NCERT Solutions for Class 9 Science Chapter 11 - Work And Energy
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NCERT Solutions for Class 9 Science Chapter 11 Work And Energy are provided here with simple step-by-step explanations. These solutions for Work And Energy are extremely popular among class 9 students for Science Work And Energy Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the NCERT Book of class 9 Science Chapter 11 are provided here for you for free. You will also love the ad-free experience on Meritnation’s NCERT Solutions. All NCERT Solutions for class 9 Science are prepared by experts and are 100% accurate.

Page No 68:

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:

When a body falls freely towards the earth, then its total energy remains constant.
As per the conservation of energy, the total energy is the sum of the kinetic energy and the potential energy wherein the sum of the two remains the same.
Hence, the correct answer is option C.

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Question 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

Answer:

The car is moving on a levelled road in the horizontal direction. It means the vertical height of the car is constant.
The potential energy depends on the height and the height from the ground remains the same. Therefore, the potential energy of the car does not change.
Hence, the correct answer is option A.

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Question 3:

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

Answer:

In case of negative work, the force and displacement will be in opposite direction. It is possible when the angle between the force and displacement is $180 °$ .
Work done, W = FS cosθ
Here, θ = 180º and cos 180º = –1
Thus, work done, W = –FS
Here, F is force and S is displacement.
Hence, the correct answer is option D.

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Question 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) momenta
(c) potential energy
(d) kinetic energy

Answer:

Acceleration due to gravity:
$g = \frac{G M_{E}}{{R_{E}}^{2}}$
Here, M_E is the mass of the Earth and R_Eis the radius of the Earth.
This means the acceleration due to gravity is always the same upto a certain height. It is independent of the mass and size of the body.
Hence, the correct answer is option A.

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Question 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 ms^–2)
(a) 6 × 10³ J
(b) 6 J
(c) 0.6 J
(d) zero

Answer:

The scientific work is considered to be done when force and the displacement along the same line. In this case, the weight of the body is acting vertically downward direction and the displacement of the girl is along the horizontal direction. It means the force and the displacement are perpendicular to each other. Therefore, the work done will be zero.
Hence, the correct answer is option D.

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Question 6:

Which one of the following is not the unit of energy?
(a) joule
(b) newton metre
(c) kilowatt
(d) kilowatt hour

Answer:

Energy is defined as the capacity to do work.
joule, newton metre and kilowatt hour are the units of energy whereas kilowatt is the unit of power.
Power is rate of energy consumption.
Hence, the correct answer is option C.

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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:

Work done is given by
W = Fs cos θ, where F is the force, s is displacement and θ is the angle between the force and displacement. Thus, work done does not depend directly on the initial velocity of an object.
Hence, the correct answer is option D.

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Question 8:

Water stored in a dam possesses
(a) no energy
(b) electrical energy
(c) kinetic energy
(d) potential energy

Answer:

The water is stored in a dam upto certain height. Potential energy is the energy stored in an object due to its position. Therefore, the water stored in a dam possesses potential energy.
Hence, the correct answer is option D.

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Question 9:

A body is falling from a height h. After it has fallen a height $\frac{h}{2}$ , 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:

According to the law of conservation of energy, the energy of the body will remain conserved. At height h, the body has potential energy and no kinetic energy. As the body moves down, the potential energy gets converted into kinetic energy. At height $\frac{h}{2}$ , half of the potential energy is converted into kinetic energy. Therefore, at height $\frac{h}{2}$ , the body will possess half potential and half kinetic energy.
Hence, the correct answer is option C.

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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:

Given:
Velocity of the rocket = v,
Let the mass of the rocket to be m.
Initial kinetic energy of the rocket:
$K E_{1} = \frac{1}{2} m v^{2}$
After increasing the velocity of the rocket three times, the kinetic energy of the rocket:
$K E_{2} = \frac{1}{2} m {(3 v)}^{2}$
Ratio of two kinetic energies:
$\frac{K E_{1}}{K E_{2}} = \frac{\frac{1}{2} m v^{2}}{\frac{1}{2} m {(3 v)}^{2}} = \frac{1}{9}$
Hence, the ratio of two kinetic energies is $\frac{1}{9}$ .

