How to Use This Page
1. Solve each exercise question yourself first — then check your method and answer below. 2. Pay close attention to worked numericals — they show the step-by-step approach examiners expect. 3. Review the Common Mistakes section — these are the most frequent errors students make on this chapter. 4. Use the Exam Checklist to confirm you have revised every concept and formula before the exam. 5. Always include SI units in your answers — marks are deducted for missing units.
Key Concepts (Quick Recap)
Work, Energy, and Power Formulas
| Quantity | Formula | SI Unit |
|---|
| Work | W = F·d·cos θ | Joule (J) |
| Kinetic energy | KE = ½mv² | Joule (J) |
| Potential energy (gravity) | PE = mgh | Joule (J) |
| Potential energy (spring) | PE = ½kx² | Joule (J) |
| Power | P = W/t = F·v | Watt (W) |
| Work-energy theorem | W_net = ΔKE = ½mv² − ½mu² | Joule (J) |
Conservation of Energy
| Principle | Statement |
|---|
| Conservation of energy | Total energy of an isolated system is constant; energy transforms from one form to another |
| Mechanical energy | E = KE + PE = constant (if only conservative forces act) |
Collisions
| Property | Elastic | Inelastic | Perfectly Inelastic |
|---|
| Momentum conserved? | Yes | Yes | Yes |
| KE conserved? | Yes | No (some lost) | No (maximum loss) |
| Objects separate? | Yes | Yes | No (stick together) |
| e (coefficient of restitution) | e = 1 | 0 < e < 1 | e = 0 |
Key Definitions
| Term | Definition |
|---|
| Work | Work is said to be done when a force displaces a body in the direction of the force. W = Fd cos θ |
| Energy | The capacity of a body to do work. It is a scalar quantity. |
| Power | The rate of doing work. P = W/t |
| Conservative force | A force for which work done is path-independent (e.g., gravity, spring force) |
| Non-conservative force | A force for which work done depends on the path (e.g., friction) |
I. Choose the Best Answer (MCQs)
1. A force of 50 N acts on a body at 60° to the horizontal. If the body moves 4 m horizontally, the work done is:
Answer
- (a) 200 J
- (b) 100 J
- (c) 173.2 J
- (d) 400 J
Answer(b) 100 J
ExplanationW = Fd cos θ = 50 × 4 × cos60° = 200 × 0.5 = 100 J.
2. A body of mass 2 kg is thrown upward with 20 m/s. Its kinetic energy at the highest point is:
Answer
- (a) 400 J
- (b) 0 J
- (c) 200 J
- (d) 100 J
Answer(b) 0 J
ExplanationAt the highest point, velocity = 0, so KE = ½mv² = ½ × 2 × 0 = 0 J. All energy is potential.
3. A motor lifts 500 kg of water through 20 m in 10 s. The power of the motor is (g = 10 m/s²):
Answer
- (a) 10 kW
- (b) 1 kW
- (c) 100 kW
- (d) 5 kW
Answer(a) 10 kW
ExplanationW = mgh = 500 × 10 × 20 = 100,000 J. P = W/t = 100,000/10 = 10,000 W = 10 kW.
4. In an elastic collision between two bodies:
Answer
- (a) Only momentum is conserved
- (b) Only kinetic energy is conserved
- (c) Both momentum and kinetic energy are conserved
- (d) Neither is conserved
Answer(c) Both momentum and kinetic energy are conserved
ExplanationBy definition, an elastic collision conserves both momentum and kinetic energy.
5. A body of mass m is moving in a circular path of radius r with constant speed v. The work done by the centripetal force in one revolution is:
Answer
- (a) mv²/r × 2πr
- (b) Zero
- (c) ½mv²
- (d) mgr
Answer(b) Zero
ExplanationCentripetal force is always perpendicular to displacement (tangential). W = F·d·cos90° = 0.
6. A spring of constant k is compressed by x. The potential energy stored is:
Answer
- (a) kx
- (b) kx²
- (c) ½kx²
- (d) ½kx
Answer(c) ½kx²
ExplanationPE of a spring = ½kx², derived from W = ∫F dx = ∫₀ˣ kx dx = ½kx².
7. Two bodies of masses m and 2m have the same momentum. The ratio of their kinetic energies is:
Answer
- (a) 1:2
- (b) 2:1
- (c) 1:4
- (d) 4:1
Answer(b) 2:1
Explanationp = mv → v₁ = p/m, v₂ = p/2m. KE₁ = p²/2m, KE₂ = p²/4m. Ratio KE₁:KE₂ = p²/2m : p²/4m = 2:1.
II. Fill in the Blanks
1. Work done is zero when the angle between force and displacement is ________. 90° 2. The SI unit of power is ________. Watt (W) 3. Kinetic energy is always ________ (positive/negative/zero). Positive 4. The work-energy theorem states that work done by net force equals change in ________. Kinetic energy 5. In a perfectly inelastic collision, the coefficient of restitution is ________. Zero 6. 1 kWh = ________ joules. 3.6 × 10⁶ 7. Gravitational potential energy at height h is ________. mgh 8. A force is conservative if work done is ________ of the path. Independent 9. 1 horsepower = ________ watts. 746 10. If momentum doubles, kinetic energy becomes ________ times. Four 11. The total mechanical energy of a freely falling body ________ (increases/decreases/remains constant). Remains constant 12. Friction is a ________ force (conservative/non-conservative). Non-conservative
III. True or False (with corrections)
1. Work and energy are scalar quantities.
- True. Both work and energy have only magnitude and no direction; they are scalars.
