Intext Exercise 1
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?
Solution 1:
Given,
Force, F = 7 N Displacement, S = 8 m
Work done = Force × Displacement
W = F × S, W = 7 × 8
= 56 Nm
= 56 J
Intext Exercise 2
Question 1:
When do we say that work is done?
Solution 1:
Two conditions need to be satisfied for work to be done: (i) a force should act on an object, and
(ii) the object must be displaced.
Question 2:
Write an expression for the work done when a force is acting on an object in the direction of its displacement.
Solution 2:
Let a constant force, F act on an object.
Let the object be displaced through a distance, s in the direction of the force (Fig. 11.1). Let W be the work done. We define work to be equal to the product of the force and displacement.
Work done = force × displacement
W = F s
Question 3:
Define 1 J of work.
Solution 3:
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 4:
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?
Solution 4:
Work done = Force × Displacement
W = F × s
Given,
Work done = Force × Displacement
W = F × s
Applied force, F = 140 N Displacement, s = 15 m
W = 140 × 15
= 2100 J
Hence, 2100 J of work is done in ploughing the length of the field.
Intext Exercise 3
Question 1:
What is the kinetic energy of an object?
Solution 1:
Kinetic energy is the energy possessed by an object due to its motion
Question 2:
Write an expression for the kinetic energy of an object.
Solution 2:
If a body of mass ‘m’ is moving with a velocity ‘v’, then its kinetic energy Ek is given by the expression,
E 1 mv2
k 2
Its SI unit is the Joule (J).
Question 3:
The kinetic energy of an object of mass, m moving with a velocity of 5 ms-1 is 25J.What will be its kinetic energy when its velocity is doubled? What will be its kinetic energy when its velocity is increased three times?
Solution 3:
We know that Expression for kinetic energy is given by m = Mass of the object
v = Velocity of the object = 5 m s−1
Given that kinetic energy, Ek= 25 J
E 1 mv2
k 2
(i) If the velocity of an object is doubled, then v = 5 × 2 = 10 m s-1. Kinetic Energy,
Kinetic energy proportional to the square of the velocity. So, v2 = (5 x 2)2 = 100 J,
Therefore its kinetic energy becomes 4 times its initial value.
(ii) If velocity is increased three times, then its kinetic energy becomes 9 times its original value, because it is proportional to the square of the velocity.
Hence, kinetic energy = 25 × 9 = 225 J.
Intext Exercise 4
Question 1: What is power? Solution 1:
Power is the rate of doing work or the rate of transfer of energy.
If W is the amount of work done in time t, then power is given by
Power = Work done Energy
Time Time
P W T
Units for power is watt (W).
Question 2:
Define 1 watt of power:
Solution 2:
1 watt is the power of an agent, which does work at the rate of 1 joule per second. We can also say that power is 1 W when the rate of consumption of energy is 1 J s–1.
1W 1J
1s
Question 3:
A lamp consumes 1000 J of electrical energy in 10 s. What is its power?
Solution 3:
Power = Work done
Time
Work done = Energy consumed by the lamp
= 1000 J Time = 10 s
Power = 1000 100Js−1
10
= 100 W
Question 4:
Define average power.
Solution 4:
Average power is obtained by dividing the total amount of work done in the total time taken to do this work.
Average Power = Total work done
Total time taken
NCERT Exercise
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.
Solution 1:
Work is done whenever the following two conditions are satisfied: (i) A force acts on the body.
(ii) And, there is a displacement of the body by the application of force in or opposite to the direction of force.
(a) While swimming, Suma applies a force to push the water backwards. Thus, Suma swims in the forward direction caused by the forward response of water. Here, the force causes a displacement. Therefore, while swimming work is done by Suma.
(b) The donkey has to apply a force in the upward direction while carrying a load. But, here the displacement of the load is in the forward direction. Since, displacement is perpendicular to force, therefore the work done is zero.
(c) A work is done by the windmill in lifting water from the well, because a windmill works against the gravitational force to lift water. (d) In this case, there is no displacement of the leaves of the plant. Therefore, the work done is zero.
