NCERT solution for class 9// Science // Chapter 11 --Work and energy
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NCERT solution for class 9// Science // Chapter 11 --Work and energy
Chapter 11- Work and
energy
Page number 148.
Q1. A force of 7 N 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 displacement what is the work done in this case.
Answer-
Force
, F = 7 N
Displacement S = 8m .
Work W = F x S
W = 7 x 8 = 56 Nm.
Page number 149 .
Q2. When do we say that work is done?
Answer- When the force is applied and object and object moves through a
distance in the direction of applied force then we say that work is done
It is equal to the product of force and
displacement
W = F x S
Q3. Write an expression for the work done
when a force is acting on an object in the direction of its displacement?
Answer- When a force F is applied on a body and it moves the body by
distance S in the direction of force then the amount of work done
is
Work done = force X distance through which
the body moves
W = F x S
Q4. Define 1j of work ?
Answer - Work = force x distance
So the unit of the work is depend upon the unit of the force and
distance in SI units the force is the show in Newton while the distance is
measured in metre so the unit of work is Newton metre and it is called joule
.The SI unit of work is joule .
Definition - The amount of work done when a force of 1 N moves a
body by a distance of 1 m in its own direction is equal to 1 Joule .
Q5. A pair of bullocks exerts a force of
140 N on a plough . The field being plough is 15 m long .How much work is done
in ploughing the length of the field .
Answer- force F = 140 N
Displacement S = 15 m
Work done = F x S
W = 140 N x 15 m
W = 2100 Nm ( J) .
Page number 152.
Q6. What is the kinetic energy of an
object? .
Answer - The energy possessed by a body by its motion is called Kinetic energy.
Moving object can do some work due to its kinetic energy
Ex-
Moving wind has a kinetic energy, flowing river water has kinetic energy
.
Q7. Write an expression for the kinetic
energy of an object?
Answer- A body moves in with accelerate with velocity = v
starting velocity is = u
it cover the distance = S
v2 - u2 = 2 as
v2- 0 = 2as
a = v2/ 2s
The force acting on a body is
F = ma
F = m x v2/2s
F x s = 1/2 mv2 ( F x
s = W )
So, W = 1/2 mv2 so it also similar
to
K.E. = 1/2 mv2
Q8 . The kinetic energy of an object of
mass m moving with a velocity of 5 ms 25J what will be its kinetic energy will
its velocity is doubled what will be its kinetic energy will its velocity is
increased by 3times .
Answer
Mass = m = 1
Velocity v = 5 m s
K.E. = 1/2 mv2
25 = 1/2 m ( 5)
m1 = 2 kg .
Case 1.
Velocity v1 = 10 ms
K.E. = 1/2 m1v12
= 1/2 m1 ( 102 )
= 1/2 100m2s2
= 1/2 x 2kg x 100
= 100 J
Thus velocity is doubled kinetic energy
becomes 4 times .
Case 2 .
Velocity = v2 = 15 ms
K.E. = 1/2 m1v22
K.E. = 1/2 x 2 kg. x (15)2
K.E. = 225 J
Thus when velocity is increased three times
its kinetic energy becomes 9 times .
Page number 156 .
Q9. What is power?
Answer- Power is the rate of doing work at the rate of utilising energy .
Q10. Define 1 watt of power?
Answer- One watt is the power of an agent which does work the rate of 1 joule
per second .
Q11. A lamp consumes 1000 J of electrical
energy in 10 seconds what is its power.
Answer- Electrical energy = 1000 J
Time = 10 second .
Power = electrical energy consumed / time
taken
P= 1000 J / 10 s
P = 100 W.
Q12. Define average power?
Answer- Average power is the total work done is dividing by total time taken
P = W / t
Exercise
Q1. Look at the
Activities listed below reasons 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 windmill 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
Answer-
a) Yes work is done in that condition but
this is negative work done because applied force is in backward direction while
the displacement is in forward direction.
b) No work is not done in this case because
the force is acting downward while the displacement takes place forward
c) Yes the work is done because the applied
force and the displacement is in same direction.
d) No work is not done in this case because
neither the force is applied nor the displacement takes place.
e) Yes work is done because the engine
applying the force on the train and train is showing displacement in the
direction of force.
f) No work is not done in this case because
neither the force is applied nor there is any displacement .
g) Yes the work is done because the energy is
applying the force and the sailboat is displace in the direction of
applied force
Q2. 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?
Answer- The work done by the force of gravity on the object will be zero .
Work done against the gravity = mgh
The height becomes zero so the mass and
acceleration due to gravity remains constant the initial and final points of
the part of the object lie on the same horizontal line .
The chemical energy of battery is converted
into electrical energy which is further converted into heat and light
energy.
Q4. Certain force acting on a 20kg mass
changes its velocity from 5 m/s to 2 m/ s calculate the work done by the
force .
Answer- mass - 20 kg .
Initial velocity - u = 5m/s
Final velocity - v = 2 m/ s
Time t = 1s .
By first equation of motion,
v = u + at
a = v-u / t
a = 2-5/1
a= -3m/S2
By third equation of motion
v2 - u2 = 2as
(2)2-(5)2= 2 x
(-3)
-21= -6m/S2
s= 7/2
Work done. = F x s
W = m x a x s
W= 20 x (-3) x 7/2
W= -210 joules .
