Download A force of 7 N acts on an object. The displacement is, say 8 m, in the

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Q1: A force of 7 N acts 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 the displacement. What
is the work done in this case?
Answer: Let F be the force acting on the object displacing it by distance S, in the same
direction of the force.
∴ F = 7N and S = 8m
Work done (W) = Force (F) × Displacement (S)
= 7 x 8 = 56 Joules
Q2(NCERT): When do we say that work is done?
Answer: Work is said to be done when the following conditions are met:
(i) a force acts on the body.
(ii) A displacment of the body occurs by virtue of applied force in the direction of applied
force.
Q3(NCERT): Define 1 J of work.
Answer: 1 J is the amount of work done by a force of 1 N on an object that displaces it
through a distance of 1 m in the direction of the applied force.
Q4: Is Work a scalar or vector quantity?
Answer: Although work is product of two vectors, it is a scalar quantity.
Q5: Define 1 Watt.
Answer: When one-joule work is done in one second by any agent, then its power is said to
be one watt. It is the unit of Power.
Q6: What is zero work?
Answer: When the force applied or the displacement is zero or when both are perpendicular
to each other, zero work is done.
Mathematically,
Q7: A stone of mass 2Kg is tied with a string and is whirled with a centripetal force of
10N in a circular path. How much work is done for making one revolution?
Answer: Zero work. Since, Centripetal Force (F) and the displacement direction are
perpendicular to each other.
Q8: Calculate the amount of work done in drawing a bucket of water weighing 10kg
from a well of depth 20m.
Answer: Given, mass m = 10 kg
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Acceleration due to gravity, g = 9.8 ms-2
Height h = 20 m
Here, work is done against gravity,
⇒ W = m.g.h = 10 × 9.8 × 20 = 1960 joules
Q9(NTSE): A box is pulled across a floor by applying a force of 100N at an angle of 60
degrees above the horizontal. How much work is done by the applying force in pulling
the object to a distance of 8m?
Answer: Consider the following figure,
A force of 100 N is applied to a box of weight W. Force is a vector quantity applied at angle
60. It has two components, horizontal component (100 cos 60) and vertical component (100
sin 60).
The horizontal component of the force should be considered here, because it causes the
displacement in horizontal direction.
∴ W = 100 cos 60 × 8 = 100 × 0.5 × 8 = 400 J
Q10: A body is thrown vertically upwards. Its velocity goes on decreasing. Write the
change in kinetic energy when its velocity becomes zero.
Answer: Zero ∵ KE = ½ mv2 and v = 0
Q11: A force of 10 N moves a body with a constant speed of 2 m/s. Calculate the power of
the body.
Answer: Self-attempt
Q12: A boy of mass 45 kg climbs up 20 steps in 20 second. If each step is 25 cm high,
calculate the power used in climbing
Answer: Self-attempt
Q13: When a player hits a football it moves along the curved path and then falls to the
ground. Calculate the work done by the force of gravity on the football.
Answer: Self-attempt
Q14: Two friends Ram and Shyam, each having weight of 40 kg, go for rock climbing. Ram
climbs to a height of 3 m in 10 s. and Shyam covers the same height in 12 s. Is the work
done by Ram and Shyam equal ? or not ? Compare the power of Ram and Shyam.
Answer: Self-attempt
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Q15: Is it possible that a force acts on a body still the work done is zero? Explain with
an example.
Answer: Yes. Consider a ball tied to a string is moving in a circular path. The centripetal
force is being applied on the ball and the centripetal acceleration is changing the direction
every instant. Yet no work is done, since centripetal force acting upon is perpendicular to the
displacement (of the moving ball).
Q16: Can a body have energy without possessing momentum?
Answer: Yes. Potential Energy is by virtue of height or position of the object. E.g. water in
the tower tank has (potential) energy though at rest. Another example is compressed spring
stores potential energy though it is at rest.
A nice presentation by an_empa (from Manila) on Work, Power and Energy.