(e) An engine applies force to pull the train. This allows the train to move in the direction of force applied. Therefore, there is a displacement in the train in the same direction. Hence, work is done by the engine on the train.
(f) During the process of food grains getting dried in the sun, no work is done because food grains do not move in the presence of solar energy.(g)To push the sailboat forward direction, a wind energy applies a force on it.. Therefore, there is a displacement in the boat along the direction of the force. Hence,work is done by wind on the boat.
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?
Solution 2:
Work done by the force of gravity on an object depends only on vertical displacement. Vertical displacement is given by the difference in the initial and final positions(height) of the object, which is zero.
Work done by gravity is expressed as , W = mgh
Where, m = mass of the object , g = acceleration due to gravity
h = Vertical displacement , which is zero . o on substituting the value of h we get, W = mg × 0 = 0 J
Therefore, the work done by gravity on the given object is zero joules.
Question 3:
A battery lights a bulb. Describe the energy changes involved in the process.
Solution 3:
When a bulb is connected to a battery, then its chemical energy is transferred into electrical energy. When the bulb receives this electrical energy, it converts it into light and heat energy. Therefore , the transformation of energy in the given situation can be shown as:
Chemical Energy → Electrical Energy → Light Energy → Heat energy
Question 4:
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.
Solution 4:
We know that change in kinetic energy is equal to the work done , so let’s calculate the kinetic energy at v = 5m/s and at v = 2m/s.
Kinetic energy is given by the expression,
(E )v 1 mv2
k 2
Where,
Ek= Kinetic energy of the object moving with a velocity, v m = Mass of the object
(i) Kinetic energy when the object was moving with a velocity of 5 m s-1
( Ek )5
1 20 (5)2 250 J
2
(ii) Kinetic energy when the object was moving with a velocity 2 m s-1
( Ek )2
1 20 (2)2 40 J
2
Therefore, work done by force = (Ek )2 − (Ek )5
= 40 − 250 = −210 J
Here, the negative sign shows that the force is acting in the direction opposite to the motion of the object.
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.
Solution 5:
We know, work done by gravity depends only on the vertical displacement of the body. It does not depend upon the path of the body. Hence, work done by gravity is given by the expression,
W = mgh
Where,
Vertical displacement, h = 0
∴W = mg × 0 = 0
So, the work done by gravity on the body is zero.
Question 6:
The potential energy of a freely falling object decreases progressively. Does this violate the law of conservation of energy? Why?
Solution 6:
No. The process does not violate the law of conservation of energy,because when the body drops from a height, its potential energy changes into kinetic energy progressively. A decrease in the potential energy is equivalent to an increase in the kinetic energy of the body. During this process, the total mechanical energy of the body remains preserved. Therefore,
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?
Solution 7:
While riding a bicycle, the rider’s muscular energy gets transferred into the bicycle'sheat energy and kinetic energy . Heat energy heats the rider’s body and kinetic energy provides a velocity to the bicycle. The transformation can be shown as:
Muscular Energy → Kinetic Energy → Heat Energy
During the transformation, the total energy remains conserved.
Question 8:
Does the transfer of energy takes place when you push a huge rock with all your might and fail to move it? Where is the energy you spend going?
Solution 8:
transfer of muscular energy to the stationary rock. Also, there is no loss of energy because muscular energy is transferred 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?
Solution 9:
1 unit of energy is equal to 1 kilowatt hour (kWh).
1 unit = 1 kWh
1 kWh = 3.6 × 106 J
Therefore, 250 units of energy = 250 × 3.6 × 106 = 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. Solution 10:
Gravitational potential energy is given by the expression, W = mgh
Where,
h = Vertical displacement = 5 m m = Mass of the object = 40 kg
g = Acceleration due to gravity = 9.8 m s-2
∴W = 40 × 5 × 9.8 = 1960 J.
At half-way down, the potential energy of the object will be
1960
2 = 980 J.
At this point, the object has an equal amount of potential and kinetic energy. This is due to the law of conservation of energy. Hence, half-way down, the kinetic energy of the object will be 980 J.
