Ncert Solutions Class 9 Science Chapter 10 Work and Energy - Param Himalaya
Example 10.1 A force of 5 N is acting on an object. The object is displaced through 2 m in the direction of the force (Fig. 10.2). If the force acts on the object all through the displacement, then work done is ?
Solution:
Given Force = 5N , Displacement = 2m
Work = Force × Displacement
W = 5 N × 2 m =10 N m or 10 J.
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1. A force of 7 N acts on an object.The displacement is, say 8 m, in the direction of the force (Fig. 10.3). Let us take it that the force acts on the object through the displacement. What is the work done in this case?
Solution :
Work done = Force × displacement
W = F × S
W = 7 × 8
W= 56 J
Example 10.2 A porter lifts a luggage of 15 kg from the ground and puts it on his head 1.5 m above the ground.Calculate the work done by him on the luggage.
Solution:
Work done = F × s = mg × s
W= 15 kg × $10 m s^{-2}$ × 1.5m
W=225 $kgms^{-2}$
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1. When do we say that work is done?
Solution :
According to science, we say that work is done when a force is applied to the object and it gets displaced under the influence of the force
2. Write an expression for the work done when a force is acting on an object in the direction of its displacement.
Solution:
If a force F acts on an object and has a displacement in the direction of force.
Then
Work = Force × displacement in the direction of force.
W = F × s
3. Define 1 J of work.
Solution:
1J in the amount of work done on an object when a force of 1N displaces the object.by 1m in the direction of the force's line of action.
W = F × S
When F = 1N , S = 1m
Then , W = 1N × 1m
W= 1J
4. 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?
Solution:
Given:
Force (F)= 140N
Displacement (s) = 15m
Work done= Force × displacement
W= F × s
W=140N × 15m
W= 2100Nm
W= 2100 J
Therefore the work done = 2100J.
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Example 10.3 An object of mass 15 kg is moving with a uniform velocity of 4 m/s. What is the kinetic energy possessed by the object?
Solution:
Mass of the object , m = 15 kg
Velocity of the object , v = 4 m /s
$E_{K} = \frac{1}{2}mv^{2}$
$E_{K} = \frac{1}{2}\times 15 \times 4 \times 4$
$E_{K}=120 J$
The kinetic energy of the object is 120J.
Example 10.4 What is the work to be done to increase the velocity of a car from 30 km/h to 60 km/h if the mass of the car is 1500 kg?
Solution:
Mass of the car , m = 1500 Kg
Initial velocity of car , u = 30 km/h
$u= \frac{30 \times 1000 m}{60 \times 60s}$
$u = \frac{25}{3} m/s$
Similarly , the final velocity of the car ,
$v = 60 km/h$
$v= \frac{50}{3} m/s$
Therefore, the initial kinetic energy of the car ,
$E_{ki} = \frac{1}{2}m.u^{2}$
$E_{ki}= \frac{1}{2}\times1500 \times(25/3)^{2}$
$E_{ki}=156250/3 J$
The final kinetic energy of the car
$E_{kf} = \frac{1}{2}mv^{2}$
$E_{kf}= \frac{1}{2}\times 1500 \times (50/3)^{2}$
$E_{kf}=625000/3 J$
Thus , the work done = change in kinetic energy
Work done = $E_{kf} - E_{ki}$
Work done = 625000/3 - 156250/3
Work done = 156250 J
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1. What is the kinetic energy of an object?
Solution :
The energy possessed by an object due to its motion is called kinetic energy.
Kinetic energy
$(E_{K}) = \frac{1}{2}m.v^{2}$
Where , m = mass of the object , v = velocity of the object
2. Write an expression for the kinetic energy of an object.
Solution:
Kinetic energy
$(E_{K}) = \frac{1}{2}mv^{2}$
Where , m = mass of the object , v = velocity of the object
3. The kinetic energy of an object of mass, m moving with a velocity of $5 m s^{1}$ 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 three times?
