>

NCERT Exemplar Class 9 Science Chapter 11 Work and Energy Solutions

NCERT Exemplar Solutions for Class 9 Science Chapter 11 Work and Energy covers all the important questions and answers as well as advanced level questions. It helps in learning about the work, unit of work, Energy, forms of energy, kinetic energy, potential energy, heat energy, chemical energy, electrical energy, light energy and mechanical energy.

The NCERT Exemplar solutions for class 9 science is very important in the examination. NCERT Exemplar Solutions for Class 9 Science Chapter 11 Work and Energy is provided by our experts. They prepared the best solutions which help the students in understanding the solutions in an easy way. This chapters also covers the other topics like law of conservation of energy, power and commercial unit of energy.


Chapter Name

Chapter 11 Work and Energy

Book Title

NCERT Exemplar for Class 9 Science

Related Study

  • NCERT Solutions for Class 9 Science Chapter 11 Work and Energy
  • Revision Notes for Class 9 Science Chapter 11 Work and Energy
  • MCQ for Class 9 Science Chapter 11 Work and Energy
  • Important Questions for Class 9 Science Chapter 11 Work and Energy

Topics Covered

  • MCQ
  • Short Answers Questions
  • Long Answers Questions

NCERT Exemplar Solutions for Chapter 11 Work and Energy Class 9 Science

Multiple Choice Questions

1. When a body falls freely towards the Earth, then its total energy :
(a) Increases
(b) Decreases
(c) Remains constant
(d) First increases and then decreases

Solution

(c) Remains constant

Body falling freely towards the earth possesses the same kinetic and potential energy. This is because the body obeys the Law of conservation of energy.


2. A car is accelerated on a levelled road and attains a velocity 4 times of its initial velocity. In this process the potential energy of the car :
(a) Does not change
(b) Becomes twice to that of initial
(c) Becomes 4 times that of initial
(d) Becomes 16 times that of initial

Solution

(a) Does not change

Potential energy is the product of height, mass and gravity. Therefore, height is a factor in determining potential energy.


3. In case of negative work the angle between the force and displacement is :
(a) 0°
(b) 45°
(c) 90°
(d) 180°

Solution

(d) 180°

Work done = FS cosθ

⇒ cos 180°= -1


4. An iron sphere of mass 10 kg has the same diameter as an aluminium sphere of mass is 3.5 kg. Both spheres are dropped simultaneously from a tower. When they are 10 m above the ground, they have the same :
(a) Acceleration
(b) Momentum
(c) Potential energy
(d) Kinetic energy

Solution

(a) Acceleration

Momentum, potential energy and kinetic energy vary with weight. But here, the acceleration is due to acceleration due to gravity which is independent of mass.


5. A girl is carrying a school bag of 3 kg mass on her back and moves 200 m on a levelled road. The work done against the gravitational force will be : (g = 10 m s–2)
(a) 6 × 103 J
(b) 6 J
(c) 0.6 J
(d) zero

Solution

(d) zero

Here, the direction of displacement is perpendicular to the direction of gravitational force. Therefore, the work done against the gravity is zero.


6. Which one of the following is not the unit of energy?
(a) Joule
(b) Newton metre
(c) Kilowatt
(d) Kilowatt hour

Solution

(c) Kilowatt

Kilowatt is SI unit of electrical power.


7. The work done on an object does not depend upon the :
(a) Displacement
(b) Force applied
(c) Angle between force and displacement
(d) Initial velocity of the object

Solution

(d) Initial velocity of the object

Work done is the product of force and displacement. But work done is not dependent on initial velocity.


8. Water stored in a dam possesses :
(a) No energy
(b) Electrical energy
(c) Kinetic energy
(d) Potential energy

Solution

(d) Potential energy

The energy stored in an object because of the position is known as the potential energy.


9. A body is falling from a height h. After it has fallen a height 2 m, it will possess :
(a) Only potential energy
(b) Only kinetic energy
(c) Half potential and half kinetic energy
(d) More kinetic and less potential energy

Solution

(c) Half potential and half kinetic energy

The body will have only potential energy when it is at height h. Similarly, when the body reaches the ground its potential energy will be zero and kinetic energy will be maximum.

At height h/2, both potential energy and kinetic energy of the body will be half.


Short Answer Questions

10. A rocket is moving up with a velocity v. If the velocity of this rocket is suddenly tripled, what will be the ratio of two kinetic energies?

