1. (a) False. A motionless stretched rubber band has potential energy, not kinetic energy.
2. (b) True.

1. (a) 1 joule
2. (b) Negative

1. We know that KE ∝ m and KE ∝ v².
2. Therefore, by doubling the mass, the kinetic energy doubles, while by doubling the velocity, the kinetic energy increases four times.

1. The girl applies a force equal to the weight of the dumbbell to lift it up. When she moves the dumbbell up, the force is in the direction of displacement, so she does positive work on it.
2. When moving the dumbbell down, the force she applies to hold it is in a direction opposite to the displacement, so she does negative work on it.

1. In order to bring the object to rest, the work done must be equal to the kinetic energy of the moving object.

1. (i) Using energy conservation: mgh = ½mv². Mass cancels: v = √(2gh) = √(2×10×10) = √200 ≈ 14.1 m/s.

(ii) Depth of depression

1. Initial energy: PE = mgh = 1.5 × 10 × 10 = 150 J.
2. Work done by sand: Work = Force × distance ⇒ 150 = 3000 × d.
3. d = 150/3000 = 0.05 m.

1. A battery has stored chemical energy. The chemical energy is converted into electrical energy during the lighting of the bulb.
2. The filament in the bulb becomes white hot and gives out light. Electrical energy is converted into heat and light energy.

1. Force of gravity = mg (vertically downward everywhere). Displacement of the object AB (say s) is in the horizontal direction.
2. Angle between force and displacement is 90°. Work done, W = Fs cos90° = 0.

1. No, it doesn't violate the law of conservation of energy.
2. The potential energy of a freely falling object decreases progressively since it keeps changing into kinetic energy of the falling object, therefore the total energy of the object remains conserved.

1. a.i. At A: Potential energy.
2. a.ii. At B: Potential energy + Kinetic energy.
3. a.iii. At C: Kinetic energy.

b.

1. The initial energy is transformed into heat energy when the pendulum bob strikes the air molecules. Thus, the amplitude of the pendulum decreases and it finally stops.

1. The man is doing work relative to the stream because he is applying force to produce relative motion between the boat and the stream.
2. But he does no work relative to the shore, as the displacement relative to the shore is zero.

1. Work done = force × displacement. For calculating work done, there must be displacement in the direction of force, but here the vertical displacement is zero (as initial and final points lie on the same horizontal line).

1. Initial velocity (u) = 60 km/h = (60×1000)/3600 = 50/3 m/s. Final velocity (v) = 0.
2. Initial kinetic energy = ½ × 1500 × (50/3)² = 208333.30 J. Final kinetic energy = 0.
3. Work done = change in kinetic energy = 208333.30 − 0 = 208333.30 J.

1. Mass of substance m = 50 g. Volume of substance V = 20 cm³.
2. Density D = M/V = 50/20 = 2.5 g cm⁻³.

1. Mass of system, m = mass of girl + mass of trolley = 35+5 = 40 kg. Initial velocity, u = 4 ms⁻¹, Final velocity, v = 0, Displacement, S = 16 m.
2. Using v²−u² = 2aS: a = (v²−u²)/2S = −16/32 = −½ ms⁻². Force = ma = 40 × (−½) = −20 N.
3. (a) Work done on the trolley = Force × Displacement = 20 N × 16 m = 320 J.
4. (b) Since the girl is just sitting on the trolley, work done by the girl = 0 J.

1. Mass of car = 1500 kg. Velocity of car = 60 kmh⁻¹ = (60×1000)/3600 = 16.67 ms⁻¹.
2. W.D. = Change in K.E. of the car = K_f − K_i. K_i = ½ × 1500 × (16.67)² ≈ 208416.68 J. K_f = 0.
3. W = 0 − 208416.68 = −208416.68 J.

1. For the horizontal motion (5 m), the angle between force and displacement is 90°. So W₁ = Fs cosθ = 10 × 5 × cos90° = 0 J.
2. For the vertical motion (10 m), the angle between force and displacement is 0°. So W₂ = Fd cos0° = 10 × 10 × 1 = 100 J.
3. Total work done = W₁ + W₂ = 0 + 100 = 100 J.

1. (a) This situation promotes saving habits and judicial use of resources, and helps save resources overall.
2. (b) The unit of energy is Joule.

(c) Three steps Mohan might have suggested

1. Switch off the power where it is not needed.
2. Use of CFL/LED lights.
3. Check the proper wiring to avoid leakage of power.

