In this chapter, we provide NCERT Exemplar Problems Solutions for Class 9 Science Chapter 11 Work and Energy for English medium students, Which will very helpful for every student in their exams. Students can download the latest NCERT Exemplar Problems Solutions for Class 9 Science Chapter 11 Work and Energy pdf, free NCERT Exemplar Problems Solutions for Class 9 Science Chapter 11 Work and Energy book pdf download. Now you will get step by step solution to each question.

Textbook | NCERT |

Class | Class 9 |

Subject | Science |

Chapter | Chapter 11 |

Chapter Name | Work and Energy |

Category | NCERT Exemplar |

**NCERT Exemplar Problems Class 9 Science Chapter 11 Work and Energy**

**Multiple Choice Questions (MCQs)**

**Question 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

**Answer:**

(c)Since, total energy of the system is always conserved, so when a body falls freely towards the earth, then its total energy remains constant i.e., the sum of the potential energy and kinetic energy of the body would be same at all points.

(c)

You can also Download **NCERT Exemplar Class 9 Science Solutions** to help you to revise complete Syllabus and score more marks in your examinations.

**Question 2:**A car is accelerated on a lavelled 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

**Answer:**

(a)Potential energy of the car don’t change and kinetic energy changes by as follows

(a)

Let, initial velocity = u

So in this process, the kinetic energy of car becomes 16 times that of initial energy.

**Question 3:**In case of negative work, the angle between the force and displacement is

(a) 0 (b) 45° (c) 90° (d) 180°

**Answer:**

**(d)**(a) Work done W = F.d cos 0

∴Work done at θ= 0°, W = F.d cos 0° (∴cos0° = 1)

=> W = F.d

For angle θ = 0°,

Work done Is positive, so it is not true.

(b) We know that work done, W = F .d cos 0

For angle 0 = 45°,

work done is positive, so it is not true.

(c) We know that work done, W = d cos θ

Work done at θ = 90°, W = F.d cos 90° (∴cos 90° = 0)

W = 0

So, it is not true.

(d) Work done at θ = 180°, W = F.d cosθ (∴ cos 180°= -1)

W = -F. d

For negative work, the angle between the force and displacement should be 180°. (/.e„ force and displacement are antiparallel to each other) So, it is true.

**Question 4:**An iron sphere of mass 10 kg has the same diameter as an aluminium sphere of mass 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) momenta

(c) potential energy (d) kinetic energy

**Answer:**

**(a)**When both spheres are dropped simultaneously from a tower, they have same acceleration. Because, during free fall acceleration of body becomes equals to (g = 9.8 m/s

^{2}) and ‘g’ depends on mass of earth and radius of earth

**Question 5:**A girl is carrying a school bag of 3 kg mass on her back and moves 200 m on a lavelled road. The work done against the gravitational force will be (9 = 10 ms

^{−2})

(a) 6 x 10

^{3}J (b) 6 J (c) 0.6 J (d) zero

**Answer:**

(d)We know that, work done = F . d cos 0

(d)

Force on school bag makes an angle 90° from the road.

i.e., 0 = 90°

W = F . d cos 90° (∴ cos 90° = 0°)

W = 0

Flence, work done against the gravitational force is zero.

**Question 6:**Which one of the following is not the unit’ of energy?

(a) Joule (b) Newton metre (c) Kilowatt (d) Kilowatt hour

**Answer:**

**(c)**We know that, joule, newton metre and kilowatt hour are the units of energy and the kilowatt is the unit of power.

**Question 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

**Answer:**

(d)We know that, W = F.d cos 0

(d)

Here, F = force applied on the object, d = displacement and 0 is angle between force and displacement. So, the work done on an object does not depend upon the initial velocity of the object.

**Question 8:**Water stored in a dam possesses

(a) no energy (b) electrical energy

(c) kinetic energy (d) potential energy

**Answer:**

**(d)**Potential energy is stored energy or the energy of position, so water stored fn a dam possesses potential energy.

**Question 9:**A body is falling from a height h. After it has fallen a height 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

**Answer:**

(c)

(c)

**Short Answer Type Questions**

**Question 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?

**Answer:**

**Question 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?

**Answer:**

Given, force applied by Avinash = 10 N

Speed of Avinash = 8 ms

^{-1}‘

Power of Avinash = F.v = 10×8 = 80W

Now, force applied by Kapil = 25 N

Speed of Kapil = 3ms

^{-1}Power of Kapil = Fv = 25×3 = 75W

Since, Avinash has more power (80 – 75) = 5 W than Kapil. So, Avinash is more powerful.

