Definition
- Work = Force × Displacement × cos θ | W = F·S cosθ
- Scalar quantity; SI unit = Joule (J) = N·m; CGS unit = erg = dyne·cm
- Dimensional formula: [ML²T⁻²]
- Work may be +ve, −ve, or zero; depends on frame of reference
- Work done by one force does NOT depend on other forces acting
| Angle θ (between F and s) | Work Done | Effect on Speed |
|---|
| θ = 0° (F parallel to s) | W = +Fs (maximum) | Speed increases |
| 0° < θ < 90° | W = +ve | Speed increases |
| θ = 90° | W = 0 | Speed constant |
| 90° < θ < 180° | W = −ve | Speed decreases |
| θ = 180° (F anti-parallel to s) | W = −Fs (minimum) | Speed decreases |
Zero Work — Important Cases
- θ = 90° between force and displacement → W = 0
- Work done by Normal force = 0 (always ⊥ to displacement)
- Work done by Centripetal force = 0 (always ⊥ to velocity)
- Work done by Magnetic force on a moving charge = 0
By constant force: W = F·S cosθ = F⃗·S⃗ (dot product)
By variable force: W = ∫F·ds = Area under F-S graph
Total work: Wtotal = W₁ + W₂ + W₃ + ... (algebraic sum)
F = slope of W-S graph
Work done in 3D: W = ∫Fxdx + ∫Fydy + ∫Fzdz
Work Done by Gravity
Work by Gravity
- Wg = mgh = mgs sinθ (where h = vertical height, s = path length, θ = angle with horizontal)
- Does NOT depend on path — only on initial and final height
- Does NOT depend on nature of motion (uniform, non-uniform)
- Wg = +ve when object moves down (gravity and displacement in same direction)
- Wg = −ve when object moves up
- Wg = 0 when object moves horizontally
Work Done by Spring Force
Spring Force
- Spring force: Fs = −kx (restoring; always towards mean position)
- Work done by spring in elongation/compression from mean (x = 0) to x: Wspring = −½kx²
- Work done by spring from x₁ to x₂: W = −½k(x₂² − x₁²)
- Work done by spring is always −ve during elongation or compression from mean
- Spring potential energy stored: U = ½kx²
⚡ Spring Block System — Sudden vs Slow Release
- Slowly released: max elongation = mg/k (equilibrium)
- Suddenly released: using COME → 0 + 0 = 0 + ½kx² − mgx → xmax = 2mg/k
- Max elongation when suddenly released = 2× max elongation when slowly released
Q: A force F = 3x² acts on a body. Find work done from x = 1 m to x = 3 m.
W = ∫F dx = ∫₁³ 3x² dx = [x³]₁³ = 27 − 1 = 26 J
Ans: 26 J ✓
Q: A spring of k = 100 N/m is compressed by 0.2 m. Find work done by spring force.
W = −½kx² = −½ × 100 × 0.04 = −2 J (spring does −ve work during compression)
Ans: −2 J ✓
2
Kinetic Energy and Momentum
Kinetic Energy
- KE = ½mv² = p²/(2m) (energy due to motion)
- Scalar quantity; always ≥ 0; never negative
- p = mv (momentum; vector); p = √(2m·KE)
| Situation | Result |
|---|
| KE = constant | |p| = const; speed = const |
| p = constant (same momentum) | KE ∝ 1/m → lighter body has more KE |
| KE = constant (same KE) | p ∝ √m → heavier body has more momentum |
| 2 identical bodies, same momentum | Must have same KE |
| 2 identical bodies, same KE | May or may not have same momentum (direction can differ) |
% Change in KE and Momentum
- For small change (≤ 5%): Use error analysis → if KE changes by x%, momentum changes by x/2%
- For large change: Use exact formula
Large Change Problems — Step by Step
KE increases by 44%: KEf = 1.44 KEi; pf = √(KEf) = √1.44 × pi = 1.2 pi → % change in p = +20%
KE decreases by 19%: KEf = 0.81 KEi; pf = √0.81 × pi = 0.9 pi → % change in p = −10%
p increases by 200%: pf = 3pi; KEf = 9 KEi → % change in KE = +800%
Q: KE of a body increased by 44.1%. Find % increase in momentum.