Chapter 6 Work And Energy
Q17 (AIEEE): A pendulum of length 1m and with a bob of mass 1 kg swings in a
vertical plane. Find the work done (from the below said choices) by weight of the bob
when it moves from position A to B (as shown in figure). (Take g = 10 m/sec^2)
a. 4J
b. -4J
c. -5 J
d. 5 J
Answer: Consider the figure below,
The work done by the bob (due to its weight) is change in PE by going up by height (h)
vertically as shown by red arrow.
∴ h = l - l.cos60 = 1 - 1/2 = ½ metres
Gain in P.E. = mgh = 1 × 10 × ½ = -5J
The bob moves from lowest position to higher position, work done is negative.
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Q18: Two bodies of masses m1 and m2 have equal kinetic energies. If p1 and p2 are
their respective momentum, then ratiop1 : p2 is ___?
Answer: As given K.E.s are equal ½ m1v12 = ½ m2v22
∴ p1/ p2 = m1v1/ m2v2
⇒ p1/ p2 = √m1/√m2
Q19: (AIEEE): An object of mass m is raised from the the earth to an height of equal to
the radius of the earth (R). Calculate the gain in potential energy by the object.
Assume gravity of the earth is g.
Answer:
Note the Potential Energy formula i.e. PE = mgh cannot be applied here. This formula (PE =
mgh) is a local approximation formula and is applicable when the height h is very small and
near to the surface of the earth.
The gravitational potential energy at any point is given by the following formula:
U = -G(m1m2)/r
where, G = is Gravitational constant, m1and m2 are the masses of the two objects and r is
the distance between them.
Gain in P.E. will be:
ΔUg = Ufinal - Uinitial = -G(m1m2)/(r + Δh) + G(m1m2)/r
This comes out to be:
⇒ ΔUg = mgΔh. r / (r + Δh)
Applying it in our case.
Δh = R
ΔUg = mgR2/ 2R = mgR/2
... (answer)
Q20(NCERT): 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?
Answer: Given, Force = 140 N
Distance covered = 15m
Work done = Force × Displacement
Force is applied in the direction of displacement.
⇒ Work done in ploughing = 140 × 15 = 2100 J
Thus 2100 joules of work is done in ploughing the field.
Q21(NCERT): What is the kinetic energy of an object?
Answer:Ability of a body to do work due to its motion is called kinetic energy. Its SI unit is
joules. Kinetic energy is directly proportional to the magnitude of velocity/speed. Higher the
speed of an object, higher will be its kinetic energy.
Q22(NCERT): Write an expression for the kinetic energy of an object.
Answer: Let a body of mass m is moving with velocity v then its kinetic energy (K.E.) can be
expressed as:
K.E. = ½ mv2
The SI unit of kinetic energy is joules.
DERIVATION: Let a body of mass m is lying stationary on a frictionless horizontal surface.
When
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a force F acts on it, in time t, it undergoes a displacement S, and it acquires velocity v
If work done W during the process is expressed as:
W = Force × Displacement = F 2kgS
Since F = ma where a is the acceleration.
⇒ W = maS
From laws of motion we know that v2- u2= 2aS
Since u = 0, ∴ v2 = 2aS
or aS = ½ mv2
Thus equation (I) becomes,
⇒ W = ½ mv2
...(I)
Q23(NCERT): The kinetic energy of an object of mass, m moving with a velocity of 5 m/s
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?
Answer: K.E. of the object = 25 J
velocity of the object (v) = 5 m/s
∵ K.E. = ½ mv2
⇒ m = 2 × K.E./v2
⇒ m = 2 × 25 / 25 = 2kg
if velocity is double, v = 2 × 5 = 10 m/s
∴ K.E. (for v = 10m/s) = ½ mv2 = ½ × 2 × 100 = 100J
if velocity is tripled, v = 3 × 5 = 15 m/s
∴ K.E. (for v = 10m/s) = ½ mv2 = ½ × 2 × 225 = 225J
Q 24(MCQ): A light and a heavy body have equal momenta. Which one has greater
kinetic energy?
