Work ,Energy and Power

Work is a very common term used in our daily life. In a general sense, when something is achieved by someone, it is called work done, but, in science, work done is expressed as the magnitude of the force applied on an object multiplied by the distance moved in the direction of the force. Work has only magnitude and no direction hence it is scalar quantity.

Work

Work done on an object is defined as the product of the magnitude of the force acting on the body and the displacement in the direction of the force which is expressed as W = F.s. The SI unit of force is Newton. If a force acting on a body causes no displacement, the work done is ZERO. For example, trying to push a wall has no net effect, hence it is considered as ZERO work done.The force component F cos θ gives the component of force along the direction in that the body is displaced. Cos θ is the angle between the force vector and displacement vector.

Conditions for work to be done:

1. A force must act on the body.
2. The body must be displaced.
3. There must be a component of force in the direction of displacement.

Types of Work:

1. Positive Work: When force and displacement are in the same direction (e.g., lifting an object).
2. Negative Work: When force and displacement are in opposite directions (e.g., friction).
3. Zero Work: When displacement is zero or force is perpendicular to displacement (e.g., circular motion)

Energy

Energy is the ability to do work. Work done by the energy may result in the change in the state , shape or direction of an object.However, this is true only if the the magnitude of energy is greater than the mass and gravity on an object. Sun is the ultimate spurce of all energies however,it exists in various forms like kinetic, potential, thermal, and chemical. The concept of energy can be clearly explained based on the law of conservation of energy, which states that energy can be transferred or converted from one form to another, but it cannot be created or destroyed. In physics, energy is a quantitative property, and its standard unit of measurement is the joule (J).SI unit of energy or work = Joule (Nm) or Kgm2s−2.

Different Forms of Energy

Energy has different forms: Light, heat, chemical, electrical or mechanical. Mechanical energy is the sum of Kinetic energy (K.E)

1. Chemical Energy: Energy stored in the bonds of molecules, found in food, fuel, and batteries.
2. Thermal Energy: Energy from the vibration of atoms and molecules. It's what we experience as heat.
3. Electrical Energy: Energy from the movement of charged particles, like electrons.
4. Radiant Energy: Energy that travels in waves, such as light and radio waves. Sunlight is a form of radiant energy.
5. Nuclear Energy: Energy stored within the nucleus of an atom.
6. Energy: Energy carried by sound waves, produced by vibrations.
7. Mechanical Energy: The total energy of motion and position in a system. A turning windmill has mechanical energy.
8. Gravitational Energy: Stored energy related to an object's height in a gravitational field.

Kinetic Energy

Kinetic energy is the energy possessed by a body due to its motion. Any object that is moving has kinetic energy. It is expressed as : KE=1/2mv² Where m is the mass of a moving body and v is the velocity. Which ,means The faster an object moves, the more kinetic energy it has. The heavier (more massive) an object is, the more kinetic energy it can store at the same speed. Kinetic energy is a scalar quantity (it has magnitude but no direction). A moving car has kinetic energy due to its speed. Similarly, a cricket ball thrown by a bowler has kinetic energy.

Work-Energy Theorem

The work done on an object is equal to the change in its kinetic energy. W = ΔKE = KE_final - KE_initial

Derivation on work done

Work done on a body, W = F · d. As per the Newton’s Second Law, F = ma, so W = ma · .Using kinematic equation v² - u² = 2ad → ad = (v² - u²)/2. Substitute in work equation W = m[(v² - u²)/2] = (1/2)mv² - (1/2)mu² . Therefore;: W = KE_final - KE_initial. This proves that: Work done on an object = Change in Kinetic Energy.

Factors affecting Kinetic Energy

1. Mass of the object : Kinetic energy is directly proportional to mass of thwe object on which the force is applied. To make it easy to understand, under a constant velocity, if we double the mass of the obeject, the kinetic energy is also doubled. Example: A truck moving at a given speed has more kinetic energy than a car moving at the same speed, because its mass is greater.

2. Velocity of the object: Kinetic energy depends on the square of the velocity.This means even a small increase in velocity results in a large increase in kinetic energy. Example: If velocity is doubled, the kinetic energy becomes four times greater.

Potential Energy

It is the energy due to the tendency of an object to inertia( rest). Suppose an object is raised to a certain height, work is done against gravity to change its position. This energy is stored as Potential Energy. It is expressed as W = F.s. Since F = ma, in the case of increasing the height, it becomes F = mg. Therefore, W (P.E) = mgh and the net potential energy , ΔPE=mg(h final−h initial). For instance, Suppose you lift a book of mass 2 kg to a height of 5 m above the ground. The potential energy is 9.8 m/s² because , it is given that the mass (m) = 2 kg and the height (h) = 5 m so the acceleration due to gravity (g) = 9.8 m/s².

