How much energy is required to launch a rocket into orbit? Why does a compressed spring store energy that can propel a pinball? What happens to kinetic energy during an automobile crash? Work, Energy and Power forms Chapter 6 of the CBSE Class 11 Physics syllabus and provides the scalar energy conservation framework essential for solving complex mechanics problems in JEE Main, JEE Advanced, and NEET.
This comprehensive guide covers work done by constant and variable forces, calculus proof of the Work-Energy Theorem, kinetic energy, conservative vs non-conservative forces, gravitational and spring potential energy, conservation of mechanical energy, vertical circular motion, instantaneous power, 1D and 2D collisions, coefficient of restitution, and five step-by-step solved entrance exam numericals.
- 1. Work Done by Constant & Variable Forces
- 2. Kinetic Energy & Calculus Proof of Work-Energy Theorem
- 3. Potential Energy & Spring Potential Energy (Us = ½kx²)
- 4. Law of Conservation of Mechanical Energy & Vertical Circle
- 5. Instantaneous Power (P = F · v) & Units
- 6. Physics of Collisions: Elastic, Inelastic & Restitution (e)
- 7. Solved Entrance Exam Numerical Problems (JEE / NEET)
- 8. Frequently Asked Questions (FAQ)
1. Work Done by Constant & Variable Forces
In physics, Work is done when a force acts on an object and causes a displacement in the direction of the force.
W = F · s = F s cosθ (SI Unit: Joule = N·m)
• Positive Work (θ < 90°): Force aids motion (e.g. gravity on falling body).
• Negative Work (θ > 90°): Force opposes motion (e.g. friction).
• Zero Work (θ = 90° or s = 0): Force perpendicular to displacement (e.g. centripetal force).
W = ∫_{x1}^{x2} F(x) dx
Geometrical Meaning: Work done equals the Area under the Force-Displacement (F-x) Graph bounded by initial x1 and final x2 positions.
2. Kinetic Energy & Calculus Proof of Work-Energy Theorem
Kinetic Energy (K) is the energy possessed by a body by virtue of its motion: K = ½m v² = p² / (2m).
Proof:
1. Work done by small displacement dx: dW = F dx = (m dv/dt) dx = m (dx/dt) dv = m v dv.
2. Integrating from initial velocity u to final velocity v:
W = ∫_{u}^{v} m v dv = m [ v² / 2 ]_u^v = ½ m v² - ½ m u².
W_net = K_final - K_initial = ΔK (Hence Proved!).
3. Potential Energy & Spring Potential Energy (Us = ½kx²)
Potential Energy (U) is the energy stored in a body due to its position, configuration, or state of strain in a conservative force field.
| Category | Conservative Force | Non-Conservative Force |
|---|---|---|
| Work Dependence | Work done depends ONLY on initial and final positions (path independent). | Work done depends on the actual path taken. |
| Closed Loop Work | Work done in a closed path is zero (∮ F · ds = 0). | Work done in a closed path is non-zero (∮ F · ds ≠ 0). |
| Examples | Gravitational force, Spring force, Electrostatic force (F = -dU/dx). | Frictional force, Viscous force, Air resistance. |
Work done against spring force to extend it by x:
W = ∫_0^x (-F_s) dx = ∫_0^x k x dx = k [ x² / 2 ]_0^x = ½ k x².
U_s = ½ k x² (Stored Elastic Potential Energy).
4. Law of Conservation of Mechanical Energy & Vertical Circle
In an isolated system under the action of conservative forces, the total mechanical energy (E = K + U) remains constant at all points during motion.
• At top point A: K_A = 0, U_A = mgh ⇒ E_A = mgh.
• At midpoint B (height h-x): K_B = mgx, U_B = mg(h-x) ⇒ E_B = mgh.
• At ground C (height 0): K_C = mgh, U_C = 0 ⇒ E_C = mgh.
E_A = E_B = E_C = mgh (Conserved!).
• Minimum Speed at Lowest Point (L):
v_L = √(5 g r) (String tension TL = 6 mg)
• Minimum Speed at Highest Point (H):
v_H = √(g r) (String tension TH = 0)
5. Instantaneous Power (P = F · v) & Units
Power is defined as the time rate of doing work or transferring energy.
- Average Power: P_avg = ΔW / Δt.
- Instantaneous Power: P = dW / dt = (F · ds) / dt = F · v = F v cosθ.