Q5. Mass of 10kg 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
Answer -
Mass = 10 kg
Acceleration due to gravity g = 10m/s2
Work done by force of gravity = m x g x
h
In above case the work done on the object by
the force of gravity is zero because the line joining A and Bis
horizontal his the height is zero there is no component of the force of gravity
in the direction of displacement.
Q6. The potential energy of a freely
falling object decreases progressively does this violate the law of
conservation of energy why?
Answer- This does not violate the laws of conservation of energy because in
the freely falling object the potential energy decreases and its kinetic energy
increases so the total energy remains same
Q7. What are the various energy
transformation that occur when you are riding a bicycle?
Answer- The various energy transformation in riding a bicycle are potential
energy to muscular energy to mechanical energy .
Q8. Does the transfer of energy take place
when you push a huge rock with all you might and fail to move it where is the
energy used spend going.
Answer- No the transfer of energy does not takes place when you push the huge
rock with all your might when we push the rock the muscles are stretched and
the blood is displaced in muscles were rapidly and these changes consume energy
and we feel tired.
Q9. A certain household has consumed 250
units of energy during a month how much energy is this in joules.
Answer- Energy consumption during a month = 250 units .
1 unit = 1 kwh
250 units = 250kWh
Now , 1kWh= 36,00,000J
250kWh = 36,00,000J x 250
= 90,00,00,000 J
Q10. 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 halfway down.
Answer- Mass m = 40 kg
Height h = 5 m
Acceleration due to gravity g = 10m
/S2
PE = mgh
= 40 x 10 x 5
= 2000J
When the object is half way down
mass m = 40 kg
height h =2.5 m
acceleration due to gravity g
= 10 m/ s 2
Initial velocity u = 0
Third equation of motion.
v2 - u2 = 2as
v2 = u2 + 2as
v2 = 0+ 2 x 10 x 2.5
= 20 m/ S2
K.E. = 1/2 mv2
=
1/2 x 40 x 20 x 2.5
=
1000J
Q11. What is the work done by the force of
gravity on a satellite moving around the earth justify your answer.
Answer- The work done on a satellite moving around the earth is zero because
when the satellite moves in a circular path then the centripetal force acts
towards the centre of the earth while the direction of the motion is tangential
to the circle and both are perpendicular to each other.
Work done = F x S x cos 90°
W. = F S cos
90° = 0
So in the case of uniform circular motion the
work done is zero.
Q12. Can there be displacement of an
object in the absence of any force acting on it think discuss this question
with your friends and teachers?
Answer- When any body is in rest then the Newton's first law of motion is
applicable to that body it continues remains at rest unless the external
unbalanced force acts on it has no displacement takes place in the absence of
force but when the body is moving then force is required to stop it in this
case the displacement takes place in absence of force
Q13. A person hold a bundle of hay over
his head for 30 minutes and gets tired has he done some work or not justify
your answer?
Answer- No the person has not done any work on the bundle of hay because there
is no displacement of the bundle .
Q14. An electric heater is rated 1500 W
.How much energy does it use in 10 hours ?
Answer- power = 1500 W
1500 W / 1000 = 1.5 kW
Time = 10 h
Power = Work done / Time taken
Electrical energy = power X Time taken
Electrical energy = 1.5 x 10
=
15 kWh
Q15. 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 happened to its energy when eventually is it a violations of a law of
conservation of energy
Answer - The bob eventually comes to rest because of the
friction at the point of pendulum and the friction of their acting on the bob
convert the mechanical energy e into heat energy slowly.
Eventually this heat energy-efficient
goes into the environment
No it is not a violent of a law of
conservation of energy .
Q16. 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
Answer - mass = m
Constant velocity = v
Energy of body = 1/2 mv2
So the same amount of work should be done on
the body in order to bring it to rest.
Work done of a object is =1/2 mv2
Q17. Calculate the work required to be
done to stop a car of 1500 kg moving at a velocity of 60 km / h
Answer -
Mass = 1500 kg
Velocity = 60km/ h
= 60 x 1000m / 3600s = 16.67ms
K.E. = 1/2 mv2
K. E. = 1/2 x 1500 kg x( 16.67 )2
= 208416.67 J
Q18. In each of the following a force F is
acting on an object of mass m the direction of displacement is from waste to
East shown by the longer Arrow observe the diagram carefully and state whether
the work done by the force F is negative positive or zero.
Answer-
a) In first condition work done is zero
because the force acting is perpendicular to the displacement .
b) In second case the work done is positive
because displacement takes place in the direction of applied force.
c) In third case the work done is negative
because the displacement takes place in the direction of opposite to the
applied force.
Q19. 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?
Answer- Yes Sony is correct because when the object is in rest the velocity is
zero and when it more in acceleration is zero the several force act on it and
they cancel to each other when the object in motion and moving with constant
velocity then its acceleration is zero in this condition a so many forces
acting at a same time and balance to each other.
Q20. Find the energy in kWh consumed in 10
hours by four devices of power 500 W each
Answer-
Power of each device = 500 W
Power of 4 device = 4 x 500
=
2000W
2000/1000 = 2kW .
Time - 10 hours
Power = work done / time taken
Electrical energy = Power x Time taken
= 2kW x 10
hrs = 20 kWh
Q21. A freely falling object eventually
stop on reaching the ground what happens to its kinetic energy .
Answer- When the object stops since the velocity becomes zero its kinetic
energy also becomes zero on reaching the ground.
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