Work and Energy

Concept: Work is done when a force acts on an object AND the object is displaced in the direction of (or having a component along) the applied force. W = F × d × cos θ.

1. Suma is swimming in a pond.
Suma applies force on water with her arms and legs, and her body moves forward (displacement occurs in the direction of applied force). Work is done.

2. A donkey is carrying a load on its back.
The donkey exerts an upward force (normal reaction) to support the load, but the displacement of the load is horizontal. The angle between force (vertical) and displacement (horizontal) is 90°, so W = F × d × cos 90° = 0. Work is NOT done by the donkey on the load (in the scientific sense).

3. A wind-mill is lifting water from a well.
The windmill applies force on water and the water is displaced upward (in the direction of force). Work is done.

4. A green plant is carrying out photosynthesis.
Photosynthesis is a biochemical process. There is no mechanical force causing displacement of an object. Work is NOT done (in the mechanical sense).

5. An engine is pulling a train.
The engine applies force on the train and the train moves in the direction of the force. Work is done.

6. Food grains are getting dried in the sun.
Drying is a physical/chemical process involving heat energy. No mechanical force causes displacement of the grains. Work is NOT done (in the mechanical sense).

7. A sailboat is moving due to wind energy.
Wind exerts force on the sail and the boat is displaced in the direction of the force. Work is done.

Given:
- Object is thrown at an angle and returns to the same horizontal level.
- Initial point and final point are on the same horizontal line.

Concept: Work done by gravity = $W = mgh$, where $h$ is the vertical displacement (change in height).

Working:
Since the initial and final points lie on the same horizontal line, the net vertical displacement of the object is:
$$h = h_{\text{final}} - h_{\text{initial}} = 0$$

Therefore, work done by gravity:
$$W = mgh = mg \times 0 = 0$$

Answer: The work done by the force of gravity on the object is zero.

Energy transformations when a battery lights a bulb:

Step 1: The battery contains stored chemical energy.

Step 2: When the circuit is complete, chemical energy of the battery is converted into electrical energy (electric current flows through the circuit).

Step 3: When electric current passes through the filament of the bulb, electrical energy is converted into heat energy and light energy.

Summary of transformation:
$$\text{Chemical Energy} \rightarrow \text{Electrical Energy} \rightarrow \text{Heat Energy + Light Energy}$$

Given:
- Mass, $m = 20$ kg
- Initial velocity, $u = 5$ m s$^{-1}$
- Final velocity, $v = 2$ m s$^{-1}$

Concept: By the work-energy theorem, the work done by a force equals the change in kinetic energy of the object.
$$W = \Delta KE = \frac{1}{2}mv^2 - \frac{1}{2}mu^2$$

Calculation:
$$W = \frac{1}{2} \times 20 \times (2)^2 - \frac{1}{2} \times 20 \times (5)^2$$
$$W = \frac{1}{2} \times 20 \times 4 - \frac{1}{2} \times 20 \times 25$$
$$W = 40 - 250$$
$$W = -210 \text{ J}$$

Answer: The work done by the force is $\mathbf{-210}$ J. The negative sign indicates that the force acts opposite to the direction of motion (it decelerates the object).

Given:
- Mass of object, $m = 10$ kg
- The object is moved from point A to point B on a horizontal surface.
- The line AB is horizontal.

Concept: Work done by gravity = $W = mgh$, where $h$ is the vertical displacement.

Explanation:
Gravitational force acts vertically downward. When the object moves horizontally from A to B, there is no change in its height, i.e., the vertical displacement $h = 0$.

The angle between the gravitational force (downward) and the displacement (horizontal) is 90°.
$$W = F \times d \times \cos 90° = mgh = mg \times 0 = 0$$

Answer: The work done on the object by the gravitational force is zero, because the displacement is perpendicular to the direction of gravitational force (there is no vertical displacement).

Answer: No, this does not violate the law of conservation of energy.