2. A body at rest has zero energy.
- False. A body at rest has zero kinetic energy but may have potential energy (e.g., a book on a shelf has gravitational PE = mgh).
3. In an inelastic collision, kinetic energy is conserved.
- False. In an inelastic collision, only momentum is conserved. Kinetic energy is partially lost (to heat, sound, deformation).
4. Work done by friction is always negative.
- False. Friction usually opposes motion (negative work), but in some cases, like a person walking, static friction does positive work on the person. Also, kinetic friction on a block on a moving conveyor can do positive work.
5. The unit of work and energy are the same.
- True. Both are measured in joules. Work is a form of energy transfer.
IV. Match the Following
| Column A | Column B |
|---|
| 1. Work done by centripetal force | (d) Zero |
| 2. Kinetic energy | (f) ½mv² |
| 3. Potential energy (gravity) | (a) mgh |
| 4. Power | (c) W/t |
| 5. Elastic collision | (e) Both KE and momentum conserved |
| 6. Coefficient of restitution | (b) e = (v₂ − v₁)/(u₁ − u₂) |
V. Answer Briefly (2-mark Questions)
1. State and explain the work-energy theorem.
Answer
The work-energy theorem states that the work done by the net force on a body equals the change in its kinetic energy: W_net = ½mv² − ½mu². This means if positive work is done, the body speeds up (KE increases); if negative work is done, the body slows down (KE decreases). The theorem holds whether forces are conservative or non-conservative.
2. Distinguish between conservative and non-conservative forces with examples.
Answer
A conservative force is one for which the work done is independent of the path taken — it depends only on initial and final positions. Examples: gravity, spring force. A non-conservative force is one for which the work done depends on the path. Example: friction. For conservative forces, total mechanical energy is conserved; for non-conservative forces, some mechanical energy is dissipated.
3. What is the coefficient of restitution? Give its values for different types of collisions.
Answer
The coefficient of restitution (e) is the ratio of relative velocity of separation to relative velocity of approach: e = (v₂ − v₁)/(u₁ − u₂). For a perfectly elastic collision, e = 1. For a perfectly inelastic collision, e = 0. For partially inelastic collisions, 0 < e < 1.
4. Define power and express it in terms of force and velocity.
Answer
Power is the rate of doing work: P = W/t. Since W = Fd and v = d/t, we get P = Fd/t = Fv. This is useful when a force moves a body at constant velocity. SI unit: watt (W), where 1 W = 1 J/s.
VI. Answer in Detail (5-mark Questions with Numerical Examples)
1. Derive the expression for kinetic energy and prove the work-energy theorem. Solve: A body of mass 5 kg initially at rest is acted upon by a force of 10 N for 4 seconds. Find the work done and the change in kinetic energy. Verify the work-energy theorem.
Answer
Derivation of KE: Work done on a body of mass m, initially at rest, by a constant force F: W = Fd = mad (from F = ma) Using v² = u² + 2as with u = 0: v² = 2as → a = v²/2s W = m × (v²/2s) × s = ½mv²
Work-energy theorem proof: W_net = F_net × d = ma × d Using v² − u² = 2as: a = (v² − u²)/2s W_net = m × (v² − u²)/2s × s = ½mv² − ½mu² = ΔKE
Numerical:
Given: m = 5 kg, u = 0, F = 10 N, t = 4 s
Step 1: Find acceleration
a = F/m = 10/5 = 2 m/s²
Step 2: Find final velocity
v = u + at = 0 + 2 × 4 = 8 m/s
Step 3: Find displacement
s = ut + ½at² = 0 + ½ × 2 × 16 = 16 m
Step 4: Work done by force
W = F × s = 10 × 16 = 160 J
Step 5: Change in kinetic energy
ΔKE = ½mv² − ½mu² = ½ × 5 × 64 − 0 = 160 J
W = ΔKE = 160 J ✓ (Work-energy theorem verified)
Answer: Work done = 160 J, ΔKE = 160 J
2. Derive expressions for velocity at any height for a body falling freely under gravity. Solve: A body is dropped from a height of 20 m. Find its speed at 5 m above the ground (g = 10 m/s²).
Answer
Derivation: A body of mass m is at height h from the ground. Total mechanical energy at the top: E = PE = mgh, KE = 0.
At any height h' from the ground: PE = mgh', and KE = ½mv'²
By conservation of energy: mgh = mgh' + ½mv'² ½mv'² = mg(h − h') v'² = 2g(h − h') v' = √(2g(h − h'))
This is the same as the kinematic equation v² = u² + 2as with u = 0, a = g, s = (h − h').