1. Kinetic energy is given by KE = ½mv². Since both objects have the same kinetic energy: ½mv_A² = ½(4m)v_B².
2. Cancelling common terms: mv_A² = 4m v_B² ⇒ v_A² = 4v_B² ⇒ v_A = 2v_B.

1. When work is done against internal conservative forces like gravity or elastic force, the energy is stored in the system as potential energy.
2. For example, when we lift an object against gravity or stretch a spring, we do work against these forces. This work is not lost but stored as energy.
3. When the object is released, this stored potential energy is converted into kinetic energy, allowing the system to perform work.

1. Yes, Sony is right, because if several forces acting on an object are balanced forces, i.e. their resultant is zero (ΣF = 0), then there is no acceleration of the object.
2. This may be seen from Newton's second law: ΣF = ma. If ΣF = 0, then ma = 0, or acceleration a = 0.

1. When a child starts from the top of a slide, they have gravitational potential energy. As they move down, this energy converts into kinetic energy.
2. Using conservation of mechanical energy: mgh = ½mv², which simplifies to v = √(2gh).

1. Total mechanical energy at O = KE + PE = 0 + 30 = 30 J (this remains constant throughout, since the track is frictionless).
2. At P (PE = 20 J): KE = 30−20 = 10 J. ½(0.5)v² = 10 ⇒ v² = 40 ⇒ v = √40 ≈ 6.32 m/s.
3. At Q (PE = 30 J): KE = 30−30 = 0 J ⇒ v = 0 m/s.
4. At R (PE = 40 J): Since PE (40 J) exceeds total energy (30 J), the ball cannot physically reach point R — KE would be negative, which is not possible. The ball would stop before reaching R.

1. (a) Chemical energy of food converts to heat and then to muscular energy, which converts into kinetic energy on pedalling the cycle.
2. (b) Anil wisely opts for the bicycle as he is concerned about the environment, to avoid noise and air pollution. Also, cycling is beneficial for health.
3. (c) By informing them about the harms of pollution and putting forward the advantages of cycling.

1. When a moving object collides with a stationary object, it transfers some or all of its kinetic energy to the stationary object.
2. During the collision, the moving object applies a force, doing work on the stationary object. This causes the stationary object to gain energy and start moving or change its shape.
3. This process shows how energy is transferred through work during an impact, following the principle of conservation of energy.

1. (i) The angle between F and S is 90°, so work done is zero.
2. (ii) The angle between F and S is 0°, so work done is positive.
3. (iii) The angle between F and S is 180°, so work done is negative.

1. (i) During upward motion: Displacement is upward, gravity acts downward — work done by gravity is negative. During downward motion: Displacement is downward, gravity acts downward — work done by gravity is positive.

(ii) Work done by air resistance

1. Initial kinetic energy: KE = ½mv² = ½ × 2 × (20)² = 400 J.
2. Potential energy at height 19.4 m: PE = mgh = 2 × 10 × 19.4 = 388 J.
3. Energy lost = Initial KE − Final PE = 400 − 388 = 12 J (lost due to air resistance).

1. Power of Avinash P_A = F_A × v_A = 10 × 8 = 80 W.
2. Power of Kapil P_K = F_K × v_K = 25 × 3 = 75 W.

1. (i) Kinetic energy = ½mv² = ½ × 900 × (30)² = ½ × 900 × 900 = 4,05,000 J.
2. (ii) Energy used = Work done against resistive force = Force × Distance = 2,000 × 30 = 60,000 J (60 kJ). (A resistive force acts opposite to the motion of the object.)
3. (iii) Minimum power = Energy used / Time taken = 60,000/1 = 60,000 W (60 kW).

OR

1. Chemical energy. The fuel in the car provides chemical potential energy, which is converted to kinetic energy when it's combusted.

1. (i) The SI unit of work is joule (J). One joule is defined as the work done when a force of one newton moves an object by one metre in the direction of the force applied.
2. (ii) Work done is zero when there is no displacement, or when the force acts perpendicular to displacement. In both cases, there is no effective movement in the direction of force, so no scientific work is done.
3. (iii) When pushing a wall, no displacement occurs despite applying force. Since scientific work requires displacement in the direction of force, the absence of movement means zero work is done, even though effort is applied and energy is used by the body.

OR

1. Work done depends directly on force and displacement. Increasing force while keeping displacement constant increases work, and similarly, increasing displacement with constant force also increases work. Both factors together determine the total work done on an object.

1. (i) The potential energy of the car remains the same, since PE (= mgh) is independent of velocity.
2. (ii) The kinetic energy of the car becomes four times, since KE = ½mv² is proportional to the square of velocity.
3. (iii) The momentum of the car will get doubled, since momentum (p = mv) is directly proportional to velocity.

