Conservation of Energy – NTSE Physics Study Notes
(Suitable for Class 10–12 students preparing for the National Talent Search Examination.)
1. Simple Explanation
Conservation of Energy states that in an isolated system (no external forces doing work) the total mechanical energy – the sum of kinetic energy (KE) and potential energy (PE) – remains constant.
- Kinetic Energy (KE): Energy possessed by a moving body.
[ KE = \tfrac12 m v^2 ] - Potential Energy (PE): Energy stored due to position or configuration.
- Gravitational PE (above a reference point):
[ PE_g = mgh ] - Elastic PE (spring):
[ PE_s = \tfrac12 k x^2 ]
- Gravitational PE (above a reference point):
If no non‑conservative forces (like friction, air resistance) are doing work, the mechanical energy (E = KE + PE) is constant:
[ E_{\text{initial}} = E_{\text{final}} ]
2. Key Points
| Concept | What to remember |
|---|---|
| Work–Energy Theorem | Net work done on a body = change in its kinetic energy: (W_{\text{net}} = \Delta KE). |
| Mechanical Energy Conservation | For a frictionless system: (KE_i + PE_i = KE_f + PE_f). |
| Potential Energy Functions | Use (PE = mgh) for gravity, (PE = \tfrac12 kx^2) for springs. |
| Non‑conservative Work | Work done by friction or air resistance appears as energy lost to heat: (W_{\text{fric}} = -\mu_k N , d). |
| Energy Loss | In real problems, account for energy dissipated: (E_{\text{lost}} = W_{\text{fric}}). |
| Units | Joule (J) = N·m = kg·m²/s². |
| Reference Point | Choose a convenient zero for PE (e.g., ground level for gravity). |
| Conservation in Systems | Even if mechanical energy isn’t conserved, total energy (including thermal, sound, etc.) is conserved. |
| Kinetic Energy of Rotational Bodies | For a rotating object: (KE_{\text{rot}} = \tfrac12 I \omega^2). |
| Energy Diagrams | Helpful to visualize energy changes (e.g., hill‑shaped KE vs. height). |
3. Worked Example
Problem
A 3 kg block is released from rest at a height of 5 m above the ground. It slides down a smooth (frictionless) incline of 30° to reach the ground. Find the speed of the block when it hits the ground.
Step‑by‑Step Solution
-
Identify energies at start (top).
- KE(_i) = 0 (block starts from rest).
- PE(_i) = (mgh = 3 \times 9.8 \times 5 = 147) J.
-
Identify energies at end (bottom).
- PE(_f) = 0 (ground level).
- KE(_f) = unknown (\tfrac12 m v^2).
-
Apply conservation of mechanical energy.
[ KE_i + PE_i = KE_f + PE_f ] [ 0 + 147 = \tfrac12 (3) v^2 + 0 ] -
Solve for (v).
[ 147 = 1.5, v^2 \quad\Rightarrow\quad v^2 = \frac{147}{1.5} = 98 ] [ v = \sqrt{98} \approx 9.9 \text{ m/s} ]
Answer: The block strikes the ground with a speed of about 9.9 m s⁻¹.