(a) The light body
(b) The heavy body
(c) Both have equal K.E.
(d) none of these
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Q1: A roller coaster is shown in figure, at points i.e. ➊ and ➋, comparatively at what
point
(i) potential energy is maximum
(ii) kinetic energy is maximum
Answer: At point ➊, comparatively kinetic energy will be maximum and at point ➋ potential
energy will be maximum.
Q2(NCERT): Look at the activities listed below. Reason out whether or not work is done
in the light of your understanding of the term ‘work’.

Suma is swimming in a pond.

A donkey is carrying a load on its back.

A wind-mill is lifting water from a well.

A green plant is carrying out photosynthesis.

An engine is pulling a train.

Food grains are getting dried in the sun.

A sailboat is moving due to wind energy.
Answer: Work = Force × Displacement × cos θ.
(i) As Suma is swimming she is applying force and also covering some distance. The
angle(θ) between the force and the distance is not 90°, so in language of physics she is
doing work.
(ii) No work is done in case the donkey is travelling on the straight road (because the angle
θ will become 90° according to the weight which is acting perpendicularly downwards). If the
donkey is travelling on a slant then the work will be positive or negative according to the
angle.
(iii) When a windmill is lifting the water the bucket covers the distance in the direction of the
force. So the angle θ will be 0° and work will be done.
(iv) No work will be done in photosynthesis as there is no distance covered by the plants in
this process. In language of physics no work is done.
(v) When an engine pulls a train the angle between distance and force becomes 0° and work
will be done.
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(vi) No work is done in this process as the grains are not moving or covering some distance
when being dried in the sun, hence no work is done.
(vii) When wind energy is applied to the boat it starts moving in the direction of the force
applied by the wind. So the angle between distance and force becomes 0° and some work
will be done.
Q3: Does the work done on a load depend on the time taken to raise the load?
Answer: No, work does not depend on time.
Q4(NCERT): 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 object is experiencing a projectile type of motion, so there would be two types
of displacements- namely vertical displacement and horizontal displacement.
CASE-I (Vertical displacement):
In vertical displacement the object finally touches the ground so the height becomes zero.
Work done by an object is given by
Work = Force due to gravity × Height(displacement)
Since Height attained by the object is zero, whole work comes out to be zero.
CASE-II (Horizontal displacement):
At all the cases horizontal displacement by the throw is acting perpendicular to the force
applied by the gravity. Since the angle is 90° the work done will be zero.
Hence in both the cases the work done by the gravity is zero.
Q5. A spring is stretched, will the work done by the stretching force positive or
negative or zero.
Answer: Since the displacement of the stretched force and the force are acting in the same
direction the angle between them becomes zero. So the work done will be positive.
Q6: Is work a scalar or vector quantity?
Answer: Work is a scalar quantity.
Q7(NCERT): A battery lights a bulb. Describe the energy changes involved in the
process.
Answer: A battery stores chemical energy which transforms into an electrical energy as
electric current. The electric current thus lights the bulb (light energy) and bulb also gets
warmer (Heat energy). The energy changes involved here are:
Chemical Energy ➼ Electrical Energy ➼ Light Energy + Heat Energy
Q8: Assuming the earth revolves around the sun in a perfect circular orbit due to
centripetal force exerted by the Sun. Is the work done by the Sun positive or negative.
Answer: Zero work is done. Since centripetal force is perpendicular to displacement (θ
=90°), no work is done.
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Q9(NCERT): Write an expression for the kinetic energy of an object in terms
(i) velocity (v) of an object
(ii) in terms of momentum (p) of an object.
Answer: Let m be the mass of the object, v is the velocity of the object and p is the
momentum of the object. Kinetic Energy (Ek) can be expressed as:
(i) Ek = ½ mv2
(ii) Ek = p2/2m
Q10: A heavy stone is lowered to the ground with the help of rope. Is the work done
by the applied force positive or negative?
Answer: Negative work. The force which holds the stone is applied in upward direction while
the displacement occurs in the downward direction (θ = 180°).