Solution

Initial velocity, u= v
New initial velocity, u' = 3v
Therefore, the ratio between initial and final kinetic energy is 1:9.


11. Avinash can run with a speed of 8 ms–1 against the frictional force of 10 N, and Kapil can move with a speed of 3 ms–1 against the frictional force of 25 N. Who is more powerful and why?

Solution

As work equals to product of force and distance, you can write the equation for power in the following way, assuming that the force acts along the direction of travel :
Power = Work done/Time
But, work done = Force × distance
Therefore, P = W/t = F×s/t = Fv
where, s is the distance travelled and v is the speed.
Power of Avinash = 10 × 8 = 80 W
Power of kapil = 25 × 3 = 75 W
Therefore, Avinash is more powerful than Kapil.


12. A boy is moving on a straight road against a frictional force of 5 N. After travelling a distance of 1.5 km he forgot the correct path at a roundabout figure of radius 100 m. However, he moves on the circular path for one and half cycle and then he moves forward upto 2.0 km. Calculate the work done by him.

Solution

F = 5 N
As work equals force times distance
W = f×s
⇒ W = 5 × [1500 + 200 + 2000]
= 18500 J
Here, 1.5 km = 1500 m; 2 km = 2000 m


13. Can any object have mechanical energy even if its momentum is zero ? Explain.

Solution

Momentum is the product of mass and velocity. The velocity of a body is zero if the body is at rest. The body at rest at certain height possess gravitational potential energy. Therefore, an object can have mechanical energy even if its momentum is zero.


14. Can any object have momentum even if its mechanical energy is zero? Explain.

Solution

No, an object with zero mechanical energy cannot have momentum asits potential and kinetic energy are zero.


15. The power of a motor pump is 2 kW. How much water per minute the pump can raise to a height of 10 m? (Given, g = 10 m s–2)

Solution

Power of the pump = 2kW = 2000 W;
t = 1 min = 60 sec;
height = 10 m;
g = 10 ms-2
Power = work done/time
(The force on an object of mass m at the surface of the earth is mg, from F = ma, when acceleration is g, the acceleration at the surface of the Earth. If the object falls through a distance h, then the work done on the object by the force of gravity is mg times h, force times distance.)
Work done = mgh
= m ×10×10
=100m
Now,
2000 W= 100m/60
Or, m = 1200 kg
Therefore, the pump can raise 1200 kg of water in one minute.


16. The weight of a person on a planet A is about half that on the Earth. He can jump up to 0.4 m height on the surface of the Earth. How high he can jump on the planet A?

Solution

As per the definition of force, the equation of force due to gravity is given by, W = mg, i.e., weight is equal to mass times gravitational acceleration In this case, the force is better known as the weight of the object. Weight of a person on Earth = w = mg1 (given);
height the person can jump (h1) = 0.4 m
U = potential energy from height,
m = mass of the object,
g = gravity and height = h
U = mgh
Therefore,
potential energy = mg1h1 ...(1)
where(h1) = 0.4 m; g1 = g
Now,
Weight of the person on planet A = W/2 = mg2/2
Let the height the person can jump = h2 :
g2 = ½ , g1 = ½ g ….(2)
Therefore,
Potential energy of planet A = mg2h2
Potential energy of the person on the Earth = mg1h1 …(3)
h2 = 0.4 × 2 = 0.8 m
Hence, he can jump double the height with the same muscular force.


17. The velocity of a body moving in a straight line is increased by applying a constant force F, for some distance in the direction of the motion. Prove that the increase in the kinetic energy of the body is equal to the work done by the force on the body.

Solution

Consider that a force ‘F’ is applied on a body having mass ‘m’ and the distance travelled be ‘s’.
Work Done (joules) = Force (Newton) × Distance (meter)
W = F × s …(1)
As,
Force (F) = ma ...(2)
By substituting (2) in (1), we get
W = ma × s …(3)
Using the Newton’s third equation of motion,
v2 – u2 = 2as


Putting the value of s in (3),
W = final kinetic energy – initial kinetic energy

Work done = change in kinetic energy


18. Is it possible that an object is in the state of accelerated motion due to external force acting on it, but no work is being done by the force ? Explain it with an example.

Solution

Yes, Force always acts in perpendicular to the displacement. If the object is moving in circular path then the force acts perpendicular to the direction of the motion and therefor will be no work done despite the action of force.


