## Chapter 11 Work and Energy

Intext Questions

Question 1: A force of 7 N acts on an object. The displacement is, say 8 m, in the direction of the force (Fig. 11.3). Let us take it that the force acts on the object through the displacement. What is the work done in this case?

Answer:

Given

Force applied = 7 N

Displacement = 8 m

Work done = ?

We know that, W = Fs

Substituting the values, we get

W = 7 × 8 = 56 J

Question 2: When do we say that work is done?

Answer: The work is said to be done when the external force displaces the object in the direction of applied force.

Question 3: Write an expression for the work done when a force is acting on an object in the direction of its displacement.

Answer: When a force F displaces a body through a distance s within the direction of the applied force, then the work done W on the body is given by 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 15 m long. How much work is done in ploughing the length of the field?

Answer:

Given

Force applied = 140 N

Displacement = 15 m

Work done = ?

We know that, W = Fs

Substituting the values, we get

W = 140 × 15 = 2100 J

Question 6: What is the kinetic energy of an object?

Answer: The energy possessed by the virtue of motion of the body is called kinetic energy.

Question 7: Write an expression for the kinetic energy of an object.

Answer: If a body of mass m is moving with a velocity v, then kinetic energy = 1/2 × m × v²

Question 8: The kinetic energy of an object of mass, m moving with a velocity of 5 m s⁻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?

Answer:

Given

K.E. of object = 25 J

Velocity of object = 5 m s⁻¹

Mass of object = ?

K.E. of object when velocity is doubled = ?

K.E. of object when velocity is tripled = ?

We know that, K.E. = 1/2 mv²

Substituting the values, we get

25 = 1/2 × m × (5)²

25 = 25m/2

m = 25 × 2/25

m = 2 kg

K.E. of object when velocity is doubled = 5 × 2 = 10 m/s

K.E. = 1/2 × 2 × (10)²

K.E. = 100 J

K.E. of object when velocity is tripled = 5 × 3 = 15 m/s

K.E. = 1/2 × 2 × (15)²

K.E. = 225 J

So, if velocity is doubled the K.E. increases 4 times, if velocity is tripled the K.E. increases 9 times.

Question 9: What is power?

Answer: Power is defined as the rate of doing work.

Question 10: Define 1 watt of power.

Answer: 1 watt is the power of an agent, which does work at the rate of 1 joule per second.

Question 11: A lamp consumes 1000 J of electrical energy in 10 s. What is its power?

Answer:

Given

Work done = 1000 J

Time taken = 10 s

Power = ?

We know that, P = W/t

Substituting the values, we get

= 1000/10

= 100 W

Question 12: Define average power.

Answer: Average power is defined as the ratio of total work done by the body to the total time taken by the body.

Exercise Questions

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’.

• Suma is swimming in a pond.

• A donkey is carrying a load on its back.

• A wind-mill is lifting water from a well.

• A green plant is carrying out photosynthesis.

• An engine is pulling a train.

• Food grains are getting dried in the sun.

• A sailboat is moving due to wind energy.

Answer:

• Work is done because the force is acting on Suma and she is swimming from one end to other.

• Work is done because the force is exerted by donkey in upward direction and there is no displacement.

• Work is done because the force is exerted by wind mill to lift water and there is displacement (by lifting water up).

• Work is not done because no force is exerted by plant to carry out photosynthesis.

• Work is done because the force is exerted by engine in forward direction and also displacement takes place.

• Work is not done because there is no force used by the sun to dry the grains.

• Work is done because the wind is applying force on sailboat causing it to move in forward direction (displacement).

Question 2: An object 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 on the object depends on the vertical displacement. It is given by final height - initial height which is 0.

Work done by gravity = mgh

Where, m is the mass of the body, h is the vertical displacement = 0

W = m × g × 0 = 0 J

Question 3: A battery lights a bulb. Describe the energy changes involved in the process.

Answer: When a bulb is connected to a battery, the chemical energy of battery is transferred to voltage. Once the bulb receives this voltage, it converts it into light and heat energy.

Chemical energy → Electrical Energy → Light energy + Heat energy

Question 4: Certain force acting on a 20 kg mass changes its velocity from 5 m s⁻¹ to 2 m s⁻¹. Calculate the work done by the force.

Answer:

Given

Mass of the object = 20 kg

Initial velocity = 5 m s⁻¹

Final velocity = 2 m s⁻¹

Work done = ?

We know that, K.E. = 1/2 × m × (v² - u²)

Substituting the values, we get

= 1/2 × 20 × (2² - 5²)

= 10 × (4 - 25)

= 10 × -21

= -210 J

Question 5: A mass of 10 kg 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 gravity depends on the vertical displacement of the body. Work done by gravity = mgh

Where, m is the mass of object, h is the vertical displacement which is 0.

W = m × g × 0 = 0 J

Question 6: The potential energy of a freely falling object decreases progressively. Does this violate the law of conservation of energy? Why?

Answer: No it doesn’t violate the law of conservation of energy because when the object falls freely from a height, potential energy gets converted into kinetic energy. A decrease in potential energy = increase in kinetic energy. So, the total mechanical energy remains conserved.

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

Answer: During riding a bicycle, the muscular energy gets converted to heat and mechanical energy.