Solution:
Mass of object = m
Velocity of the object = 5 m /s
Kinetic energy
$E_{K} = 25J$
$E_{K} = \frac{1}{2}mv^{2}$
$25= \frac{1}{2}m \times 5^{2}$
$25= \frac{1}{2}m \times 25$
$m = \frac{25 \times 2}{25}=2Kg$
Case (i) when the velocity is doubled.
Mass of the object , m= 2 kg
And , v = 2 × 5 m/s = 10 m/s
Kinetic energy
$E_{K} = \frac{1}{2}mv^{2}$
$E_{K}=\frac{1}{2}\times2\times10^{2}$
$E_{K} = 100J$
Case (ii) when the velocity is increased three times
$v = 3 \times 5 = 15 m/s$
m = 2 Kg
Kinetic energy
$E_{K} = \frac{1}{2}mv^{2}$
$E_{K}=\frac{1}{2}\times2\times15^{2}$
$E_{K} = 225J$
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Example 10.5 Find the energy possessed by an object of mass 10 kg when it is at a height of 6 m above the ground. Given, g = $9.8 m s^{–2}$.
Solution:
Mass of the object , m = 10 kg
Displacement ( height ) h = 6m
Acceleration due to gravity ,
$g = 9.8 m/s^{2}$
Potential energy = mgh
$P.E= 10\times 9.8\times6$
$P.E=588 J$
The potential energy is 588J.
Example 10.6 An object of mass 12 kg is at a certain height above the ground.If the potential energy of the object is 480 J, find the height at which the object is with respect to the ground.Given, g = $10 m s^{–2}$
Solution:
Mass of the object , m = 12 kg
Potential energy , $E_{p} = 480 J$
$E_{p} = mgh$
$480= 12 \times 10 \times h$
$h= \frac{480}{120}$
$h = 4m$
The object is at height of 4 m.
Example 10.7 Two girls, each of weight 400N climb up a rope through a height of 8m. We name one of the girls A and the other B. Girl A takes 20 s while B takes 50 s to accomplish this task. What is the power expended by each girl?
Solution:
(i) Power expended by girl A:
Weight of the girl , mg = 400 N
Displacement (height),h = 8m
Time taken , t = 20s
$Power = \frac{Work}{time}$
$P = \frac{mgh}{t}$
$P = \frac{400 \times 8}{20}$
$P = 160W$
(ii) Power expended by girl B
Weight of the girl , mg = 400N
Displacement (height),h = 8m
Time taken , t = 50s
$Power = \frac{Work}{time}$
$P = \frac{mgh}{t}$
$P = \frac{400 \times 8}{50}$
$P = 64W$
Power expended by girl A is 160W
Power expended by girl B is 64 W
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Example 10.8 A boy of mass 50 kg runs up a staircase of 45 steps in 9 s. If the height of each step is 15 cm, find his power. Take g = $10 m s^{–2}$.
Solution:
Weight of the boy ,
W= mg = 50 kg × $10 m s^{–2}$
W = 500N
Height of the staircase ,
$h = 45 \times \frac{15}{100} m$
$h = 6.75 m$
Time taken to climb , t =9s
$Power = \frac{Work}{time}$
$P = \frac{mgh}{t}$
$P=\frac{500N \times 6.75}{9s}$
$P=375W$
Power is 375W
1. What is power?
Solution :
The rate of doing work or the rate of conversion of energy is called power.
$Power = \frac{Work}{time}$
$P = \frac{W}{t}$
Where , P = power , W = work , t = time
2. Define 1 watt of power.
Solution:
1 watt is the power of the agent which does 1J of work n one second . In other words , if the rate of energy used is 1 J/s then the power will be 1 W.
When
Work W = 1 J , Time t= 1 sec
Then
$P = \frac{W}{t}$
$P = \frac{1}{1}$
$P = 1 W$
3. A lamp consumes 1000 J of electrical energy in 10 s. What is its power?
Solution:
W= 1000 J , t = 10 sec
i.e; $P = \frac{W}{t}$
$P= \frac{1000J}{10Sec}$
$P= 100 J/sec$
$P = 100 W$
4. Define average power.
Solution :
$Average Power = \frac{Total\:energy\:used}{Total\:time\:taken}$
Exercise Solution :
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’.
a. Suma is swimming in a pond.
b. A donkey is carrying a load on its back.