- SI Unit: Watt (1 W = 1 J/s = 1 kg·m²/s³). Commercial unit: Kilowatt-hour (1 kWh = 3.6 × 10⁶ J). Horsepower (1 hp = 746 W).
6. Physics of Collisions: Elastic, Inelastic & Restitution (e)
A Collision is an isolated event in which two or more colliding bodies exert relatively strong forces on each other for a relatively short time.
| Collision Type | Momentum Conserved? | Kinetic Energy Conserved? | Coefficient of Restitution (e) |
|---|---|---|---|
| Elastic Collision | YES (Σp_i = Σp_f) | YES (ΣK_i = ΣK_f) | e = 1 (e.g. subatomic particle collisions). |
| Inelastic Collision | YES (Σp_i = Σp_f) | NO (KE converted to heat/sound) | 0 < e < 1 (Real-world collisions). |
| Perfectly Inelastic | YES (Σp_i = Σp_f) | NO (Maximum KE Loss) | e = 0 (Bodies stick together, v = (m1 u1 + m2 u2) / (m1 + m2)). |
For 1D Elastic Collision (e = 1):
• v1 = [ (m1 - m2) / (m1 + m2) ] u1 + [ (2 m2) / (m1 + m2) ] u2
• v2 = [ (2 m1) / (m1 + m2) ] u1 + [ (m2 - m1) / (m1 + m2) ] u2
Special Case: If m1 = m2, the colliding bodies INTERCHANGE THEIR VELOCITIES! (v1 = u2, v2 = u1).
7. Solved Entrance Exam Numerical Problems (JEE / NEET)
Initial velocity at x = 0: u = 5(0)^(3/2) = 0.
Final velocity at x = 2: v = 5(2)^(3/2) = 5 × 2√2 = 10√2 m/s.
By Work-Energy Theorem: W = ΔK = ½ m v² - ½ m u² = ½ × 0.5 × (10√2)² - 0 = 0.25 × 200 = 50 J.
Initial PE U1 = ½ k x1² = ½ × 800 × (0.05)² = 400 × 0.0025 = 1.0 J.
Final PE U2 = ½ k x2² = ½ × 800 × (0.10)² = 400 × 0.01 = 4.0 J.
Additional Work Required ΔW = U2 - U1 = 4.0 - 1.0 = 3.0 J.
Time t = 15 min = 15 × 60 = 900 s.
Useful output power P_out = (m g h) / t = (30,000 × 10 × 40) / 900 = 12,000,000 / 900 = 13,333.33 W = 13.33 kW.
Since Efficiency η = (P_out / P_in) × 100% ⇒ 30% = 0.30.
Electric Input Power P_in = P_out / 0.30 = 13.33 kW / 0.30 = 44.44 kW.
(a) Perfect Inelastic Collision: v = (m1 u1 + m2 u2) / (m1 + m2) = (0.02 × 500 + 0) / (0.02 + 0.98) = 10 / 1.0 = 10 m/s.
(b) Initial KE Ki = ½ m1 u1² = ½ × 0.02 × (500)² = 0.01 × 250,000 = 2500 J.
Final KE Kf = ½ (m1 + m2) v² = ½ × 1.0 × (10)² = 50 J.
Loss in KE ΔK = Ki - Kf = 2500 - 50 = 2450 J (98% energy lost!).
Velocity of separation after rebound v = √(2 g h2) = √(2 × 9.8 × 6.4) = √(125.44) = 11.2 m/s.
Coefficient of Restitution e = v / u = √(h2 / h1) = √(6.4 / 10) = √(0.64) = 0.8.
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8. Frequently Asked Questions (FAQ)
The Work-Energy Theorem states that the net work done by all forces (conservative, non-conservative, internal, and external) acting on a particle equals the change in its kinetic energy: W_net = ΔK.
A force is conservative if work done by it depends only on initial and final positions, independent of path (e.g. Gravity, Spring force). A non-conservative force depends on the path taken and dissipates mechanical energy into heat/sound (e.g. Friction, Air resistance).
The elastic potential energy stored in a spring stretched or compressed by displacement x is Us = ½k x², where k is the spring stiffness constant.
To complete a full vertical circle of radius r without string slackening, the minimum initial speed required at the lowest point is v_L = √(5 g r).
In an Elastic collision (e = 1), total kinetic energy is completely conserved. In a Perfectly Inelastic collision (e = 0), the colliding bodies stick together, resulting in the maximum possible loss of kinetic energy.
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