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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:
Speed of Avinash, v₁ = 8 ms^-1
Frictional force offered to Avinash, $F_{1} = 10 N$
Speed of Kapil, v₂ = 3 ms^-1
Frictional force offered to Kapil, $F_{2} = 25 N$
Power offered by Avinash,
$\begin{array}{rcl} P_{1} & = & F_{1} \times v_{1} \\ = & 10 \times 8 \\ = & 80 W \end{array}$
Power offered by Kapil,
$\begin{array}{rcl} P_{2} & = & F_{2} \times v_{2} \\ = & 25 \times 3 \\ = & 75 W \end{array}$
Hence, Avinash is more powerful than Kapil.

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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 of a round having a radius of 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.

Answer:

Given: Frictional force $(F) = 5 N$ ,

Total displacement(includes radius of a round) of the boy is,
$\begin{array}{rcl} D & = & 1500 + 2000 + (1.5 \times 2 \times 3.14 \times 100) \\ = & 4442 m \end{array}$
Work done by the boy is,
$W = F \times D = 5 \times 4442 = 22210 J$
Hence, the work done by him is 22210 J.

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Question 13:

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

Answer:

Mechanical energy comprises of potential energy and kinetic energy. If the momentum of the object is zero, it means the velocity of the object is zero and an object is not moving. So, the kinetic energy of the object is zero.
But if the object is kept stationary at certain height then it will have potential energy. Therefore, an object can have mechanical energy even if its momentum is zero.

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Question 14:

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

Answer:

Mechanical energy comprises of potential energy and kinetic energy. If the mechanical energy of an object is zero, it means the sum of kinetic energy and potential energy is also zero.
Kinetic energy:
$K . E . = \frac{1}{2} m v^{2} \Rightarrow v = 0$
Momentum, p = mv = 0
Therefore, an object cannot have momentum if its mechanical energy is zero.

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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:

Given:
Power of a motor pump, (P) = 2 kW
Height (h) = 10 m
Work done by the pump in raising m mass of the water to a height of 10 m is,
$\begin{array}{rcl} W & = & m g h \\ = & m \times 10 \times 10 \\ = & 100 m \end{array}$
Water per minute the pump can raise to a height of 10 m,
$P = \frac{W}{1 minute} = \frac{W}{60 s} \Rightarrow 2000 = \frac{100 m}{60} \Rightarrow m = \frac{60 \times 2000}{100} \Rightarrow m = 1200 kg$
Hence, 1200 kg water can be raised per minute by the pump to a height of 10 m.

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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:

Given:
Weight on planet A = $\frac{1}{2}$ Weight on Earth
Height (h) = 0.4 m
Acceleration due to gravity on planet A (g') = $\frac{1}{2}$ Acceleration due to gravity on Earth (g)
Energy on both the planet will remain the same,
$P E_{A} = P E_{E} \Rightarrow m g' h_{A} = m g h \Rightarrow h_{A} = \frac{g h}{g'} \Rightarrow h_{A} = \frac{g \times 0.4}{\frac{1}{2} g} \Rightarrow h_{A} = 0.8 m$
Hence, the person can jump 0.8 m high on planet A.

Page No 70:

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:

Given:
Constant force applied on the body is F.
Acceleration of the body due to constant force can be written as:
$a = \frac{F}{m}$
$Work done (W) = Force (F) \times Displacement (s) W = F s$
Applying third equation of motion, $v^{2} = u^{2} + 2 a s$
Substituting the values of acceleration and workdone in above equation of motion,
$v^{2} = u^{2} + 2 (\frac{F}{m}) s \Rightarrow v^{2} - u^{2} = 2 \frac{F s}{m} \Rightarrow \frac{v^{2} - u^{2}}{2} = \frac{W}{m} \Rightarrow m \frac{v^{2} - u^{2}}{2} = W \Rightarrow W = \frac{1}{2} m (v^{2} - u^{2})$
Hence, the above equation indicates that the increase in the kinetic energy of the body is equal to the work done by the force on the body.

