1. Work-Energy Theorem Statement: The work done on an object (or system) equals the change in its energy: W = ΔE. When positive work is done on an object, its energy increases; when negative work is done, its energy decreases.
2. Setup for Derivation: Consider an object of mass m starting from rest (initial velocity u = 0) and reaching a final velocity v under a constant force F. Let the displacement be s in the direction of force.
3. Step 1 — Using kinematic equation: v² = u² + 2as. Since u = 0: v² = 2as, so s = v²/(2a).
4. Step 2 — Work done by the force: W = F × s = ma × s (using Newton’s 2nd law F = ma).
5. Step 3 — Substituting s: W = ma × v²/(2a) = ½mv²
6. Step 4 — Applying Work-Energy Theorem: Since the object started from rest, all this work appears as kinetic energy. Therefore: K = ½mv²
7. Conclusion: The kinetic energy of an object of mass m moving with velocity v is K = ½mv². Its SI unit is joule (J). Kinetic energy is always positive and has no direction.

1. Definition: Gravitational potential energy is the energy stored in a system of an object and the Earth due to the position (height) of the object above the Earth’s surface. It is equal to the work done against gravity to raise the object to that height.
2. Derivation of U = mgh: Consider an object of mass m lying on the ground (potential energy = 0 at ground level). To raise it slowly to height h, we apply an upward force equal to the gravitational force: F = mg.
3. Work done: W = F × h = mg × h = mgh
4. By Work-Energy Theorem: This work appears as potential energy of the object. Therefore: U = mgh
5. Activity 7.1 — Sand and Ball: A heavy ball is dropped from heights of 1 m, 2 m, and more onto a bed of loose sand. The depression in the sand is deepest when the ball is dropped from the greatest height. This is because a greater height means greater potential energy (U = mgh), which is converted to kinetic energy during the fall. A faster-moving ball creates a deeper depression, confirming: greater height → greater potential energy.

1. Law of Conservation of Mechanical Energy: When an object moves under the gravitational force and no other external force acts on it, the total mechanical energy (kinetic energy + potential energy) of the object remains constant throughout its motion.
2. Setup: Consider an object of mass m dropped from rest at Point A at height h above the ground. Let Point B be an intermediate point (height h’), and Point C be the ground (height = 0).
3. At Point A (start): KE = 0 (at rest), PE = mgh. Total ME = 0 + mgh = mgh
4. At Point B (intermediate, height h’, velocity v): Using kinematics: v = gt and h’ = h − ½gt². KE = ½mv² = ½mg²t². PE = mgh’ = mgh − ½mg²t². Total ME = (mgh − ½mg²t²) + ½mg²t² = mgh
5. At Point C (just before hitting ground): All PE is converted to KE. PE = 0. KE = mgh. Total ME = 0 + mgh = mgh
6. Conclusion: At every point (A, B, C), the total mechanical energy = mgh = constant. The lost potential energy is exactly converted to kinetic energy. This proves the law of conservation of mechanical energy.

1. At the top of the slide: The child is at rest. KE = 0, PE = mgh. Total ME = mgh.
2. At the bottom of the slide: Height = 0, so PE = 0. All energy is kinetic. KE = ½mv². Total ME = ½mv².
3. Applying conservation: Total ME at top = Total ME at bottom.
4. Equating: mgh = ½mv²
5. Solving for v: Mass m cancels from both sides: gh = ½v² → v² = 2gh → v = √(2gh)

Explanation: Since mass m cancels out in the energy equation, the velocity at the bottom is the same for all children of different masses sliding down the same height h. Similarly, the shape of the slide does not matter — only the vertical height h determines the final velocity (when friction is neglected).

1. Definition: An inclined plane is a simple machine — a flat surface inclined at an angle to the ground — that helps move heavy loads to a higher or lower level with less effort than lifting directly.
2. Setup for Derivation: Let the mass of the object = m. Load = mg (weight). Let F’ = effort (force needed to push object up the incline). Let L = length of the inclined plane. Let h = vertical height to be reached.
3. Work done by effort (along incline): W_effort = F’ × L
4. Potential energy gained by object (vertical height): PE = mgh
5. By Work-Energy Theorem (ignoring friction): F’ × L = mgh, so mg/F’ = L/h
6. Mechanical Advantage: MA = Load/Effort = mg/F’ = L/h

Why longer/shallower is better: Since MA = L/h, a longer inclined plane (larger L) for the same height h gives a greater mechanical advantage. This means a smaller effort F’ is needed to raise the same load. However, the object has to be pushed over a greater distance — total work done remains constant. This is why loading ramps for trucks, roads on hills, and wheelchair ramps are made long and gentle rather than short and steep.

1. Working of a Lever: A lever is a rigid bar that rotates about a fixed point called the fulcrum. When a small force (effort F₁) is applied at one end over a large distance (d₁), it creates a large force (F₂) at the other end over a small distance (d₂).
2. Principle: Work input = Work output (no energy loss). F₁ × d₁ = F₂ × d₂
3. Mechanical Advantage: MA = Load/Effort = F₂/F₁ = d₁/d₂ = effort arm / load arm. By increasing the effort arm, the lever applies a larger force on the load.

Three Classes of Levers:

Conclusion: The inclined plane reduces the effort from 150 N to about 90 N — but the person must push the object along the 50 cm ramp instead of lifting it over 30 cm directly. Total work done remains the same.