Question 11:
What is the work done by the force of gravity on a satellite moving round the earth? Justify your answer.
Solution 11:
If the direction of force is perpendicular to displacement, then the work done is zero.When a satellite moves around the Earth, then the direction of force of gravity on the satellite is perpendicular to its displacement. Hence, 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.
Solution 12:
Yes. object moving with uniform velocity
Suppose an object is moving with constant velocity. The net force acting on it is zero. Hence, there can be a displacement without a force.
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.
Solution 13:
When a person holds a bundle of hay over his head gravitational force acts on the hay downwards. But, there is no displacement in the bundle of hay in the direction of force so no work is done..
Question 14:
An electric heater is rated 1500 W. How much energy does it use in 10 hours?
Solution 14:
Energy consumed by an electric heater can be obtained with the help of the expression,
P W T
Where,
P = 1500 W = 1.5 kW T = 10 hrs
Work done = Energy consumed by the heater
Therefore, energy consumed = Power × Time
= 1.5 × 10 = 15 kWh
Hence, the energy consumed by the heater in 10 h is 15 kWh.
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?
Solution 15:
The law of conservation of energy states that energy can be neither created nor destroyed. It can only be converted from one form to another.Consider the case of an oscillating pendulum.
When a pendulum moves from its mean position P to either of its extreme positions A or B, it rises through a height h above the mean level P. At this point, Bob comes to rest momentarily and the kinetic energy of the bob changes completely into potential energy. As it moves towards point P, its potential energy decreases progressively. Accordingly, the kinetic energy increases. As the bob reaches point P, its potential energy converted to kinetic energy at this point bob possesses only kinetic energy. This process is repeated as long as the pendulum oscillates.
The bob does not oscillate forever. It comes to rest because there is some friction in the air. . The pendulum loses energy to overcome this friction After all the energy is lost, it comes to a stop.
The law of conservation of energy is not violated because the energy lost by the pendulum to overcome friction is gained by its surroundings. Hence, the total energy of the pendulum and the surrounding system remain conserved.
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?
Solution 16:
A body in motion has kinetic energy. Kinetic energy of an object of mass, m moving with a velocity, v is given by the expression,
E 1 mv2
k 2
1 mv 2
To bring the object to rest, 2
amount of work is required to be done on the object.
Question 17:
Calculate the work required to be done to stop a car of 1500 kg moving at a velocityof 60 km/h?
Solution 17:
Ek
Kinetic energy,
Where,
1 mv2
2
Mass of car, m = 1500 kg
Velocity of car, v = 60 km/h =
60 5 ms −1
18
E 1 1500 60 5
20.8 104 J
k 2
18
So 20.8 × 10 4 J needs to be done to stop the car. .
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.
Solution 18: Case I
In this case, the direction of force acting on the block is perpendicular to the displacement. So, work done by force on the block will be zero.
Case II
In this case, the direction of force acting on the block is in the direction of displacement. So, work done by force on the block will be positive.
Case III
In this case, the direction of force acting on the block is opposite to the direction of displacement. So,, work done by force on the block 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?
Solution 19:
Acceleration of an object zero when net force acting on an object is zero. Net force can be zero even when there are multiple forces are acting on the body. For a uniformly moving object, the net force acting on the object is zero. Hence, the acceleration of the object is zero. Hence, Soni is right.
Question 20:
Find the energy in kW h consumed in 10 hours by four devices of power 500 W each.
Solution 20:
Energy consumed by an electric device can be obtained with the help of the expression for power,
P W T
Where,
P = 500 W = 0.50 kW T = 10 hrs
Work done = Energy consumed by the device
Energy consumed by each device is = Power × Time
= 0.50 × 10 = 5 kWh
Hence, the energy consumed by four equal rating devices in 10 hrs will be 4 × 5 kWh
= 20 kWh.
Question 21:
A freely falling object eventually stops on reaching the ground. What happens to its kinetic energy?
Solution 21:
As the object hits the hard ground, all its kinetic energy gets converted into heat energy and sound energy and it is lost to the environment. It can also deform the ground depending upon the nature of the ground and the amount
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