = 5 m^(−1/2)·s^(−1). Find work done from x = 0 to x = 2 m.

3 marks · Section B Q39 (OR) · Marking Scheme · Answer: 50 J

Model Answer

F = ma; a = dv/dt = (3/2)ax^(1/2) × dx/dt = (3/2)a²x²

W = ∫₀² F dx = ∫₀² 0.5 × (3/2)a²x² dx = (3/4)a² × [x³/3]₀²

= (3/4) × 25 × 8/3 = 50 J

Explanation

Variable force: W = ∫F(x)dx. Chain rule gives acceleration in terms of x. Alternatively use work-energy theorem W = ΔK from v(0) and v(2).

PYQ18. Define work, energy and power. Give their SI units. Give two events where force acts and the body moves but no mechanical work is done.

5 marks · Section B Q42/Q43 · 312/MAY/204A/B/C

Model Answer

Work: Product of force component along displacement and displacement; SI unit joule (J).

Energy: Capacity to do work; SI unit joule (J).

Power: Rate of doing work; SI unit watt (W) = J/s.

Zero-work examples (any two):

  • Centripetal force in uniform circular motion (F ⊥ displacement)
  • Carrying load horizontally at constant height (gravity ⊥ horizontal displacement)
  • Pushing a wall that does not move (zero displacement)
  • Coolie walking on platform with load on head (load not displaced vertically)

Explanation

W = Fd cos θ. When θ = 90°, W = 0 even if force and motion exist. Distinguish "effort" from physics work.

PYQ19. Give an example of a variable force. Derive an expression for work done under such a force.

5 marks · Section B Q42/Q43 (OR) · 312/MAY/204A/B/C

Model Answer

Example: Spring force F = −kx, or gravitational force near Earth F = −mgĵ.

Derivation: Divide path into small displacements Δx. Work on each segment ≈ F(x)Δx. Total:

W = ∫x₁x₂ F(x) dx

For spring from 0 to x: W = ∫₀ˣ kx dx = ½kx² (area under F–x graph).

Explanation

Constant-force formula W = Fs is a special case. Graphically, work = area under F–x curve. Leads to elastic PE U_s = ½kx².

Problem Solving — L6 Work, Energy and Power

Six problems spanning this chapter’s NIOS syllabus. Every question is built from the notes and formula sheet: solve with equations first, then read the formal textbook-style write-up, the easy explanation, and the topic in depth (formulas, meaning, exam tips). Explanations open by default.

Question 1 of 6Work

Draw a diagram showing force F at angle θ = 60° to displacement d. A constant force of 20 N acts at 60° to the displacement of 4.0 m. Calculate work done by the force. When is work zero even if force and displacement are both non-zero?

W = F d cos θ
W = F · d
1 J = 1 N·m

Pencil sketch (labelled)

Work: W = F d cos θ d F θ along d: F cos θ
Pencil sketch: force at angle θ to displacement

Solution — step by step with formulas

  1. NIOS formula: W = F d cos θ.
  2. W = 20 × 4.0 × cos 60° = 80 × ½ = 40 J.
  3. Work is zero if θ = 90° (F perpendicular to d), e.g. ideal centripetal force in uniform circular motion, or if d = 0.

Final answer: W = 40 J; zero when θ = 90° (or d = 0).

Formulas used in this problem

W = F d cos θ
W = F · d
1 J = 1 N·m

Textbook formal language

In the NIOS treatment, work done by a constant force is W = F d cos θ, equivalently the scalar (dot) product of force and displacement. Work is a scalar: it may be positive (force component along displacement), negative (opposite), or zero (perpendicular). The SI unit is the joule: 1 J = 1 N·m.

Working formula set for this problem: W = F d cos θ; W = F · d; 1 J = 1 N·m. In the NIOS presentation, physical quantities must be expressed in SI units and the relevant law or definition stated before substitution. Vector quantities require an explicit choice of positive direction; scalar work and energy require attention to sign conventions of the textbook.

Easy language (same idea, plain words)

Only the part of the push that points along the motion counts. At 60°, cos 60° = ½, so half of 20 N is “useful” along the 4 m path: 10 × 4 = 40 J. If you push sideways while the object moves forward, the angle is 90° and you do no work—even though you get tired!

Read the question once for the story, once for the numbers. Write the formula, plug in values with units, then simplify. If a result looks huge or tiny, re-check powers of ten and whether you used sin/cos of the correct angle.

