NCERT Solutions for Class 9 Science Chapter 11 Work and Energy
NCERT Solutions for Class 9 Science Chapter 11 Work and Energy on this step-by-step Work and Energy answer guide. In some of State Boards and CBSE schools, students are taught thru NCERT books. As the chapter comes to an end, students are requested few questions in an exercising to evaluate their knowledge of the chapter. Students regularly want guidance dealing with those NCERT Solutions for Class 9 Science Chapter 11 Work and Energy.
It’s most effective natural to get stuck withinside the exercises while solving them so that you could assist students rating better marks, we have provided grade by grade NCERT answers for all exercises of Class 9 Science so you can be searching for assist from them. Students must solve those exercises carefully as questions withinside the very last exams are requested from those so these exercises at once have an impact on students’ final score. Find all NCERT Solutions for Class nine Science Work and Energy below and prepare for your tests easily.
NCERT Solution Class 9
Chapter 11: Work and Energy
Q.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 ‘ .
( i ) Suma is swimming in a pond .
( ii ) A donkey is carrying a load on its back .
( iii ) A wind – mill is lifting water from a well .
( iv ) A green plant is carrying out photosynthesis .
( v ) An engine is pulling a train .
( vi ) Food grains are getting dried in the sun .
( vii ) A sailboat is moving due to wind energy .
Ans . ( i ) Suma is doing work as she is able to move herself by applying force with the movement of her arms and legs in the water .
( ii ) Donkey is not doing any work ( in the sense of physics ) because the weight it is carrying ( in the direction of force ) and displacement are perpendicular to each other .
( iii ) Wind mill is lifting water from a well and doing work against the gravity .
( iv ) There is no force and displacement is present in this situation , so work done is zero .
( v ) During the pulling of a train , an engine does the work against the friction present between the railway track and wheels .
( vi ) During the drying grains , there is no force as well or displacement is present . So , no work is done here .
( vii ) Work is done by the wind as it moves the sailboat towards the direction of the force ( force of blowing air ) .
Q. 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 ?
Ans . When the object moves upwards , the work done by gravity is negative ( as the direction of gravitational force is towards the Earth’s centre ) and when the object come downwards , there is a positive work done .
Vertical displacement is given by the difference in the initial and final positions / heights of the object which is zero . So , the total work done is zero for the entire motion .
Q.3 . A battery lights a bulb . Describe the energy changes involved in the process .
Ans . Battery converts chemical energy into electrical energy . When a bulb is connected to a battery , the chemical energy of the battery is transferred into electrical energy .
When the bulb receives this electrical energy , then it converts it into light and heat energy . Hence , the transformation of energy in the given situation can be shown as :
Chemical energy ⇒ Electrical energy ⇒ Light energy + Heat energy
Q. 4. Certain force acting on a 20 kg mass changes its velocity from 5 m / s to 2 m / s . Calculate the work done by the force .
Ans . Given that ,
Mass of the body = 20 kg
Initial velocity = 5 m / s
Final velocity = 2 m / s
We know that ,
Work done = Change in kinetic energy
So , the work done by the force is 210 J. The negative sign indicates that the force applied is opposite to the motion .
Q. 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 .
Ans . The work down by the gravitational force acting on the body is zero .
Because the direction of applied force is vertically downwards and the displacement is horizontal , that is , force and displacement are perpendicular to each other . There is no displacement in the direction of the force .
Q. 6. The potential energy of a freely falling object decreases progressively . Does this violate the law of conservation of energy ? Why ?
Ans . No. This process does not violate the law of conservation of energy . Because when the body falls from a height , its potential energy changes into kinetic energy progressively .
A decrease in the potential energy is equal to an increase in the kinetic energy of the body . During the process , total mechanical energy of the body remains conserved . Therefore , the law of conservation of energy is not violated .
Q. 7. What are the various energy transformations that occur when you are riding a bicycle ?
Ans . While riding a bicycle , the muscular energy of the cyclist is converted into kinetic ( rotational ) energy of wheels of cycle which is further converted into kinetic energy . This kinetic energy provides a velocity to the bicycle . Some energy is lost as beat due to the friction of the ground .
During the transformation , the total energy remains conserved .
The transformation can be shown as :
Muscular energy → Kinetic energy + Heat energy
Q. 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 ?
Ans . When we push the rock and fail to move it , there is 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 and the rock to become hot .