Law of Conservation of Energy

The law states that, energy can neither be created nor destroyed; it can only change from one form to another. The total energy of an isolated system always remains constant. I,e P.E+KE=0. In any physical or chemical process, energy may transform from one form to another (for example, potential energy to kinetic energy, or electrical energy to heat). However, the total energy before and after the transformation remains the same.

Examples

1. Pendulum Motion

At the highest point: all energy is Potential Energy (PE). At the lowest point: all energy is Kinetic Energy (KE). At any point: PE+KE=Constant PE + KE =Constant

2. Free Fall of an Object

When an object falls from a height, its potential energy decreases but its kinetic energy increases. The sum of PE and KE remains constant throughout the fall (ignoring air resistance).

3. Hydroelectric Power Plant

Water stored at height has potential energy.As water falls, PE → KE.Turbines convert KE → Mechanical energy → Electrical energy. Total energy is conserved in the process.

Power

The rate of doing work or the rate of transfer of energy is called power. It is denoted by P. The expression for power is written as , P=W/t. Where W is the work done and t is the time at which work is finished. Average power = Total energy consumed/Total time taken. The SI Unit of power is Watt (W). 1 Watt = 1 Joule of work done in 1 second. 1000 joules is written as 1 kilojoule. However, the commercial unit of power is kWh, i.e. energy used in 1 hour at 1000 Joules/second. 1kWh=3.6×106J.

 

Summary Table

Quantity	Definition	Formula	Unit	SI Unit Symbol
Work	Force × displacement in the direction of force	W = F × s × cos θ	Joule	J
Kinetic Energy	Energy possessed by a body due to its motion	K = ½ m v²	Joule	J
Potential Energy	Energy possessed by a body due to its position or configuration	U = m g h	Joule	J
Power	Rate at which work is done or energy is transferred	P = W / t	Watt	W

TEXTBOOK EXERCISES

Question 1. A force of 7N acts on an object. The displacement is, say 8m, in the direction of the force. Let us take it that they force acts on the object through the displacement. What is the work done in this case?

Answer: Work done on an object = 7N × 8m = 56 Nm or 56 J.

Question 2. When do we say that work is done?
Answer: Two conditions need to be satisfied for work to be done: A force should act on an object and the object must be displaced.

Question 3. Write an expression for the work done when a force is acting on an object in the direction of its displacement.
Answer: Let a constant force F act on an object. Let the object be displaced through a distance S in the direction of the force. Let W be work done. We define work to be equal to the product of the force and displacement. Therefore, work done = force × displacement W = F × S.

Question 4. Define 1 J of work.
Answer: 1 J is the amount of work done on an object when a force of 1 N displaces it by 1 m along the line of action of the force.

Question 5. A pair of bullocks exerts a force of 140 N on a plough. The field being ploughed is 15m long. How much work is done in ploughing the length of the field?
Answer:
Force exerted, F = 140N.
Displacement, S = 15m
Work done in ploughing the field
W = F × S
= 140 × 15
= 2100J
= 2.1 × 10³ J.Free Kinetic Energy Calculator – calculate kinetic energy step by step.

Question 6. What is the kinetic energy of an object?
Answer: Objects in motion possess energy, we call this energy kinetic energy.

Question 7. Write an expression for the kinetic energy of an object.
Answer:
KE = 1/2 mv².

Question 8. The kinetic energy of an object of mass, m moving with a velocity of 5 ms_-1 is 25 J. What will be its kinetic energy when its velocity is doubled? What will be its kinetic energy when its velocity is increased three times?

Formula

KE = 1/2 m v²

Step 1: Find the Mass of the Object

25 = 1/2 × m × (5²)
25 = 1/2 × m × 25
25 = 12.5m
m = 2 kg

Step 2: When Velocity is Doubled (v = 10 m/s)

KE = 1/2 × 2 × (10²)
KE = 1 × 100
KE = 100 J

Step 3: When Velocity is Tripled (v = 15 m/s)

KE = 1/2 × 2 × (15²)
KE = 1 × 225
KE = 225 J

Final Answer

• Kinetic Energy at double velocity = 100 J
  • Kinetic Energy at triple velocity = 225 J

Key Point

Kinetic energy depends on the square of velocity.
- If velocity is doubled → KE becomes 4 times.
- If velocity is tripled → KE becomes 9 times.