OR

1. The sum of the kinetic energy and potential energy of an object is called mechanical energy. The total amount of mechanical energy is simply the sum of the potential energy and the kinetic energy.

1. (a) Kinetic
2. (b) Positive
3. (c) Units (kWh)
4. (d) Friction
5. (e) Potential

1. Work will be done if a force acts on an object and displacement occurs in the direction of that force.
2. a. Suma is doing work; she applies force to move horizontally.
3. b. Donkey is not doing work; the displacement and force are at 90°.
4. c. Work is done by the windmill; the water is lifted against gravity.
5. d. No work is done by a green plant during photosynthesis.
6. e. The engine applies a pulling force on the train, and the train moves in the direction of this force. Therefore, work is done.
7. f. During drying of food grains in the sun, no work is done.
8. g. Work is done by the air; the sailboat moves in the direction of the force exerted by the wind.

1. (i) Between A and B, the graph is a horizontal straight line. Observation: The speed remains constant at 35 m/s from 0 to 1 second. Description: The car is moving with uniform motion (constant velocity), likely while the driver reacts to the obstruction before actually hitting the brakes.
2. (ii) KE = ½mv² = ½ × 1000 × (35)² = 500 × 1225 = 612,500 J (612.5 kJ).
3. (iii) By the Work-Energy Theorem, work done by brakes = change in KE. K_B = 612,500 J, K_C = 0 J. W = K_final − K_initial = 0 − 612,500 = −612,500 J (negative sign shows work is done opposite to motion, to stop the car).
4. (iv) When the brakes are applied, kinetic energy transforms into heat energy (due to friction between brake pads/discs and tires/road) and sound energy (the screeching sound often heard).

1. (i) No, mechanical work is not the only method of energy transfer. Energy can also be transferred as heat, through radiation like sunlight, by electric currents, and through sound waves, showing that energy transfer can occur in multiple ways.
2. (ii) When positive work is done on an object, it gains energy. This increases its capacity to do work. For example, when a force moves an object, it may gain kinetic energy. This stored energy allows the object to apply force on other objects and perform work in different situations.
3. (iii) The gained energy can be transferred when the object interacts with another object. For example, a moving bowling ball has kinetic energy. When it strikes pins, it transfers this energy to them, causing them to move.

1. (i) Energy stored through deformation involves changing the shape of an object, like stretching or compressing it. In contrast, energy stored through relative position depends on how objects are arranged in a system without changing their shape.
2. (ii) When two magnets with unlike poles are pulled apart, work is done against the attractive magnetic force. This work gets stored as potential energy due to their changed positions. When released, the magnets move toward each other, converting stored potential energy into kinetic energy.
3. (iii) No, work done against internal forces does not always increase potential energy. For example, when work is done against friction, the energy is not stored but lost as heat. Friction is a non-conservative force, so it dissipates energy instead of storing it as potential energy in the system.

1. (i) Chemical energy → Kinetic energy + Potential energy
2. (ii) Potential energy (elastic) → Kinetic energy
3. (iii) Light energy → Chemical energy
4. (iv) Potential energy → Kinetic energy
5. (v) Chemical energy → Heat energy + Light energy
6. (vi) Chemical energy → Heat energy + Light energy + Sound energy
7. (vii) Sound energy → Electrical energy
8. (viii) Electrical energy → Light energy + Heat energy
9. (ix) Light energy → Electrical energy

Body falling from height H, showing positions A, B, C with distances x and (H-x)

1. The mechanical energy (KE + PE) of a freely falling object remains constant. Suppose a body of mass m falls from point A, at height H from the surface. Initially at A, KE = 0, so total energy at A = 0 + mgH = mgH …(i)
2. Suppose during the fall, the body is at position B, having fallen a distance x. Using v² = u² + 2as: v² = 0 + 2gx = 2gx. So KE at B = ½mv² = ½m(2gx) = mgx.
3. PE at B = mg(H−x). Total energy at B = mgx + mg(H−x) = mgH …(ii)
4. At point C (just above the ground), PE = 0. Using v² = 0 + 2gH: KE at C = ½mv² = ½m(2gH) = mgH. Total energy at C = mgH + 0 = mgH …(iii)

1. (i) If no work is done on an object and its velocity does not change, its energy remains constant. If its position or shape also does not change, its potential energy stays the same.
2. (ii) Gravitational potential energy increases with height. The higher an object is above the ground, the more potential energy it has, because more work is required to lift the object to a greater height.
3. (iii) The formula U=mgh assumes that the value of gravitational acceleration (g) remains constant. This is true only near the Earth's surface. At greater heights, the value of g changes, so the formula is no longer accurate.