Q.9 . A certain household has consumed 250 units of energy during a month . How much energy is this in joules ?
Ans . Given that , energy consumed in a month = 250 unit
We know that , 1 unit = 1 kWh = 3.6 x 106 J
So , 250 units = 250 x 3.6 × 106 J
= 900 × 106 J
= 9 × 108 J
Hence , the energy consumed = 9 × 108 J
Q.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 .
Ans . Given that , m = 40 kg
g = 9.8 m / s²
h = 5 m
We know that , potential energy = mgh
Putting the values ,
Potential energy = 40 x 9.8 x 5 = 1,960 J
According to law of conservation of energy , the total mechanical energy ( kinetic and potential energy ) of an object remains constant .
Therefore , when the object is half – way down , its potential energy becomes half the original energy and remaining half converted into kinetic energy .
Q. 11. What is the work done by the force of gravity on a satellite moving round the earth ? Justify your answer .
Ans . Work is done whenever the given two conditions are satisfied :
( i ) A force acts on the body .
( ii ) There is a displacement of the body by the application of force in or opposite to the direction of force .
If the direction of force is perpendicular to displacement , then the work done is zero .
So , in this case , as the satellite moves around the Earth , the displacement in a short interval is along the tangential direction and the force ( gravitational force ) is towards the centre of the Earth . Since the force and displacement are perpendicular to each other , the work done by gravitational force is zero .
Q. 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 .
Ans . Yes , it is possible . There may be displacement in the absence of force .
We know that , F = mass x acceleration = m.a
In the condition of absence of force , F = 0 ,
Then , ma = 0 ⇒ a = 0 [ as mass , m ≠ 0 ]
If a = 0 , the object is either at rest or in a state of uniform motion in a straight line . In case , the object is moving in a straight line , there must be displacement . So , in the absence of force , there may be displacement of the object .
Q. 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 .
Ans . The person who is holding a bundle of hay gets tired because his muscular energy is converting into thermal energy .
And there is no displacement in the bundle of hay . So , there is no displacement at all . So , work done by the person on the bundle is zero .
Q. 14. An electric heater is rated 1,500 W. How much energy does it use in 10 hours ?
Ans . Given that ,
Power = 1500 W = 1.5 kW
Time 10 hours
We know that , Energy = Power x Time
The energy used by heater = Power x Time
= 1.5 kW x 10 h
= 15 kWh
Q. 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 ?
Ans . 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 these points A and B , the kinetic energy the bob transfers completely into the potential energy . The kinetic energy becomes zero and the bob possesses only potential energy .
As it moves towards point P , its potential energy decreases progressively and accordingly , the kinetic energy increases . As the bob reaches at point P , its potential energy becomes zero and the 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 air resistance resists its motion . The pendulum loses its kinetic energy to overcome this friction and stops after some time . The law of conservation of energy is not violated because the energy lost by the pendulum to overcome as heat friction is gained by its surroundings . Hence , the total energy of the pendulum and the surrounding system remain conserved .
Q. 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 ?
Ans . The object is in motion , so its energy is kinetic energy . The kinetic energy of an object of mass m , moving with a velocity v , is given by :
The kinetic energy of the object , when it comes to rest is zero because its final velocity becomes zero .
So , work done on object = Change in kinetic energy
Q. 17. Calculate the work required to be done to stop a car of 1,500 kg moving at a velocity of 60 km / h ?
Ans . Given ,
Mass of the car , m = 1,500 kg
Hence , the work required to stop the car is 20.8 x 104 J.
Q. 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 .
Ans . There are following conditions which are needed to fulfil the work to be done in physics :
( i ) A force should act on object and it must cause the displacement
( ii ) The object must be displaced in the direction of the force .
Case I : In the first case , the force and displacement are perpendicular to each other , so work done is zero .
Case II : In the second case , the force and displacement are in the same directions , so the work done is positive .
Case III : In the third case , the force and displacement are in the opposite direction , so the work done is negative .
Q. 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 ?
Ans . Acceleration in an object could be zero even when several forces are acting on it . This happens when all the forces cancel out each other , that is , 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 . Hence , Soni is right .
Mathematically , if the resultant force acting on a body in different directions is zero , then the acceleration will be zero .