**Question 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 round about of radius 100 m as shown in figure. 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.

**Answer:**Given, force applied by boy against friction = 5 N

Displacement on the circular path = One cycle + Half cycled = 0 + Half cycle

= 0 + Diameter of circular path (∴Displacement depends on initial and final point)

= 0+ 2r = 0+2×100 [∴r = 100m]

= 0 + 200 = 200 m

∴ Total displacements = 1.5 km + 200 m + 2.0 km

= 1.5 x 1000 + 200 + 2 x 1000km (1 km= 1000 m)

= 3700 m

Work done by boy = F . s cos θ

= 5 x 3700 x cos 0 = 18500 J

**Question 13:**Can any object have mechanical energy even if its momentum is zero?

**Answer:**

Since, mechanical energy is the sum of kinetic energy and potential energy.

And as given that, momentum of the body is zero, it means velocity of the body is zero, so it has kinetic energy equals to zero. But it may have potential energy.

So, even if the momentum of the body is zero, it may have

**mechanical energy.**

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

**Answer:**

Since, mechanical energy = potential energy + kinetic energy

If mechanical energy = 0

So, PE+ KE= 0

⇒ PE=-KE

So, we can say that body may have momentum, in case mechanical energy is zero

**Question 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 ms

^{-2}]

**Answer :**

**Question 16:**

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

**Answer:**

It is given that weight of person on the earth = w {i.e.,w = mg)

And as he can jump upto height (h, = 0.4m)

So, potential energy at this point = mgh = mg x .04 …(i)

And it is given that

**Question 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.

**Answer:**

Consider an object of mass m moving with a uniform velocity u.

Let, it now be displaced through a distance s, when a constant force F acts on it in the direction of its displacement.

From the third equation of motion,

v

^{2}= u

^{2}+ 2as

v

^{2}– u

^{2}= 2as

It is clear that the work done is equal to the change in the kinetic energy of an object.

**Question 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.**Answer:**

Yes, when force acts in a direction perpendicular to the direction of displacement.

e.g., earth revolves around the sun under gravitational force of sun on earth, but no work is done by the sun, though earth has centripetal acceleration.

**Question 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 ms-2]

**Answer:**

If the energy of the ball reduces by 40% after striking the ground, then remaining energy of the ball will

be 60% of initial energy.

Let initial energy of the body of mass (m) at height(h) is (mgh).

According to the question, mgh’ = 60% of mgh [given,h = 10m and g = 10ms

^{-2}]

**Question 20:**

If an electric iron of 1200 W is used for 30 min everyday, find electric energy consumed in the month of April.**Answer:**

**Long Answer Type Questions**

**Question 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?

**Answer:**

**Question 22:**An automobile engine propels a 1000 kg car A along a lavelled road at a speed of 36 kmh

^{-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 6 just after the collision.

**Answer:**

**Question 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 appling 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?**Answer:**

**Question 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? [given = 10 ms

^{-2}].

**Answer:**

Given, m = 250kg, height (h) = 1 m and acceleration due to gravity g = 10 ms

^{-2}(a) Work done by the man in lifting the box

W = Potential energy of box W = mgh

W = 250 x 1×10= 2500 J

(b) Work done is zero in holding a box, because displacement is zero.

(c) In holding the box, the energy of man loses. Due to loss of energy he felt tired.

**Question 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 utilised? [g= 10 ms-2]

**Answer:**

(i) Power is defined as the rate of doing work or the rate of transfer of energy. The unit of . power is watt or kilowatt. [1 kW= 1000W]

(ii) Kilowatt is the unit of power while kilowatt hour is bigger unit of energy

1 kWh = 1000 x 3600 ⇒ 1 kilowatt hour = 3.6x 10

^{6}J

**Question 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 ms-1 vertically? [g – 10 ms-2]**Answer:**

**Question 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 speed of 20 ms-1?**Answer:**

**All Chapter NCERT Exemplar Problems Solutions For Class 9 Science**

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