1. (i) A moving object acquires energy when work is done on it. When a force is applied over a distance, energy is transferred to the object, usually as kinetic energy, enabling it to move and perform further work.
2. (ii) When a moving object collides with another object, it exerts a force on it, causing displacement. This means work is done on the second object. At the same time, energy is transferred from the first object to the second, changing its motion or position.
3. (iii) The work-energy theorem states that the total work done on an object is equal to the change in its energy. This remains true even if the force is not constant. In such cases, total work is calculated over the displacement, and it still equals the overall change in the object's energy.

1. (i) The principle of conservation of mechanical energy states that the total mechanical energy (potential + kinetic) remains constant during free fall if only gravity acts.
2. (ii) When an object falls, it initially has maximum potential energy and no kinetic energy. As it falls, potential energy decreases while kinetic energy increases. By the time it reaches the ground, most of the energy has converted into kinetic energy.
3. (iii) In real-life conditions, air resistance and friction act on the falling object. These forces convert some mechanical energy into heat and sound, so total mechanical energy is not conserved.

1. (i) The work-energy theorem states that the work done on an object equals the change in its energy. This is based on the law of conservation of energy, which says energy cannot be created or destroyed. When work is done, energy is transferred, and the total energy remains constant, only changing forms.
2. (ii) Understanding energy transfer is important because it helps explain many real-life processes. In electricity, it helps in designing efficient circuits and devices. In sound, it explains how vibrations travel. This knowledge supports technological development, improves energy use, and helps us understand how different physical systems are connected through energy.

1. (i) Initial kinetic energy K₁ = ½mv² …(1)
2. New kinetic energy (mass = 4m, velocity = v/2): K₂ = ½ × 4m × (v/2)² = ½ × 4m × v²/4 = ½mv² …(2)
3. From (1) and (2): K₁ : K₂ = 1 : 1.
4. (ii) When a sparkle is lighted, chemical energy is converted into heat and light energy.

1. (i) The energy possessed by a body by virtue of its motion is called its kinetic energy. No, the kinetic energy of an object cannot be negative because both m and v² are always positive: KE = ½mv².

(ii) Work done by the car engine

1. Given: mass m = 1200 kg, displacement s = 40 m, time t = 5 s, initial velocity u = 0.
2. Using s = ut + ½at²: 40 = 0 + ½ × a × (5)² = 12.5a ⇒ a = 40/12.5 = 3.2 ms⁻².
3. W = Fs = ma × s = 1200 × 3.2 × 40 = 153,600 J.

1. (i) Work is said to be done only when the point of application of a force moves in the direction of the applied force. Mathematically, work = Force × Displacement.

Two factors on which work depends

1. The magnitude of applied force.
2. Displacement in the direction of the force.

(ii) Ratio of kinetic energy

1. Let masses of car and truck be 2m and 3m. KE of car = ½(2m)(30)² = 900m. KE of truck = ½(3m)(30)² = 1350m.

1. According to the law of conservation of energy, energy is neither created nor destroyed; it can only be changed from one form to another.
2. Consider a body of mass m at rest at A, at an elevation h from the ground. Total energy at A = PE + KE = mgh + 0 = mgh.
3. As the body falls and strikes the ground at B, its elevation becomes 0, so PE = 0. Using v² = u² + 2gh (with u=0): v² = 2gh.
4. KE at B = ½mv² = ½m(2gh) = mgh. Total energy at B = mgh + 0 = mgh.
5. At an intermediate point C, having fallen through x: PE at C = mg(h−x). Using v² = 2gx: KE at C = ½mv² = ½m(2gx) = mgx. Total energy at C = mg(h−x) + mgx = mgh.

Derivation

1. Kinetic energy is the energy possessed by virtue of the motion of a body. Let a body of mass m be at rest. A force F N acts on it and produces an acceleration a m/s², and it acquires a velocity v m/s after moving a distance s metres.
2. Using the third equation of motion: v² = u² + 2as. Since u=0: v² = 2as ⇒ a = v²/2s.
3. From Newton's second law: F = ma = m(v²/2s) = mv²/2s.
4. Work done on the moving body = Kinetic energy = Force × Distance = (mv²/2s) × s = ½mv².

Calculation

1. Given: Mass = 5 kg, Velocity = 2.5 m/s.
2. K.E. = ½mv² = ½ × 5 × (2.5)² = ½ × 5 × 6.25 = 3125/200 = 15.625 J.