Question 9. What is power?
Answer:Power is defined as the rate of doing work or the rate of transfer of energy w
P = w/t

Question 10. Define 1 watt of power.
Answer:
Power is 1 W when the rate of consumption of energy is 1 JS^-1.

Question 11. A lamp consumes 1000 J of electrical energy in 10s. What is its power?
Answer:
Power = 1000 J Work w = 1000J
Time = 10 s
Power of lamp, P = w/t
= 1000/10 = 100W
∴ Power of a lamp = 100 KW.

Question 12. Define average power.
Answer: We obtain average power by dividing the total energy consumed by the total time taken.

Textbook Exercises-II

Question 1. Look at the activities listed below. Reason out whether or not work is done in the light of your understanding of the term ‘work’

1. Suma is swimming in a pond.
2. A donkey is carrying a load on its back.
3. A windmill is lifting water from a well.
4. A green plant is carrying out photosynthesis.
5. An engine is pulling a train.
6. Food grains are getting dried in the sun.
7. A sailboat is moving due to wind energy.

Answer:

1. Work is being done by Seema because she displaces the water by applying the force.
2. No work is being done by the donkey because the direction of force i.e the load is vertically downward and displacement is along the horizontal. If displacement and force are perpendiculars then no work is done.
3. Work is done because the windmill is lifting the water i.e., it is changing the position of water.
4. No work is done because there are no force and displacement.
5. Work is done because the engine is changing the position of the train.
6. No work’ is done because there is no force and no displacement.
7. Work is done because the force acting on the boat is moving it.

Question 2. An object was thrown at a certain angle to the ground moves in a curved path and falls back to the ground the initial and the final points of the path of the object lie on the same horizontal line. What is the work done by the force of gravity on the object?
Answer:
Work done by the force of gravity, W = mgh.
Where h = difference in height of initial and final positions of the object.
According to the question, the initial and final positions of the object lie in the same horizontal line. So h = 0.
∴ Work done W = mg × 0 = 0

Question 3. A battery lights a bulb. Describe the energy changes involved in the process.
Answer: In the case given in the question, the battery has chemical energy which is converted into energy. Electric energy provided to the bulb further converted into light energy.

Question 4. A certain force acting on a 20 kg mass changes its velocity from 5 ms_-1 to 2 ms_-1. Calculate the work done by the force.
Answer:

Concept: Work-Energy Theorem. Work done (W) = ΔKE = KE_final - KE_initial

Step 1: Initial Kinetic Energy

KE_initial = 1/2 m v²
KE_initial = 1/2 × 20 × (5²)
KE_initial = 10 × 25 = 250 J

Step 2: Final Kinetic Energy<

KE_final = 1/2 m v²
KE_final = 1/2 × 20 × (2²)
KE_final = 10 × 4 = 40 J

Step 3: Work Done

W = KE_final - KE_initial
W = 40 - 250
W = -210 J

The work done by the force is –210 J. –Note that the negative sign indicates that the force is acting opposite to the direction of motion (retarding force).

Question 5. A mass of 10kg is at a point A on a table. It is moved to a point B. If the line joining A and B is horizontal what is the work done on the object by the gravitational force? Explain your answer.
Answer:
Work done by gravitational force W = mgh
Where h = Difference in the heights of initial and final positions of the object.
Here both the initial and final positions are on the same horizontal line.
So there is no difference in height i.e., h = 0.
∴ work done W = mg × 0 = 0.

Question 6. The potential energy of a freely falling object decreases progressively. Does this violate the law of conservation of energy? Why?
Answer: The total mechanical energy remains constant / as P.E. of the freely falling object decreases. Its kinetic energy and the kinetic energy increases on account of an increases its velocity) the law of conservation of energy is not violated.

Question 7. What are the various energy transformations that occur when you are riding a bicycle?
Answer:

1. Muscular energy into kinetic energy
2. the kinetic energy of the rotation of the which into kinetic energy of the bicycle.

Question 8. Does the transfer of energy take place when you push a huge rock with all your might and fail to move it? Where is the energy you spend going?
Answer: When we push a huge rock and fall to move it. The energy spent in doing so is absorbed by the.rock. This energy is converted into potential energy of the configuration of the rock which results in its deformation. However t his deformation is not visible on account of the huge size of the rock.

Question 9. A certain household has consumed 250 units of energy during a month. How much energy is this in Joules?
Answer: Energy consumed w = 250 units
= 250 kwh
= 250 × 1000 W × 3600 S
= 250 × 1000 J/S × 3600 S
= 9 × 10⁸ J.

Question 10. An object of mass 40 kg is raised to a height of 5m above the ground what is its potential energy? If the object is allowed to fall, find its kinetic energy when it is halfway down.