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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, it is 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. When a body is rotating uniformly in a circular path, the direction of centripetal force and the displacement are perpendicular to each other. In this condition, the work done will be zero.

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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:

Given:
Height (h) = 10 m
Acceleration due to gravity, g = 10 ms^–2
Initial energy of the ball,
$E_{1} = m g h = m \times 10 \times 10 = 100 m$
After striking the reduced energy is,
$\begin{array}{rcl} E_{2} & = & E_{1} - 0.4 E_{1} \\ = & 0.6 E_{1} \\ = & 0.6 \times 100 m \\ = & 60 m \end{array}$
The height upto which the ball will rise,
$E_{2} = 60 m \Rightarrow m g h_{2} = 60 m \Rightarrow h_{2} = \frac{60}{10} \Rightarrow h_{2} = 6 m$
Hence, the ball will bounce back upto 6 m height.

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Question 20:

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

Answer:

Given:
Power of an electric iron, P = 1200 W
Time taken, t = 30 minutes
Number of days in the month of April is 30 days.
Electric energy consumed in the month of April:
$\begin{array}{rcl} E & = & P \times 30 t \\ = & 1200 \times 30 \times 60 \times 30 \\ = & 64800000 J \end{array}$
Hence, the electric energy consumed in the month of April is 64800 kJ.

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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:

Let mass of heavy object = $m_{2}$
Mass of light object = $m_{1}$
Momentum (p) of both the objects are the same.
Ratio of the kinetic energy of the light and heavy objects:
$\begin{array}{rcl} \frac{K E_{1}}{K E_{2}} & = & \frac{(\frac{p^{2}}{2 m_{1}})}{(\frac{p^{2}}{2 m_{2}})} \\ = & \frac{m_{2}}{m_{1}} \end{array}$
Here, $m_{2} > m_{1}$ , it means $K E_{1} > K E_{2}$ .
Hence, the lighter object will have more kinetic energy.

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Question 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.

Answer:

Given:
Mass of car A $(m_{A})$ = 1000 kg
Initial speed of car A $(u_{A})$ = 36 km h^–1 = 10 m s^–1
Final speed of car A $(v_{A})$ = 0,
Friction force (F) = 100 N
Mass of car B $(m_{B})$ = 1000 kg
Initial speed of car B $(u_{B})$ = 0
Power of car A is,
$\begin{array}{rcl} P & = & F \times v_{A} \\ = & 100 \times 10 \\ = & 1000 W \end{array}$
Applying the conservation of momentum,
$m_{A} u_{A} + m_{B} u_{B} = m_{A} v_{A} + m_{B} v_{B} \Rightarrow (1000 \times 10) + (1000 \times 0) = (1000 \times 0) + (1000 \times v_{B}) \Rightarrow v_{B} = \frac{10000}{1000} \Rightarrow v_{B} = 10 {ms}^{- 1}$
Hence, the power is 1000 W and the speed of the car B is 10 ms^–1 .

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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 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?

Answer:

Given: Mass of girl, $m_{1}$ = 35 kg
Mass of trolley, $m_{2}$ = 5 kg,
Initial velocity of the trolley, $u = 4 {ms}^{- 1}$
Final velocity of the trolley, $v = 0 {ms}^{- 1}$
Displacement, s = 16 m
Applying the third equation of motion,
$v^{2} - u^{2} = 2 a s \Rightarrow a = \frac{v^{2} - u^{2}}{2 s} \Rightarrow a = \frac{- 4^{2}}{2 \times 16} \Rightarrow a = - 0.5 {ms}^{- 2}$
Force applied on trolley:
$\begin{array}{rcl} F & = & m a \\ = & 40 \times (- 0.5) \\ = & - 20 N \end{array}$
(a) The work done on the trolley is,
$\begin{array}{rcl} W & = & F \times s \\ = & 20 \times 16 \\ = & 320 J \end{array}$
Hence, the work done on the trolley is 320 J.
(b) The work done by the girl is,
$W_{2} = 0$
Since the girl does not move w.r.t. the trolley (as she is sitting on it).
Hence, the work done by the girl is zero.