19. A ball is dropped from a height of 10 m. If the energy of the ball reduces by 40% after striking the ground, how much high can the ball bounce back? (g = 10 m s–2)

Solution

Given, g = 10 ms-2 ; h = 10 m
energy possessed by the ball = mgh
= m × 10 × 10 = 100 m joules
Energy left in the ball after striking the ground
= (100 – 40)/100 = 60/100
( As energy is reduced by 40% after striking the ground)
Therefore,
remaining energy = 60 m joules ….(1)
Let the height at which the ball bounces back = h1 ….(2)
Therefore,
Energy possessed by the ball = mgh
Using (1) and (2), we get
60 m = m × 10 × h1
h1 = 6 metres.
Thus, the height at which the ball bounces back = 6 m


20. If an electric iron of 1200 W is used for 30 minutes every day, find electric energy consumed in the month of April.

Solution

Power of electric iron = 1200W
Usage per day = 30min
= 0.5hrs
Number of days in the month of April =30 days
Electrical energy consumed, E = P× t
= 1200×0.5×30
= 18000 Wh
= 18 kWh
= 18 units
Therefore, the total electricity consumed in April is 18 units.


Long Answer Questions

21. A light and a heavy object have the same momentum. Find out the ratio of their kinetic energies. Which one has a larger kinetic energy?

    Solution

    Consider the mass and velocity of the light object to be m1 and v1 respectively. Similarly, let the mass and velocity of the heavy object to be m2 and v2 respectively
    Momentum = mass × velocity
    i.e., p = mv ...(1)
    Momentum of light object = m1v1
    Momentum of heavy object = m2v2
    Given,
    light and a heavy object have the same momentum
    Therefore, m1v1 = m2v2
    But, we know that
    Kinetic energy = ½ mv2
    Thus,
    Kinetic energy of light object (KE1)

    But, V1 > V2 [from (4)]
    Therefore, (K.E.)l > (K.E.)h
    ‘The lighter one will have more kinetic energy than the heavy one. Moreover, Kinetic energy is directly proportional to the mass of the object.


    22. An automobile engine propels a 1000 kg car (A) along a levelled road at a speed of 36 km h–1. Find the power if the opposing frictional force is 100 N. Now, suppose after travelling a distance of 200 m, this car collides with another stationary car (B) of same mass and comes to rest. Let its engine also stop at the same time. Now, car (B) starts moving on the same level road without getting its engine started. Find the speed of the car (B) just after the collision.

    Solution
    Given, Final velocity of ear
    A(vA) = 0 (since it comes to rest after colliding with car B);  (uA) = 36 km/h = 10 m/s (since, 1 km/hr = 518 m/s);
    Frictional force = 100 N
    Since, the car A moves with a uniform speed, it means that the engine of car applies a force equal to the frictional force

    mA = mB = 1000 kg
    Initial velocity of the car, v = 36 km/h = 10 m/s
    Frictional force = 100 N
    Car A moves with a uniform speed, which means engine of car A applies a force equal to the frictional force.
    Power = (Force ×distance)/time
    = F×v
    = 100 N × 10 m/s
    = 1000 W
    After collision
    mA uA + mB uB = mA vA + mB vB
    1000 × 10 + 1000 × 0= 1000 × 0 + 1000 × vB
    Therefore, vB = 10 m s–1


    23. A girl having mass of 35 kg sits on a trolley of mass 5 kg. The trolley is given an initial velocity of 4 ms–1 by applying a force. The trolley comes to rest after traversing a distance of 16 m.
    (a) How much work is done on the trolley?
    (b) How much work is done by the girl?

    Solution

    (a) Given,
    Mass of the girl = 35 kg;
    Mass of the trolley = 5 kg;
    Initial velocity(u) = 4 ms-1 ;
    Final velocity (v) = 0 (as it comes to rest) ;
    distance (s) = 16 m
    By using equation of motion, we get
    v2 - u= 2as
    ⇒ v2 = u2 + 2as
    ⇒ 0 = (4)2 + 2 × a × (16)
    ⇒ 0 = 16 + 32a
    ⇒ −16 = 32a
    ⇒ −16/32 = a
    ⇒ −1/2 = a
    ⇒ a = − 0.5 m/s2

    (Acceleration is negative, therefore, retardation)
    Total mass of the trolley = mass of girl + mass of trolley
    = 36 + 5 = 40 kg
    Force (frictional) acting on the trolley = ma
    = 40 × (-0.5) = –20 N
    Work done on the trolley = Fs
    = 20 N × 16 m
    = 320 J
    (b) As the girl does not move w.r.t. the trolley, the work done by the girl = 0.