Q11: What is the relationship between kinetic energy and work?
Answer: Work done by the resultant force acting on a moving object is equal to change in
kinetic energy of the particle.
i.e. if v is the final velocity and u is the initial velocity then work done (W)
W = ΔEk = Ek (final vel.) - Ek (initial vel.) = ½ mv2 - ½ mu2
It is also called as work - energy theorem.
Q12(NCERT): 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.
Answer: Given, mass of the object (m) = 20kg
initial velocity of the object (u) = 5m/s
final velocity of the object (v) = 2 m/s
According to Work Energy theorem, W = ½ mv2 - ½ mu2
∴ W = ½ × 20g × ( 22- 52) = 10 × (4 - 25) = 10 × -21 = -210J
The negative sign shows that the force is acting in the direction opposite to the motion of the
object.
Q13: 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.
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Answer: The gravitational force (F) = 10 × 9.8 = 98N or 10 kg Wt acts downwards.
The block moves from A to B horizontally. Thus the direction of motion is perpendicular to
the direction of gravitational force (θ = 90°), therefore work done is zero.
Q14: Can kinetic energy of an object be negative?
Answer: No it can't be negative. Since mass m and v2 are always positive, thereforekinetic
energy is always positive.
Q15: The magnitude of momentum of a moving object is equal to its kinetic energy.
What will be the velocity of the object considering SI units for the system?
Answer: Since momentum (mv) = Kinetic energy ( ½ mv2)
⇒ mv = ½ mv2
⇒ v = 2 m/s
Q16: What is mechanical energy? What are its types?
Answer: Energy can exist in many forms like chemical energy, electrical energy, mechanical
energy etc. Mechanical energy means the ability of to do work due to its position,
configuration or motion. Thus mechanical energy is the sum of potential energy (PE) and
kinetic energy (KE).
Q17: What is law of conservation of energy?
Answer: Law of conservation of energy states that energy can neither be created nor be
destroyed, but it can be transformed from one form to another.
As shown in figure, the ball is dropped from height h. It reaches ground with velocity (vf )
At Top: y = h, velocity v = 0, Ek = 0 and Ep = mgh
At position y:velocity = v, Ek = ½ mv2 and Ep = mgy
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At bottom: y = 0, velocity = vf , Ek = ½ mvf2 and Ep =0
Total Energy = Ek + Ep = Constant
(neglecting air resistance)
⇒ mgh = ½ mv2 +mgy = ½ mvf2
Q18(NCERT): The potential energy of a freely falling object decreases progressively.
Does this violate the law of conservation of energy? Why?
Answer: No. The process does not violate the law of conservation of energy. When the body
falls from a height, its potential energy transforms into kinetic energy progressively. A
decrease in the potential energy is equal to an corresponding increase in the kinetic energy
of the body. In the process, total mechanical energy of the body remains constant.
Therefore, the law of conservation of energy is obeyed.
Q19: From where the falling raindrops gains kinetic energy?
Answer: The potential energy or gravitational potential energy of the falling raindrops gets
converted into kinetic energy.
Q20: What is a conservative forces? Give an example.
Answer: Conservative force is the one which does a work moving an object from one point to
another but is independent of path. Generally the work done is reversible.
e.g. gravitational force is a conservative force. When an object rises up to a certain height
(h), its potential energy increases by a value of mgh. This increase in potential energy is
independent of path being followed i.e. when object is lifted up or pushed up through a slant.
When the object falls back to same ground level, its potential energy also decreases to initial
value.
Q21: A person throws a stone of mass 3kg. It crosses the top of a wall of 2m high at a
speed of 4m/s. What is the total mechanical energy possessed by the stone when it
crosses the wall?
(Take g = 10 m/s)
Answer: Total Mechanical energy = K.E. + P.E.