C. A wind-mill is lifting water from a well.
d. A green plant is carrying out photosynthesis.
e. An engine is pulling a train.
f. Food grains are getting dried in the sun.
g. A sailboat is moving due to wind energy.
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 :
No, work is done by the object or the force of gravity. The reason is that the displacement of the object is taking place in the horizontal direction, whereas, the force of gravity is acting downwards in the vertical direction.
3. A battery lights a bulb. Describe the energy changes involved in the process.
Solution :
First the chemical energy of the battery is converted into electrical energy. Then the bulb converts the electrical energy into heat and light energy.
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.
Solution:
Mass of the object m = 20 kg
Initial velocity u = 5 m/s
Final velocity v = 2 m/s
Work done by the force = change in kinetic energy
$$work= \frac{1}{2}m (v^{2})- \frac{1}{2}m (u^{2})
$$work= \frac{1}{2}m (v^{2} - u^{2})$$
$$work= \frac{1}{2}20 (2^{2} - 5^2)$$
$$work= 10 \times(4-25)$$
$$work= 10 \times (-21)$$
$$Work \ done = -210 J$$
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:
Ans.Mass m = 10 kg, acceleration due to gravity $g = 10 ms^{2}$,
work done by the force of gravity = mgh.
In this case the amount of work done by gravitational force on the object will
be zero because the line joining the points A and B lie in same horizontal plane.
Its gravitational force does not act any component of displacement. Because it is
vertical W = Fs cos 90° = 0.
6. The potential energy of a freely falling object decreases progressively. Does this violate the law of conservation of energy? Why?
Solution :
No, it does not violate the law of conservation of energy here. The reason is that there is a proportional and equal relationship between potential energy and kinetic energy. That is, the total energy of the object is always conserved.
7. What are the various energy transformations that occur when you are riding a bicycle?
Solution :
Muscular energy is used in riding the bicycle. Then the muscular energy is transforms into mechanical energy and kinetic energy.
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?
Solution:
In pushing a huge rock no transfer of muscular energy takes place because when we fail to push it, then that muscular energy is spent in working against the friction between the earth and the rock and no work is done as no displacement occurs.
9. A certain household has consumed 250 units of energy during a month. How much energy is this in joules?
Solution :
1 unit = 1 kWh
Energy consumed by a certain household = 250 kWh
since
$1 kWh = 3.6 \times 10^{6} J$
therefore
$250 kWh= 250 \times 3.6 \times 10^{6}$
$Energy = 9 \times 10^{8} J$
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 :
Mass of the object m = 40 kg
Height h = 5 m
Acceleration due to gravity $g = 10 m/s^{2}$
Potential energy
P= m × g × h = 40 × 10 ×5
P= 2000 J
When the object is half way down the height of the object is = 2.5 m
initial velocity (u) = 0 (thrown from ground/rest)
$v^{2} = u^{2} + 2gh$
$v^{2} = 0 + 2 \times 10 \times 2.5$
$v^{2}= 50$
$Kinetic \:energy = \frac{1}{2}m \times v^{2}$
$K.E= \frac{1}{2} \times 40 \times 50$
$K.E= 1000 J$
11. What is the work done by the force of gravity on a satellite moving round the earth? Justify your answer.
Solution:
The satellite is moving on a round path, displacement in the object is perpendicular to the direction of force.
$W = F \times S \times cos \theta$
$W = F \times S \times cos90^{o}$
$W = F \times S \times 0$
$W = 0$
Therefore, work done is zero.
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 :
No
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
Solution :
Of course, the person does not do any work on a bundle of hay. For this the reason is, no displacement is taking place in the bundle. There is no work in standing with a burden on head. There must be displacement for work.
$$W = F \times S \times cos \theta$$
$$W = F \times S \times cos90^{o}$$
$$W = F \times S \times 0$$
$$W = 0$$
14. An electric heater is rated 1500 W. How much energy does it use in 10 hours?
Solution:
Electric heater’s power (P) = 1500 W
P= 1.5 kW
Energy = power × time
Energy = 1.5 KW × 10 h = 15 kWh
15. Illustrate the law of conservation of energy by discussing the energy changes which occur when we draw a pendulum bob to one side and allow it to oscillate. Why does the bob eventually come to rest? What happens to its energy eventually? Is it a violation of the law of conservation of energy?