Answer: Mass m = 40 kg.
height h = 5 m.
Potential energy PE = mgh = 40 × 9.8 × 5 = 1960J.
KE at half way down = PE at halfway down = mgh/2
= 40 × 9.8 × 5/2 = 980 J.

Question 11. What is the work done by the force of gravity on a satellite moving round the earth? Justify your answer.
Answer:

The satellite round the earth moves in a circular orbit. Here the force of gravity acts towards the centre of the earth and displacement of the satellite is along the tangent of the circular path that means therefore and displacement are perpendicular to each other.
So, work done, W = F.S cosθ
= F × S cos 90°
= F × S × 0
= 0
That is, no work is done by the force of gravity.

Question 12. Can there be displacement of an object in the absence of any force acting on it? think. Discuss this question with your friends and teacher.
Answer:
If an object moves with a constant velocity (i.e, there is no acceleration) then no force acts on it. As the object is moving ie it is displaced from one position to another position.

Question 13. A person holds a bundle of hay over his head for 30 minutes and gets tired. Has he done some work or not? Justify your answer.
Answer:
The person has no movement ie his displacement is zero. So the person had done no work (? work is done only when the object is displaced).

Question 14. An electric heater is rated 1500 W. How much energy does it use in 10 hours?
Answer:
Time = 10 h = 10 × 60 min
= 10 × 60 × 60 S

Power =Energy or Work/Time

Energy = power × time
= 1500 × 10 × 60 × 60
= 5.4 × 10⁷ J.

Question 15. Illustrate the law of conservation of energy by discussing the energy changes which occur when we draw a pendulum bob to one side and allow it to oscillate why does the bob eventually comes to rest? What happens to its energy eventually? Is it a violation of the law of conservation of energy?
Answer: When the pendulum oscillates in the air, the air friction opposes its motion. So some part of the kinetic energy of the pendulum is used to overcome this friction. With the passage of time, the kinetic energy of the pendulum goes on decreasing and finally becomes zero. The kinetic energy of the pendulum is transferred to the atmosphere. So energy is being transferred ie it is converted into one form to another. So here is no violation of the law of conservation of energy.

Question 16. An object of mass, m is moving with a constant velocity, v. How much work should be done on the object in order to bring the object to rest?
Answer:
The work done on the object to bring the object to rest
= change in kinetic energy
= Final kinetic energy – Initial Kinetic energy
Here final kinetic energy is zero because the object is brought to rest.

= 0-1/2m^v2

= -1/2mv²

Question 17. Calculate the work required to be done to stop a car of 1500 kg moving at the velocity of 60km/h?
Answer: Mass, m = 1500 kg.
Initial velocity, u = 60 kmh^-1
= 60 × 5/18 = 16.67 ms^-1

Final velocity, v = 0
(? the car comes to rest)
Work done to stop the car = change in kinetic energy
= Final kinetic energy – Initial Kinetic energy

=1/2m^v2-1/2m^u2

= 208333.3J

Question 18.In each of the following a force, F is acting on an object of mass M. The direction of displacement is from west to east shown by the longer arrow. Observe the diagram carefully and state whether the work done by the force is negative, positive or zero.

Answer:
Cas i) The force and displacement are perpendicular to each other. So S = 90°
Work done = FScos θ°
= FScos 90°
= FS × 0 = 0
(? cos 90° = 0)
Cas ii) The force and displacement are in the same direction so θ = 0°
Work done = FScos θ°
= FScos 90°
= FS × 1 = FS
(? cos θ° = 1)
That is the work done is positive.
Cas iii) The force and displacement in opposite direction θ = 180°
Workdone = FScos θ°
= FScos 180°
= FS × -1 = FS
(? cos 180° = -1)
That is workdone is negative.

Question 19. Soni says that the acceleration in an object could be zero even when several forces are acting on it. Do you agree with her? Why?
Answer: Yes, I agree with Soni, the acceleration of an object can be zero even when several forces are acting on it if the resultant of all the forces acting is zero.

Question 20. Find the energy in kWh consumed in 10 hours by four devices of power 500 w each.
Answer:
Total power p = 500w × 4 = 2000w.
Time, t = 10h
Energy = p × t.
= 2000w × 10h
= 2kwh × 10h = 20kwh.
∴ Energy = 20kwh.

Question 21. A freely falling object eventually stops on reaching the ground. What happens to its kinetic energy?
Answer:
P.E. of the configuration of the body and the ground (the body may be deformed and the ground may at the place of collision).
This process energy in which the kinetic energy of a freely falling body is lost in an unproductive chain of energy charges is called dissipation of energy