Given, velocity (v) = 4 m/s
mass of the stone = 3kg
height (h) of wall = 2 m
K.E. of stone = ½ mv2 = ½ × 3 × 42
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K.E. = 24 J
P.E. = mgh = 3 × 10 × 2 = 60J
∴ Total Mechanical energy = 24 J + 60J = 84J
Q22: What are other forms of energies apart from mechanical energy? Does Law of
Conservation of Energy apply to all of them?
Answer: Energy can exist in various forms such as:
✶solar energy, ✶light energy ✶heat energy ✶hydro energy ✶wind energy, ✶chemical
energy, ✶electric energy, ✶magnetic energy ✶geothermal energy, ✶nuclear energy
✶sound energy ✶biomass energy ✶ocean thermal energy ✶mechanical energy
All forms of energy obey law of conservation of energy.Following figure illustrates an
example how electrical energy can be transformed to another form of energy and vice-versa.
Q23(NCERT): What are the various energy transformations that occur when you are
riding a bicycle?
Answer: The muscular energy of the cyclist transforms into kinetic (mechanical) energy. Due
to movement of muscles, cyclist also produces heat. Thus muscular energy is converted into
heat energy. The tyres of the cycle are in contact with road and experience friction. Due to
friction, tyres get heated. Thus kinetic energy of tyres get converted to heat energy.
Muscular energy of cyclist ➟ Kinetic Energy + Heat energy
Kinetic Energy of Tires + Energy of Friction ➟ Heat Energy in tyres
Q24(NCERT): 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?
Answer: Yes the transfer of energy takes place. The muscular energy is converted to heat
energy which is dissipated to the surrounding and partly absorbed by the rock as well.
Q25(MCQ): If the momentum of a body is increased n times, its kinetic energy increases
(a) n times
(b) 2n times
(c) √n times
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(d) n2 times
Answer: (d) n2 times
(Hint: Kinetic energy in terms of momentum(p) is Ek = p2/2m)
Q26: Define potential energy? What are different types of potential energy?
Answer: The energy stored in a body or a system due to its position in a force field or due to
its state of strain or configuration is called potential energy.
Different types of potential energy are:
1. Gravitational Potential Energy: Due to its position (height) of an object, it
possesses gravitational potential energy.
2. Elastic Potential Energy: Due to its configuration or strain a stretched or
compressed spring contains elastic potential energy.
3. Electrostatic Potential Energy: Energy possessed by a charge due to its
position in electric field is called electrostatic potential energy.
In mechanics, we deal with gravitational potential energy and elastic potential energy.
Q27(CBSE 2011): Name the type of energy possessed by a raised hammer?
Answer: Potential energy (due to its position).
Q28(CBSE 2011): Identify the kind of energy possessed by a running athlete.
Answer: Kinetic energy (due to motion of athlete).
Q29(CBSE 2011): What is the form of energy possessed by a running car?
Answer: Kinetic energy (due to motion of car).
Q30(CBSE 2011): Name the type of energy possessed by the following?
(i) Stretched slinky (ii) Speeding car
Answer:
(i) stretched slinky : potential energy due to its configuration.
(ii) speeding car: kinetic energy due to its speed.
Q31: Mathematically show that law of conservation of energy holds true for a free
falling object. Consider negligible air resistance.
Answer: According to law of conservation of energy,
Total Mechanical Energy (E) = Kinetic energy (Ek)+ Potential Energy (Ep) = constant
Let a body of mass m falls from a point A, which is at a height h from the ground as shown.
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At A:
velocity = 0
height = h
Kinetic Energy Ek = 0
Potential Energy Ep = mgh
Total Energy (E) = Ek + Ep = 0 + mgh = mgh
....(I)
At B:
Total Displacement (S) = x
Height from ground = h - x
According to law of motion, v2 - u2 = 2aS
⇒ v2 - 0 = 2gx
(initial velocity = 0)
v2 = 2gx
Kinetic Energy Ek = ½ mv2 = ½ × m × 2gx
⇒ Ek = mgx
Potential Energy Ep = mg (h -x)
∴ Total Energy (E) = Ek + Ep = mgx + mg(h -x) = mgx + mgh - mgx = mgh ...(II)
At C:
Height w.r.t. ground = 0
Displacement from top (S) = h
According to law of motion, v2 - u2 = 2aS
⇒ v2 - 0 = 2gh
v2 = 2gh
Kinetic Energy Ek = ½ mv2 = ½ × m × 2gh
⇒ Ek = mgh
Potential Energy Ep = mg × 0 = 0
∴ Total Energy (E) = Ek + Ep = mgh + 0 = mgh
...(III)
Since equations I and II and III are same, we conclude under the force of gravity, the
mechanical energy of a body remains constant.