KEf = 144.1% KEi; pf = √(144.1/100) × pi = 1.2pi; % increase = 20%
Ans: 20% ✓
Q: KE changed by 1%. Find % change in momentum (small change).
Using error method: KE = p²/2m → ΔKE/KE = 2(Δp/p) → Δp/p = ½(ΔKE/KE) = 0.5%
Ans: 0.5% ✓
Wtotal = ΔKE = KEf − KEi = ½mvf² − ½mvi²
Total work done by ALL forces on a body = Change in its kinetic energy. Always valid.
Implications
- W = +ve → ΔKE = +ve → KE increases → speed increases
- W = −ve → ΔKE = −ve → KE decreases → speed decreases
- W = 0 → ΔKE = 0 → KE constant → speed constant
- If a constant force acts: KE ∝ t² (since v = at, KE = ½ma²t²)
- Does not depend on mass → speed of collision with ground is same for objects dropped from same height regardless of mass
Q: A body of mass 2 kg starts from rest. A force F = 4t N acts on it. Find KE at t = 2 s.
a = F/m = 4t/2 = 2t; v = ∫a dt = t²; at t = 2: v = 4 m/s; KE = ½×2×16 = 16 J
Ans: 16 J ✓
Q: A 10 kg block moves from rest to 6 m/s in 3 s on a smooth surface. Find work done by net force.
W = ΔKE = ½×10×36 − 0 = 180 J
Ans: 180 J ✓
4
Conservative and Non-Conservative Forces
| Property | Conservative Force | Non-Conservative Force |
|---|
| Work depends on path | No — only initial & final position | Yes — depends on path |
| Work in closed loop | Always zero (∮F·dr = 0) | May or may not be zero |
| Potential energy defined | Yes | No |
| Work reversible | Yes | No |
| Examples | Gravity, spring force, electrostatic force | Friction, normal force, tension, air drag |
Fcf = −dU/dr (Conservative force = negative gradient of PE)
ΔU = −Wcf = −(work done by conservative force)
If Wcf = +ve → ΔU = −ve → PE decreases
If Wcf = −ve → ΔU = +ve → PE increases
Key Points
- PE = energy stored due to shape, size, and position
- Scalar; may be +ve, −ve, or zero
- Absolute value of PE is not defined — only change in PE (ΔU) is defined
- PE depends on reference point; ΔU does NOT depend on reference
- In every natural phenomenon, PE always decreases (tendency)
Gravitational Potential Energy
For point object: U = mgh (taking ground as reference; h above ground)
For extended object: U = Mg × hCOM (height of centre of mass from reference)
Chain Problems (Extended Body)
For a uniform chain of mass M, length L hanging from table — when pulled up or let down, use COM of hanging/lying portion:
Uhanging part = −(hanging mass) × g × (Lhang/2)
Total PE = PE of lying part + PE of hanging part
Spring Potential Energy
Uspring = ½kx²
Change in spring PE from x₁ to x₂: ΔU = ½k(x₂² − x₁²)
Spring PE is always ≥ 0 (always positive regardless of compression or elongation)
6
Conservation of Mechanical Energy (COME)
Wtotal = WCF + WNCF = ΔKE
WCF = −ΔU (work by conservative force = −change in PE)
So: −ΔU + WNCF = ΔKE
WNCF = ΔKE + ΔU = Δ(KE + U) = ΔEmech
Law of Conservation of Mechanical Energy
- Valid ONLY when WNCF = 0 (no non-conservative force like friction)
- (KE + U)initial = (KE + U)final
- Total mechanical energy (KE + PE) remains constant
- When NCF (friction) acts: WNCF = ΔEmech (mechanical energy changes)
Important Application — Leaving Circular Path
- Ball on outer surface of sphere — loses contact when N = 0
- At angle θ from top: mg cosθ = mv²/R → v² = gR cosθ
- Using COME from top: ½mv² = mgR(1 − cosθ) → v² = 2gR(1 − cosθ)
- Combining: gR cosθ = 2gR(1 − cosθ) → cosθ = 2/3 → θ ≈ 48.2°
- Speed at that point: v = √(2gR/3)
- Height from bottom: h = R + R cosθ = R(1 + 2/3) = 5R/3
Q: A ball is released from the top of a smooth hemisphere of radius R. At what angle from vertical does it leave the surface?