Topic in depth — Work by a constant force

From the notes: if d = 0, W = 0 (pushing a wall that does not move). Friction often does negative work and removes kinetic energy. Always measure θ between the force vector and the displacement vector, not between force and some other line on the diagram.

Link to chapter notes (L6 — Work by a constant force): this idea sits with the definitions and worked examples in the detailed notes and formula sheet. Memorise: W = F d cos θ; W = F · d; 1 J = 1 N·m. In multi-step questions, keep a free-body diagram or energy flow sketch before algebra; most errors are missing forces or wrong signs, not hard maths.

Exam tip

Quote the law in one line, then write W = F d cos θ; W = F · d; 1 J = 1 N·m before numbers. Box the final answer with unit. For numericals, keep at least three significant figures until the last step unless the data are coarse.

Common mistakes

  • Mixing up scalar and vector quantities (e.g. treating momentum as unsigned).
  • Using the wrong sign convention for work/heat/force direction.
  • Forgetting to convert units (g↔kg, cm↔m, minutes↔seconds).
  • Applying a formula outside its assumptions (e.g. F = ma when mass is not constant).
Question 2 of 6KE

A 2.0 kg body speeds from 3.0 m·s⁻¹ to 5.0 m·s⁻¹. Find the change in kinetic energy and the net work done on the body.

K = ½ m v²
W_net = ΔK (work–energy theorem)

Solution — step by step with formulas

  1. K_i = ½ m u² = ½×2.0×(3.0)² = 9.0 J.
  2. K_f = ½ m v² = ½×2.0×(5.0)² = 25.0 J.
  3. ΔK = K_f − K_i = 16.0 J.
  4. By the work–energy theorem (NIOS): W_net = ΔK = 16.0 J.

Final answer: ΔK = 16 J = W_net

Formulas used in this problem

K = ½ m v²
W_net = ΔK (work–energy theorem)

Textbook formal language

Kinetic energy of a particle is K = ½mv². The work–energy theorem states that the net work done by all forces acting on the particle equals the change in its kinetic energy. Thus W_net = K_f − K_i without needing each force separately if only ΔK is required.

Working formula set for this problem: K = ½ m v²; W_net = ΔK (work–energy theorem). In the NIOS presentation, physical quantities must be expressed in SI units and the relevant law or definition stated before substitution. Vector quantities require an explicit choice of positive direction; scalar work and energy require attention to sign conventions of the textbook.

Easy language (same idea, plain words)

Kinetic energy is the energy of motion: ½ × mass × speed². At 3 m/s the body has 9 J; at 5 m/s it has 25 J. The jump of 16 J is exactly the net work that went into speeding it up.

Read the question once for the story, once for the numbers. Write the formula, plug in values with units, then simplify. If a result looks huge or tiny, re-check powers of ten and whether you used sin/cos of the correct angle.

Topic in depth — Kinetic energy

KE depends on the reference frame through v. The theorem is a scalar energy bookkeeping tool that complements Newton’s laws. If friction does −10 J and you need +16 J of KE change, other forces must supply +26 J of work.

Link to chapter notes (L6 — Kinetic energy): this idea sits with the definitions and worked examples in the detailed notes and formula sheet. Memorise: K = ½ m v²; W_net = ΔK (work–energy theorem). In multi-step questions, keep a free-body diagram or energy flow sketch before algebra; most errors are missing forces or wrong signs, not hard maths.

Exam tip

Quote the law in one line, then write K = ½ m v²; W_net = ΔK (work–energy theorem) before numbers. Box the final answer with unit. For numericals, keep at least three significant figures until the last step unless the data are coarse.

Common mistakes

  • Mixing up scalar and vector quantities (e.g. treating momentum as unsigned).
  • Using the wrong sign convention for work/heat/force direction.
  • Forgetting to convert units (g↔kg, cm↔m, minutes↔seconds).
  • Applying a formula outside its assumptions (e.g. F = ma when mass is not constant).
Question 3 of 6PEg

A 5.0 kg mass is raised slowly by 2.0 m. Taking g = 10 m·s⁻², find increase in gravitational PE. Who does work against gravity?

U = mgh
ΔU = mg Δh

Solution — step by step with formulas

  1. ΔU = mgh = 5×10×2 = 100 J.
  2. The external agent does +100 J work against gravity (if raised slowly, KE≈0).