We know that , F = m x a ,
If the net force is zero ,
F = 0 , then ma = 0 and a = 0 [ as m ≠ 0 ]
Q. 20. Find the energy in kW – h consumed in 10 hours by four devices of power 500 W each .
Ans . Given that , power rating of each device , P = 500 W
Time for which each device runs , t = 10 h
So , the power of four devices = 4 x 500 W = 2 kW
We know that , Power Energy consumed / Time
Energy consumed = Power x Time
= 2 kW x 10 h
= 20 kW h
= 20 units [ ∵ 1 unit = 1 kWh ]
Hence , the energy consumed by four devices of power 500 W each in 10 h will be 20 units .
Q. 21. A freely falling object eventually stops on reaching the ground . What happens to its kinetic energy ?
Ans . 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 .
As the object hits the hard ground , all its kinetic energy gets converted into heat energy and sound energy .
Q. 1. A force of 7 N acts on an object . The displacement is , say 8 m , in the direction of the force . Let us take it that the force acts on the object through the displacement . What is the work done in this case ?
[ NCERT Q. 1 , Page 148 ]
Ans . Given that , Displacement = 8 m
Force = 7 N
Work done , W = Force x Displacement
W = 7 x 8 = 56 J
Q. 2. When do we say that work is done ? [ NCERT Q. 1 , Page 149 ]
Ans . Work is said to be done when a force causes displacement of an object in the direction of the applied force .
Necessary conditions for work done :
( i ) A force should act on object and it must cause the displacement .
( ii ) The object must be displaced in the direction of the force .
Q. 3. Write an expression for the work done when a force is acting on an object in the direction of its displacement . [ NCERT Q. 2 , Page 149 ]
Ans . Mathematically , work done is expressed as the product of force and displacement in the direction of force .
W = Fd
Where , W = Work done on an object
F = Applied force on the object
d = Displacement of the object
W = Fd cos θ is not in the NCERT textbook , but useful for competitive exams .
Where , θ = the angle between the force and the displacement vector .
Q4 . Define 1 J of work . [ NCERT Q. 3 , Page 149 ]
Ans . 1 Joule is defined as the amount of work done by force of 1 N when displacement is 1 m in the direction of the force .
Q. 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 ? [ NCERT Q. 4 , Page 149 ]
Ans . Given that , Displacement = 15 m
Force = 140 N
Work done , W = Force x Displacement
= 140 x 15 = 2100 J
Q. 6. What is the kinetic energy of an object ? [ NCERT Q. 1 , Page 152 ] [ KVS , Patna Region , SA – II ]
Ans . Kinetic energy is the energy due to motion of the object .
Q. 7. Write an expression for the kinetic energy of an object . [ NCERT Q. 2 , Page 152 ]
Ans . If an object of mass m is moving with a speed v , then its kinetic energy ( KE ) is given by the expression ,
Q. 8. The kinetic energy of an object of mass , m moving with a velocity of 5 m / s 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 3 times ? [ NCERT Q. 3 , Page 152 ]
Ans . Given that ,
K.E. of the object = 25 J
Velocity of the object , v = 5 m / s
( ii ) If velocity is tripled , v = 3 × 5 = 15 m / s
Q. 9. What is power ? [ NCERT Q. 1 , Page 156 ] [ KVS , Patna Region , SA – II ]
Ans . The rate of doing work is called power .
Since power is the rate of doing work
Power = work / time
P = W / T
Where , P = power
W = work done
T = Time taken in doing work .
Q. 10. Define 1 watt of power . [ NCERT Q. 2 , Page 156 ] [ KVS , Patna Region , SA – II ]
Ans . One watt is the power that it takes to do one joule of work in one second .
Q. 11. A lamp consumes 1,000 J of electrical energy in 10 s . What is its power ? [ NCERT Q. 3 , Page 156 ]
Ans . Given that ,
W = 1,000 J
t = 10 s
As we know that power is rate of doing work , P = W / t
P = 1,000 / 10 = 100 W
Q. 12. Define average power . [ NCERT Q. 4 , Page 156 ]
Ans . When an object does different amounts of work or uses energy in different intervals of time then the ratio between the total work or energy consumed to the total time is called average power .
So , average power is defined as the ratio of total energy delivered to the total time .
Power = Total Energy / Total Time
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