Numerical:
Given: h = 20 m, height above ground h' = 5 m, g = 10 m/s², u = 0
Using energy conservation:
mgh = mgh' + ½mv'²
mg × 20 = mg × 5 + ½mv'²
Dividing by m:
g × 20 = g × 5 + ½v'²
10 × 20 = 10 × 5 + ½v'²
200 = 50 + ½v'²
150 = ½v'²
v'² = 300
v' = √300 = 10√3 ≈ 17.32 m/s
Verification using v² = u² + 2gs:
Distance fallen = 20 − 5 = 15 m
v'² = 0 + 2 × 10 × 15 = 300 ✓
Answer: Speed at 5 m above ground = 10√3 ≈ 17.32 m/s
3. Discuss elastic collision in one dimension and derive expressions for final velocities. Solve: A ball of mass 2 kg moving at 6 m/s collides elastically with a ball of mass 4 kg at rest. Find the velocities of both balls after collision.
Answer
Derivation: For elastic collision between two bodies m₁ and m₂ with initial velocities u₁ and u₂:
By conservation of momentum: m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂ ... (i)
By conservation of kinetic energy: ½m₁u₁² + ½m₂u₂² = ½m₁v₁² + ½m₂v₂² ... (ii)
From (ii): m₁(u₁² − v₁²) = m₂(v₂² − u₂²) m₁(u₁ − v₁)(u₁ + v₁) = m₂(v₂ − u₂)(v₂ + u₂) ... (iii)
Dividing (iii) by (i) after rearranging: u₁ − v₁ = v₂ − u₂ (relative velocity of approach = relative velocity of separation) Or: v₂ − v₁ = u₁ − u₂ ... (iv)
Solving (i) and (iv) simultaneously:
v₁ = [(m₁ − m₂)u₁ + 2m₂u₂] / (m₁ + m₂)
v₂ = [(m₂ − m₁)u₂ + 2m₁u₁] / (m₁ + m₂)
Numerical:
Given: m₁ = 2 kg, m₂ = 4 kg, u₁ = 6 m/s, u₂ = 0
v₁ = [(m₁ − m₂)u₁ + 2m₂u₂] / (m₁ + m₂)
= [(2 − 4) × 6 + 2 × 4 × 0] / (2 + 4)
= [−12 + 0] / 6
= −12/6 = −2 m/s
(Negative: m₁ rebounds)
v₂ = [(m₂ − m₁)u₂ + 2m₁u₁] / (m₁ + m₂)
= [(4 − 2) × 0 + 2 × 2 × 6] / (2 + 4)
= [0 + 24] / 6
= 24/6 = 4 m/s
Verification — Momentum conservation:
Before: m₁u₁ + m₂u₂ = 2 × 6 + 4 × 0 = 12 kg·m/s
After: m₁v₁ + m₂v₂ = 2 × (−2) + 4 × 4 = −4 + 16 = 12 kg·m/s ✓
KE conservation:
Before: ½ × 2 × 36 + 0 = 36 J
After: ½ × 2 × 4 + ½ × 4 × 16 = 4 + 32 = 36 J ✓
Answer: v₁ = −2 m/s (rebounds), v₂ = 4 m/s (forward)
Common Mistakes to Avoid
1. Forgetting cos θ in W = Fd cos θ — when force and displacement are not in the same direction, you must include the angle. Work can be positive, negative, or zero depending on θ.
2. Confusing KE and momentum — KE = ½mv² (depends on v²), while p = mv (depends on v). Doubling velocity quadruples KE but only doubles momentum.
3. Applying conservation of energy with friction present — mechanical energy is conserved only when all forces are conservative. If friction is present, some energy is lost as heat; use W_friction = −fₖd on the energy side.
4. Not checking both conservation laws in collisions — always verify both momentum conservation AND kinetic energy conservation for elastic collisions. For inelastic collisions, only momentum is conserved.
5. Sign errors in potential energy — PE = mgh requires choosing a reference level (h = 0). Be consistent with this choice throughout the problem.
6. Mixing up units — convert mass to kg, distance to m, speed to m/s before calculating. Using grams, cm, or km/h without conversion gives wrong answers.
7. Forgetting that work is a scalar — work has no direction, but it can be positive (force along displacement), negative (force opposite to displacement), or zero (force perpendicular to displacement).
8. Confusing perfectly inelastic with inelastic — in perfectly inelastic collisions, the bodies stick together (e = 0). In general inelastic collisions, they separate but KE is not fully conserved (0 < e < 1).
Exam Checklist
- [ ] Can define and calculate work (W = Fd cos θ) with proper angle
- [ ] Can derive KE = ½mv² and PE = mgh
- [ ] Know and can prove the work-energy theorem (W_net = ΔKE)
- [ ] Can solve conservation of energy problems (E = KE + PE = constant)
- [ ] Know the difference between conservative and non-conservative forces
- [ ] Can solve elastic collision problems (both momentum and KE conserved)
- [ ] Can solve perfectly inelastic collision problems (bodies stick together)
- [ ] Know power formulas: P = W/t and P = Fv
- [ ] Remember unit conversions: 1 kWh = 3.6 × 10⁶ J, 1 hp = 746 W
- [ ] Can handle problems involving friction with energy methods (W_friction = fₖ × d)