Final answer: ΔU = 100 J; external agent supplies the work

Formulas used in this problem

U = mgh
ΔU = mg Δh

Textbook formal language

Near Earth’s surface, gravitational PE relative to a reference level is mgh. Change in PE equals work done against gravity for quasistatic lift.

Working formula set for this problem: U = mgh; ΔU = mg Δh. In the NIOS presentation, physical quantities must be expressed in SI units and the relevant law or definition stated before substitution. Vector quantities require an explicit choice of positive direction; scalar work and energy require attention to sign conventions of the textbook.

Easy language (same idea, plain words)

Lifting stores energy in the “height account.” 5 kg up 2 m stores 100 J. You pay that energy with your muscles (or a machine).

Read the question once for the story, once for the numbers. Write the formula, plug in values with units, then simplify. If a result looks huge or tiny, re-check powers of ten and whether you used sin/cos of the correct angle.

Topic in depth — Gravitational potential energy

Only differences in PE matter physically; choose a convenient zero (floor, ground).

Link to chapter notes (L6 — Gravitational potential energy): this idea sits with the definitions and worked examples in the detailed notes and formula sheet. Memorise: U = mgh; ΔU = mg Δh. In multi-step questions, keep a free-body diagram or energy flow sketch before algebra; most errors are missing forces or wrong signs, not hard maths.

Exam tip

Quote the law in one line, then write U = mgh; ΔU = mg Δh before numbers. Box the final answer with unit. For numericals, keep at least three significant figures until the last step unless the data are coarse.

Common mistakes

  • Mixing up scalar and vector quantities (e.g. treating momentum as unsigned).
  • Using the wrong sign convention for work/heat/force direction.
  • Forgetting to convert units (g↔kg, cm↔m, minutes↔seconds).
  • Applying a formula outside its assumptions (e.g. F = ma when mass is not constant).
Question 4 of 6Conservation

A 1.0 kg stone is dropped from rest from 5.0 m height (g = 10). Find speed just before hitting ground using energy conservation. State when this method fails.

K + U = constant (conservative forces only)

Solution — step by step with formulas

  1. mgh = ½mv² ⇒ v = √(2gh) = √100 = 10 m·s⁻¹.
  2. Fails if non-conservative work (air drag, friction) is significant: then ΔE_mech = W_nc.

Final answer: v = 10 m·s⁻¹; fails when friction/drag does work

Formulas used in this problem

K + U = constant (conservative forces only)

Textbook formal language

If only conservative forces act, total mechanical energy is conserved. Loss in PE equals gain in KE for free fall from rest.

Working formula set for this problem: K + U = constant (conservative forces only). In the NIOS presentation, physical quantities must be expressed in SI units and the relevant law or definition stated before substitution. Vector quantities require an explicit choice of positive direction; scalar work and energy require attention to sign conventions of the textbook.

Easy language (same idea, plain words)

Height energy turns into speed energy. From 5 m you hit at 10 m/s if air doesn’t steal energy.

Read the question once for the story, once for the numbers. Write the formula, plug in values with units, then simplify. If a result looks huge or tiny, re-check powers of ten and whether you used sin/cos of the correct angle.

Topic in depth — Mechanical energy conservation

Conservative force: work independent of path (gravity, ideal spring). Friction is path-dependent.

Link to chapter notes (L6 — Mechanical energy conservation): this idea sits with the definitions and worked examples in the detailed notes and formula sheet. Memorise: K + U = constant (conservative forces only). In multi-step questions, keep a free-body diagram or energy flow sketch before algebra; most errors are missing forces or wrong signs, not hard maths.

Exam tip

Quote the law in one line, then write K + U = constant (conservative forces only) before numbers. Box the final answer with unit. For numericals, keep at least three significant figures until the last step unless the data are coarse.

Common mistakes

  • Mixing up scalar and vector quantities (e.g. treating momentum as unsigned).
  • Using the wrong sign convention for work/heat/force direction.
  • Forgetting to convert units (g↔kg, cm↔m, minutes↔seconds).
  • Applying a formula outside its assumptions (e.g. F = ma when mass is not constant).
Question 5 of 6Power

An engine pulls a train with constant force 5000 N at steady 10 m·s⁻¹. Find instantaneous power delivered by the force.