Solution :
Ans.When the bob of a simple pendulum is draw to one side and released, it starts oscillating. Let the bob be released from point B, then the mean passes through position. A to reach point C. Again reaches back to B from C. Thus, point B has maximum potential energy. There is no kinetic energy here. Again when the bob reaches the mean position A, it has maximum kinetic energy. There is no potential energy.
In the same way, when the bob reaches again from point A to point C, then this point has maximum potential energy. There is no kinetic energy. Hence, maximum potential energy is converted into maximum kinetic energy and maximum kinetic energy into maximum potential energy which explains the law of conservation of energy. Due to the friction of air, the bob gradually comes to rest and all the kinetic energy is converted into heat energy. That is, energy is transformed from one form to another.
16. An object of mass, m is moving with a constant velocity, v. How much work should be done on the object in order to bring the object to rest?
Solution:
Mass = m , u = v , v = 0
From the work energy theorem
Work done = change in kinetic energy
= $\frac{1}{2}mv^{2} - \frac{1}{2}mu^{2}$
= $\frac{1}{2}m(v^{2} - u^{2})$
= $\frac{1}{2}m(0 - u^{2})$
$work \:done= -\frac{1}{2}mu^{2})$
The work done on the object = $-\frac{1}{2}mu^{2})$
17. Calculate the work required to be done to stop a car of 1500kg moving at a velocity of 60 km/h?
Solution:
Initial velocity of the car
$$u= 60 \frac{km}{h}$$
$$u = \frac{(60 \times 1000)} {60 \times 60}$$
$$u = \frac{50}{3}\frac{m}{s}$$
Final velocity (v) = 0
Initial kinetic energy
$$K_{i}= \frac{1}{2}mu^{2} $$
$$K_{i}=\frac{1}{2} \times 1500 \times (\frac{50}{3})^{2}$$
$$K_{i}=208333.30 J$$
Final kinetic energy
$$K_{f}= \frac{1}{2}mv^{2} $$
$$K_{f}= \frac{1}{2} \times 1500 \times 0$$
$$K_{f}=0$$
Therefore , work done = change in kinetic energy
$$W = K_{i} - K_{f}$$
$$W= 208333.30 - 0 = 208333.30 J$$
18. In each of the following a force, F is acting on an object of mass, m. The direction of displacement is from west to east shown by the longer arrow. Observe the diagrams carefully and state whether the work done by the force is negative,positive or zero.
Solution:
$$W = F.S.cos \theta$$
(i) Since in this diagram displacement is perpendicular to the direction of force, so work done is zero.
$$W=F.S.cos90$$
$$W=0$$
(ii) Since in this diagram displacement is in the direction of force, so work done is positive.
$$W=F.S.cos0$$
$$W=F.S$$
(iii) Since in this diagram displacement is in the opposite direction of the force applied hence work done is negative.
$$W=F.S.cos180$$
$$W=-F.S$$
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:
Soni is right because when the object is at rest and its speed is zero, the acceleration becomes zero. An object can have many forces acting simultaneously, but they can cancel or neutralize each other, when the object is in motion with the same velocity, in that case the acceleration is zero. Even in such a situation many balancing forces can act on an object simultaneously.
20. Find the energy in joules consumed in 10 hours by four devices of power 500 W each.
Solution:
Energy consumed by each device
Energy = Power × time
Energy consumed by four device
Energy = 4 × power × time
E= 4 × 500 × 10= 20000 Wh
E= 20 kWh
Energy in joule
$1 kWh = 3.6 \times 10^{6}j$
So
$E = 20 \times 3.6 \times 10^{6}j$
$E= 7.2 \times 10^{7}j$
21. A freely falling object eventually stops on reaching the ground. What happenes to its kinetic energy?
Solution:
A free falling object eventually stops on reaching the ground since on striking the ground its kinetic energy is transmitted to the ground and converted into sound energy and heat energy.