Explanation:
When an object falls freely from a height, its potential energy decreases. However, as it falls, its velocity increases, so its kinetic energy increases progressively.

According to the law of conservation of energy, energy can neither be created nor destroyed; it can only be transformed from one form to another.

During free fall:
- The potential energy (= $mgh$) decreases as height $h$ decreases.
- The kinetic energy (= $\frac{1}{2}mv^2$) increases as velocity $v$ increases.
- The total mechanical energy (KE + PE) remains constant at every point.

$$KE + PE = \text{constant}$$

Thus, the decrease in potential energy is exactly equal to the increase in kinetic energy. The law of conservation of energy is fully obeyed.

Energy transformations while riding a bicycle:

Step 1: The rider's body uses chemical energy (stored in food/muscles).

Step 2: The rider's muscles convert chemical energy into mechanical energy (muscular effort applied to the pedals).

Step 3: The mechanical energy of the pedals is transferred to the wheels through the chain, causing the bicycle to move — this is kinetic energy of the bicycle.

Step 4: Some energy is also lost as heat energy due to friction between the tyres and the road, and in the chain and bearings.

Summary:
$$\text{Chemical Energy} \rightarrow \text{Mechanical (Muscular) Energy} \rightarrow \text{Kinetic Energy of Bicycle + Heat Energy (due to friction)}$$

Answer: In the scientific sense, no work is done on the rock because the rock does not move (displacement = 0). Therefore, no energy is transferred to the rock.

However, the person does spend energy. This energy is used by the muscles of the body. When we push the rock, our muscle fibres undergo repeated contraction and relaxation. The chemical energy stored in the body (from food) is used up and converted into heat energy, which causes the body to feel warm and tired.

Conclusion: The energy spent by the person is converted into heat energy within the body (muscles). No mechanical work is done on the rock since there is no displacement.

Given:
- Energy consumed = 250 units

Concept: 1 unit of electrical energy = 1 kilowatt-hour (kWh)
$$1 \text{ kWh} = 1000 \text{ W} \times 3600 \text{ s} = 3.6 \times 10^6 \text{ J}$$

Calculation:
$$\text{Energy} = 250 \text{ units} = 250 \text{ kWh}$$
$$= 250 \times 3.6 \times 10^6 \text{ J}$$
$$= 9 \times 10^8 \text{ J}$$

Answer: The energy consumed is $\mathbf{9 \times 10^8}$ J.

Given:
- Mass, $m = 40$ kg
- Height, $h = 5$ m
- $g = 10$ m s$^{-2}$

Part (i): Potential Energy at height 5 m
$$PE = mgh = 40 \times 10 \times 5 = 2000 \text{ J}$$

Part (ii): Kinetic Energy at half-way down (i.e., at height = 2.5 m)

At half-way down, height = $\frac{5}{2} = 2.5$ m

Using the law of conservation of energy:
$$\text{Total Energy} = KE + PE = \text{constant} = 2000 \text{ J}$$

Potential energy at half-way point:
$$PE_{\text{half}} = mgh' = 40 \times 10 \times 2.5 = 1000 \text{ J}$$

Therefore, kinetic energy at half-way point:
$$KE = \text{Total Energy} - PE_{\text{half}} = 2000 - 1000 = 1000 \text{ J}$$

Answer:
- Potential energy at height 5 m = 2000 J
- Kinetic energy at half-way down = 1000 J

Answer: The work done by the force of gravity on a satellite moving around the earth is zero.

Justification:
For a satellite in a circular orbit around the earth, the gravitational force acts towards the centre of the earth (centripetal direction), which is always perpendicular to the direction of motion (displacement) of the satellite at every point.

Since the angle $\theta$ between the gravitational force and the displacement is always 90°:
$$W = F \times d \times \cos 90° = 0$$

Also, in a circular orbit, the satellite returns to the same point after one revolution, so the net displacement in the direction of gravity is zero.

Therefore, the work done by gravity on the satellite is zero.