P = W/t
P = F v (constant F along v)

Solution — step by step with formulas

  1. P = F v = 5000 × 10 = 5.0 × 10⁴ W = 50 kW.

Final answer: P = 50 kW

Formulas used in this problem

P = W/t
P = F v (constant F along v)

Textbook formal language

Power is the time rate of doing work. For constant force collinear with velocity, P = F·v.

Working formula set for this problem: P = W/t; P = F v (constant F along v). In the NIOS presentation, physical quantities must be expressed in SI units and the relevant law or definition stated before substitution. Vector quantities require an explicit choice of positive direction; scalar work and energy require attention to sign conventions of the textbook.

Easy language (same idea, plain words)

Power is how fast you deliver energy. Force times speed gives watts when they point the same way.

Read the question once for the story, once for the numbers. Write the formula, plug in values with units, then simplify. If a result looks huge or tiny, re-check powers of ten and whether you used sin/cos of the correct angle.

Topic in depth — Power

1 horsepower ≈ 746 W (if used in problems). Average power is total work over total time.

Link to chapter notes (L6 — Power): this idea sits with the definitions and worked examples in the detailed notes and formula sheet. Memorise: P = W/t; P = F v (constant F along v). In multi-step questions, keep a free-body diagram or energy flow sketch before algebra; most errors are missing forces or wrong signs, not hard maths.

Exam tip

Quote the law in one line, then write P = W/t; P = F v (constant F along v) before numbers. Box the final answer with unit. For numericals, keep at least three significant figures until the last step unless the data are coarse.

Common mistakes

  • Mixing up scalar and vector quantities (e.g. treating momentum as unsigned).
  • Using the wrong sign convention for work/heat/force direction.
  • Forgetting to convert units (g↔kg, cm↔m, minutes↔seconds).
  • Applying a formula outside its assumptions (e.g. F = ma when mass is not constant).
Question 6 of 6SpringElastic PE

Draw a labelled sketch of a mass attached to a compressed spring on a smooth track. A spring of stiffness k = 200 N·m⁻¹ is compressed by 0.10 m from natural length. Find elastic PE stored. If released against a 0.50 kg mass, find maximum speed of the mass.

U = ½ kx²
F = −kx

Pencil sketch (labelled)

Spring–mass k m x U = ½kx² → ½mv²
Pencil sketch: spring and mass labelled

Solution — step by step with formulas

  1. U = ½×200×(0.10)² = 1.0 J.
  2. Energy → KE: ½mv² = 1.0 ⇒ v = √(2/0.50) = 2.0 m·s⁻¹.

Final answer: U = 1.0 J; v_max = 2.0 m·s⁻¹

Formulas used in this problem

U = ½ kx²
F = −kx

Textbook formal language

Hookean spring stores U = ½kx². On a smooth horizontal surface, elastic PE converts fully to kinetic energy of the attached mass at the mean position.

Working formula set for this problem: U = ½ kx²; F = −kx. In the NIOS presentation, physical quantities must be expressed in SI units and the relevant law or definition stated before substitution. Vector quantities require an explicit choice of positive direction; scalar work and energy require attention to sign conventions of the textbook.

Easy language (same idea, plain words)

Squashing the spring banks 1 J. On ice (no friction) that becomes speed: 2 m/s for half a kilogram.

Read the question once for the story, once for the numbers. Write the formula, plug in values with units, then simplify. If a result looks huge or tiny, re-check powers of ten and whether you used sin/cos of the correct angle.

Topic in depth — Elastic potential energy of a spring

x is displacement from natural length. Amplitude in SHM relates energy to ½kA².

Link to chapter notes (L6 — Elastic potential energy of a spring): this idea sits with the definitions and worked examples in the detailed notes and formula sheet. Memorise: U = ½ kx²; F = −kx. In multi-step questions, keep a free-body diagram or energy flow sketch before algebra; most errors are missing forces or wrong signs, not hard maths.

Exam tip

Quote the law in one line, then write U = ½ kx²; F = −kx before numbers. Box the final answer with unit. For numericals, keep at least three significant figures until the last step unless the data are coarse.

Common mistakes

  • Mixing up scalar and vector quantities (e.g. treating momentum as unsigned).
  • Using the wrong sign convention for work/heat/force direction.
  • Forgetting to convert units (g↔kg, cm↔m, minutes↔seconds).
  • Applying a formula outside its assumptions (e.g. F = ma when mass is not constant).