    24. Four men lift a 250 kg box to a height of 1 m and hold it without raising or lowering it.
    (a) How much work is done by the men in lifting the box?
    (b) How much work do they do in just holding it?
    (c) Why do they get tired while holding it? (g = 10 ms–2)

    Solution

    (a) Given,
    mass = 250 kg
    height(s) = 1 m
    g = 10 ms-2
    F = mg(g = gravity)
    ⇒ F = 250 kg × g (g = 10 ms-2) = 2500 N
    ⇒ s = 1 m
    ⇒ W = F.s = 2500 × 1 N m = 2500 J

    (b) The men did not do any work in just holding it because there is no displacement while holding the box. Therefore, the work done is zero.

    (c) There is no displacement while holding the box. Therefore, the work done is zero.


    25. What is power? How do you differentiate kilowatt from kilowatt hour? The Jog Falls in Karnataka state are nearly 20 m high. 2000 tonnes of water falls from it in a minute. Calculate the equivalent power if all this energy can be utilized? (g = 10 ms–2)

    Solution

    Power is defined as the rate of doing work. Power measures the rate of work done.
    Power = work/time

    Difference between kilowatt and kilowatt hour

    Kilowatt

    Kilowatt hour

    kW defines how much energy a device uses or generates in a given amount of time.

    kWh defines how much energy that device actually uses or generates.

    kW is a measure of power.

    kWh is a measure of energy.

    kilowatt hour = kilowatt × hour
    or, kWh = kW × hr

    kilowatt = kilowatt hour/hour
    or, kW = kWh/ hr

    Given,
    h = 20 m, and
    mass, m = 2000 × 103 kg = 2 × 106 kg


    26. How is the power related to the speed at which a body can be lifted? How many kilograms will a man working at the power of 100 W, be able to lift at constant speed of 1 m s–1 vertically? g = 10 m/s–2 .

    Solution

    Power = work/time
    Work = force × displacement
    Force = mass ×acceleration
    Acceleration = velocity/time
    Therefore,
    Power = velocity × mass × distance/time × time

    Given,
    Power, P = 100W
    Velocity, v = 1m/s
    Time = 1s
    Displacement, s = 1m
    Acceleration, a = 10m/s2
    From equation,
    P = m × a × s/t
    ⇒ 100 = m×10×1
    ⇒ m = 10kg


    27. Define watt. Express kilowatt in terms of joule per second. A 150 kg car engine develops 500 W for each kg. What force does it exert in moving the car at a speed of 20 m s–1?

    Solution

    One watt is defined as the energy consumption rate of one joule per second. The power is said to be one watt, when a work of 1 joule is done in 1 s
    1 W =1 J/ 1 s
    One watt is also defined as the current flow of one ampere with voltage of one volt.
    1 kilowatt = 1000 Js-1
    Given,
    m = 150 kg;
    power = 500 w/kg:
    speed = 20 m/s
    Power is 500 W/kg.
    So, total power developed by the engine of 150 kg.
    Total power = 150 × 500 = 7.5 × 104 W
    As Power = force × speed
    Force = Power/Speed = (7.5 × 104)/20
    = 3.75 × 103 N


    27. Compare the power at which each of the following is moving upwards against the force of gravity? (given g = 10 ms–2)
    (i) A butterfly of mass 1.0 g that flies upward at a rate of 0.5 ms–1.
    (ii) A 250 g squirrel climbing up on a tree at a rate of 0.5 ms–1.

    Solution

    (i) Given,
    mass of butterfly = 1 g = (1/1000) kg :
    g = 10 ms-2 ;
    speed (v) = 0.5 ms-1
    Power = force × speed
    But, Force = mg
    Therefore, Power = mg × v
    P = 1/1000 × 10 × 0.5
    ⇒ P = 0.5/100 = 5 × 103 W

    (ii) Given,
    mass of squirrel = 250 g = 250/1000 g = ¼ kg
    g = 10 ms-2 ;
    speed(v) = 0.5 ms-1
    Power = force × speed
    But, Force = mg
    Therefore, Power = mg × v
    P = ¼ × 10 × 0.5
    ⇒ P = 1.25 W
    Thus, the power with which the squirrel is climbing is more than that of a butterfly flying.

    Previous Post Next Post