Q32(NCERT): 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.
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Answer: Given mass pf the object = 40 kg
acceleration du to gravity (g) = 10 m/s2
Height w.r.t ground h = 5m
Potential Energy (Ep) = mgh = 40 × 10 × 5 = 2000J
Velocity of the object (at h = 5m) = 0 m/s
Kinetic Energy Ek = ½ mv2 =0
Under the force of gravity, the mechanical energy of a body remains constant.
∴ Total Energy (E) = Ek + Ep = 0 + 2000J = 2000J
At mid-way, the height of the object (x) = h/2 = 5/2 = 2.5m
Displacement from top (S) = h/2 = 2.5m
Using Laws of motion equation, v2 - u2 = 2aS
⇒ v2 - 0 = 2gS
Kinetic Energy Ek at mid-way = ½ mv2 = ½ × m × 2gS = mgS = 40 × 10 × 2.5 =1000J
Potential Energy (Ep) at mid-way = mgx = 40 × 10 × 2.5 = 1000J
Alternately, Total Energy = Ek + Ep = 2000J
⇒ Ep = 2000 - Ek = 2000 - 1000 = 1000J
Q33(Guj Board): Calculate potential energyof a person having 60 kg mass on the summit
of Mt. Everest. Height of Mt. Everest is 8848 m from sea level. (g = 9.8 m/s2)
Answer: Given mass of the person (m) = 60 kg
Height of the person (h) = 8848 m
Potential Energy possessed by person at the summit = mgh
= 60 × 9.8 × 8848 = 5202624 J = 5.2 × 106 J
Q34(NCERT): What is the work done by the force of gravity on a satellite moving round
the earth? Justify your answer.
Answer: Work done by force of gravity on a satellite moving round the earth = 0 J.
Force of gravity (F, centripetal force) acts towards the center of the earth.
The displacement/motion of the satellite (s) is perpendicular to the force of gravity i.e. acting
tangentially.
Therefore θ = 90°.
Work done = F × s × cos θ = F × s × cos 90° = 0
(∵ cos 90°)
Q35(NCERT): Can there be displacement of an object in the absence of any force acting
on it?
Answer: Yes. If an object is already moving with uniform motion, in that case net external
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force is zero, still there is displacement.
Q36: Define power.
Answer: The rate at which work is done by a force is called power. Work done per unit time
is called power.
Power = Work(W) ÷ Time (T)
The SI unit of power is Watt. 1 Watt = 1 J per second.
Q37: A motor pump does same amount of water by lifting underground water in
shorter time than a hand pump. What do you infer from this statement?
Answer: Both motor pump and hand pump does same amount of work but the motor pump
does it in shorter time. It indicates, rate of doing work by motor is higher. Or power of motor
pump is higher than that of hand pump.
Q38: What is a horsepower?
Answer: A horsepower (hp) is another unit of power. 1 hp = 746 Watts
Q39: What is Kilowatt hour?
Answer: Kilo watt hour is a unit of energy. It define the relation between work and power. 1
kW hour means one killowatt of power is supplied for an hour.
⇒
1 kWh = 1000J/s × 3600 s = 3.6 × 106 J
Q40(NCERT): A certain household has consumed 250 units of energy during a month.
How much energy is this in joules?
Answer: 1 unit of energy = 1 kilowatt hour (kWh).
6
∵ 1 kWh = 3.6 × 10 J
∴ 250 units of energy = 250 × 3.6 × 106 = 9 × 108 J