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, there is no transfer of muscular energy to the stationery rock. There is no loss of energy as muscular energy get s converted into heat energy, which causes our body to become hot.

Question 9: A certain household has consumed 250 units of energy during a month. How much energy is this in joules?

Answer:

Given

Energy consumed = 250 units (250 kWh)

Energy consumed in joules = ?

We know that, 1 kWh = 3600000 J

So, 250 kWh = 250 × 3600000 J

= 900000000 J

= 9 × 10⁸ J

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

Answer:

Given

Mass of the object = 40 kg

Height = 5 m

Potential Energy = ?

Kinetic energy when it is half-way down = ?

We know that, P.E. = mgh

Substituting the values, we get

P.E. = 40 × 9.8 × 5 = 1960 J

When the object is half-way down, h = 5/2 m

P.E. = 40 × 9.8 × 5/2 = 980 J

According to Law of Conservation of Energy, Total M.E. = P.E. at half way down + K.E.

1960 = 980 + K.E.

K.E. = 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: Work is said to be done when a force acts on a body and the object must be displaced in or opposite direction of the force. If the direction of force is perpendicular to displacement, then the work done is zero. When a satellite is moving around the Earth, the force of gravity on the satellite is perpendicular to its displacement. Therefore, the work done on the satellite by the Earth is zero.

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: Yes, even if there is no force acting on the object, it can move. An object accelerates when a single force is applied to it. A force is acting on object when it accelerates. Assume that an object travels at a constant velocity. It is subjected to zero net force. However, the object’s motion is accompanied by a displacement. As a result, displacement can occur without the use of force. Hence, an object can have a displacement in the absence of any external force acting on it.

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: Work is said to be done when a force acts on a body and the object must be displaced in or opposite direction of the force. When an individual holds a bundle of hay over his head, there is no displacement in the hay bundle. Although the force of gravity is acting on the bundle, the person isn’t applying any force on it. Therefore, in the absence of force, work done by the person on the bundle is zero.

Question 14: An electric heater is rated 1500 W. How much energy does it use in 10 hours?

Answer:

Given

Power of heater = 1500 W = 1.5 kWh

Time = 10 hr

Energy = ?

We know that, Energy consumed = Work done = Power × time

Substituting the values, we get

= 1.5 × 10

= 15 kW

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 come to rest? What happens to its energy eventually? Is it a violation of the law of conservation of energy?

Answer: When an apparatus moves from its mean position P to either of its extreme positions A or B, it rises through a height h on top of the mean level P. At this time, the K.E. of the bob changes fully into P.E. The K.E. thus becomes zero, and the bob possesses solely P.E. Because it moves towards purpose P, its P.E. decreases increasingly. Consequently, the K.E. will increase. Because the bob reaches purpose P, its P.E. becomes zero and also the bob possesses solely K.E. This method is perennial as long because the apparatus oscillates.

The bob doesn’t oscillate forever. It involves rest as a result of air resistance resists its motion. The apparatus loses its K.E. to beat this friction and stops once a while. The law of conservation of energy isn’t violated. Hence, the overall energy of the apparatus stays preserved.

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:

Given

Mass of the body = m

Velocity of the body = v

We know that, K.E. = 1/2 × m × v²

Substituting the values, we get

= 1/2mv²

In order to bring it to rest, its velocity has to be reduced to zero. An external force has to absorb energy from the object, negative work equal to its kinetic energy -1/2mv².

Question 17: Calculate the work required to be done to stop a car of 1500 kg moving at a velocity of 60 km/h?

Answer:

Given

Mass of car = 1500 kg

Velocity of car = 60 km/hr (50/3 m s⁻¹)

Work done to stop car = ?

We know that, K.E. = Work done = 1/2mv²

Substituting the values, we get

= 1/2 × 1500 × 50/3 × 50/3

= 625000/3

= 208333.33 J

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 diagrams carefully and state whether the work done by the force is negative, positive or zero.

Answer:

a) In this case, the direction of force functioning on the block is perpendicular to the displacement, so the work done will be zero.

b) In this case, the direction of force on the block is in the same direction of displacement. So, the work done will be positive.

c) In this case, the direction of force on the block is in opposite direction of displacement. So, the work done will be 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: Acceleration of an object might be zero even if many force are acting on it. This happens when all the forces cancel out each other i.e. the net force acting on the object is zero. For a uniformly moving object, the net force acting on the object is zero. Hence, the acceleration of the object is zero.

Question 20: Find the energy in kW h consumed in 10 hours by four devices of power 500 W each.

Answer:

Given

Power consumed by each device = 500 W

No. of devices = 4

Time = 10 hr

Energy consumed (in kWh) = ?

1 kW = 1000 W

So, 500 W = 0.5 kW

Energy consumed by 4 devices in 1 hr = 0.5 × 4 = 2 kWh

Energy consumed by 4 devices in 10 hr = 2 × 10 = 20 kWh

Question 21: A freely falling object eventually stops on reaching the ground. What happens to its kinetic energy?

Answer: When an object falls freely towards the ground, its potential energy decreases and kinetic energy increases; as the object touches the ground, all its potential energy gets converted into kinetic energy. Since, the object hits the ground, kinetic energy gets converted to heat and sound energy.

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