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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? (g = 10 ms^–2)

Answer:

Given: Mass of box, m = 250 kg
Height or Displacement, h = 1 m
(a) Work done by the men (W) = $Force (F) \times Displacement (h)$
$\begin{array}{rcl} W & = & F \times h \\ = & m g h \\ = & 250 \times 10 \times 1 \\ = & 2500 J \end{array}$
Hence, the work done by the men is 2500 J.
(b) In holding the box, no work is done. Because in holding the box, the displacement is zero.
(c) They get tired while holding the box because they have to apply force against the gravity to keep it stationary.

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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 utilized? (g = 10 ms^–2)

Answer:

The rate of doing work is called power.
Kilowatt is the unit of power whereas kilowatt hour is the unit of energy. Kilowatt hour is also termed as 'unit', to indicate energy consumption, it is called commercial unit of energy.
Given: Height, h = 20 m
Mass, m = 2000000 kg
Time, t = 1 min = 60 sec
Equivalent power (P) = $\frac{Work done (W)}{time (t)}$

$\begin{array}{rcl} \Rightarrow & P = \frac{W}{t} \\ = & \frac{m g h}{t} \\ = & \frac{2000000 \times 10 \times 20}{60} \\ = & 6666666.67 W \\ = & 6666.67 kW \end{array}$
Hence, the equivalent power is 6666.67 kW.

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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:

Power is also defined as the product of force and velocity.
Mathematically, $Power (P) = Force (F) \times Velocity (V)$
Given: Power, $P = 100 W$
Velocity, $v = 1 {ms}^{- 1}$
Acceleration due to gravity, $g = 10 {ms}^{- 2}$
Relation between power and velocity is,
$P = F \times v P = m g \times v$
$\begin{array}{rcl} m & = & \frac{P}{g \times v} \\ = & \frac{100}{10 \times 1} \\ = & 10 kg \end{array}$
Hence, a man can lift 10 kg.

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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 a speed of 20 ms^–1?

Answer:

One watt is defined as one joule of work done in one second. It is the SI unit of power.
One kilowatt in terms of joule per second,
$1 kilowatt = 1000 watt = 1000 \times \frac{1 joule}{1 second} = 1000 {Js}^{- 1}$
Given: Mass of car engine, $m = 150 kg$
$Power per kg = 500 W$
Speed of moving car, $v = 20 {ms}^{- 1}$
Total power developed by the car engine = $150 \times 500 = 75000 W$
Force developed in moving the car:
$P = F \times v 75000 = F \times 20$
$\begin{array}{rcl} \Rightarrow & F = \frac{75000}{20} \\ = & 3750 N \end{array}$
Hence, the force exerted by the engine to move the car is 3750 N.

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Question 28:

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.

Answer:

(i) Given: $Mass, m = 1.0 g = 10^{- 3} kg$
Velocity, $v = 0.5 {ms}^{- 1}$
The power with which butterfly is moving upward,
$\begin{array}{rcl} P_{1} & = & Force \times Velocity \\ = & m g \times v \\ = & 10^{- 3} \times 10 \times 0.5 \\ = & 0.005 W \end{array}$
Hence, the power with which butterfly is moving upward is 0.005 W.
(ii) Given: $Mass, m = 250 g = 0.25 kg$
$Velocity, v = 0.5 {ms}^{- 1}$
The power with which squirrel is climbing up on a tree,
$\begin{array}{rcl} \Rightarrow & P = m g \times v \\ = & 0.25 \times 10 \times 0.5 \\ = & 1.25 W \end{array}$
Hence, the power with which squirrel is climbing up on a tree is 1.25 W.

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