N = 0 at leave point; mg cosθ = mv²/R; COME: ½mv² = mgR(1−cosθ) → v² = 2gR(1−cosθ); combining: cosθ = 2−2cosθ → 3cosθ = 2 → cosθ = 2/3 → θ = cos⁻¹(2/3)
Ans: θ = cos⁻¹(2/3) ≈ 48.2° ✓
Q: A block slides down a frictionless incline of height 5 m. Find speed at bottom.
COME: ½mv² = mgh → v = √(2gh) = √(2×10×5) = 10 m/s
Ans: v = 10 m/s ✓
Q: A ball swings on a string of length 1 m. At the bottom speed = 4 m/s. Find max height reached.
COME: ½mv² = mgh → h = v²/2g = 16/20 = 0.8 m; since h < L(=1), it doesn't complete full circle
Ans: h = 0.8 m ✓
7
Equilibrium — F-r and PE-r Graphs
Key Concept
- F = −dU/dr → F is negative slope of U-r graph
- At equilibrium: F = 0 → dU/dr = 0 (slope of U-r graph = 0, i.e. at minima or maxima)
| Type of Equilibrium | PE-r Graph | F-r Graph | Stability |
|---|
| Stable | At minima (U is minimum) | F changes from +ve to −ve | Restored to equilibrium |
| Unstable | At maxima (U is maximum) | F changes from −ve to +ve | Moves away from equilibrium |
| Neutral | Constant (flat region) | F = 0 throughout region | Stays at new position |
Definition
- Power = rate of doing work = dW/dt
- Scalar quantity; SI unit = Watt (W) = J/s
- 1 HP (Horsepower) = 746 W; 1 kW = 1000 W
- Dimensional formula: [ML²T⁻³]
Instantaneous Power: Pinst = dW/dt = F·v = Fv cosθ
Average Power: Pavg = Total work / Total time = W/t = ΔKE/t
Power by gravity: Pgravity = Mgv sinθ (θ = angle with horizontal)
Slope of W-t graph = Power | Area of P-t graph = Work done
Power and Angle (Between F and v)
- θ = 90° (F ⊥ v) → P = 0 (e.g. centripetal force, normal on horizontal surface)
- θ < 90° → P = +ve → energy of body increases
- θ > 90° → P = −ve → energy of body decreases
Efficiency
- η = Poutput / Pinput = Woutput / Winput
- η = Puseful / Pengine
- Always ≤ 1 (or ≤ 100%)
Q: A pump lifts 200 kg of water per minute to a height of 5 m. Find power of pump.
W = mgh = 200×10×5 = 10000 J per minute; P = W/t = 10000/60 ≈ 167 W
Ans: P ≈ 167 W ✓
Q: A vehicle of mass 1000 kg moves at constant velocity 20 m/s against a frictional force of 500 N. Find engine power.
At constant velocity, engine force = friction = 500 N; P = Fv = 500×20 = 10,000 W = 10 kW
Ans: P = 10 kW ✓
Q: An engine of 40% efficiency produces 1000 J useful work. Find energy consumed.
η = Wout/Win → Win = Wout/η = 1000/0.4 = 2500 J
Ans: 2500 J ✓
Q: A body of mass 2 kg has speed v = 3t² m/s. Find instantaneous power at t = 2 s.
a = dv/dt = 6t; F = ma = 2×6t = 12t; at t=2: F = 24 N, v = 3(4) = 12 m/s; P = Fv = 24×12 = 288 W
Ans: P = 288 W ✓