Selina Concise Physics Class 10 ICSE Solutions Work, Energy and Power

Selina Concise Physics Class 10 ICSE Solutions Work, Energy and Power

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Selina ICSE Solutions for Class 10 Physics Chapter 2 Work, Energy and Power

Exercise 1(A)

Solution 1.

Work is said to be done only when the force applied on a body makes the body move. It is a scalar quantity.

Solution 2.

(i) When force is in direction of displacement, then work done , W = F x S
(ii) When force is at an angle θ to the direction of displacement, then work done, W= F S cos θ

Solution 3.

(a) When force is at an angle θ to the direction of displacement, then work done, W= F S cos θ
(b) (i) For zero work done, the angle between force and displacement should be 90o as cos 90= 0
W = FS cos90o= FS x 0 = 0
(ii) For maximum work done, the angle between force and displacement should be 0o as cos0= 1
Hence, W=FScos 0o=FS

Solution 4.

Two conditions when the work done is zero are:

  1. When there is no displacement (S = 0) and,
  2. When the displacement is normal to the direction of the force (θ = 90o).

Solution 5.

(i) If the displacement of the body is in the direction of force, then work done is positive.
Hence, W= F x S
For example: A coolie does work on the load when he raises it up against the force of gravity. The force exerted by coolie (=mg) and displacement, both are in upward direction.

(ii) If the displacement of the body is in the direction opposite to the force, then work done is negative.
Hence, W =- F x S
For example: When a body moves on a surface, the frictional forces between the body and the surface is in direction opposite to the motion of the body and so the work done by the force of friction is negative.

Solution 6.

Work is done against the force.

Solution 7.

When a body moves in a circular path, no work is done since the force on the body is directed towards the centre of circular path (the body is acted upon by the centripetal force), while the displacement at all instants is along the tangent to the circular path, i.e., normal to the direction of force.

Solution 8.

Work done by the force of gravity (which provides the centripetal force) is zero as the force of gravity acting on the satellite is normal to the displacement of the satellite.

Solution 9.

Work is done only in case of a boy climbing up a stair case.

Solution 10.

When a coolie carrying some load on his head moves, no work is done by him against the force of gravity because the displacement of load being horizontal, is normal to the direction of force of gravity.

Solution 11.

Force applied by the fielder on the ball is in opposite direction of displacement of ball. So, work done by the fielder on the ball is negative.

Solution 12.

When a coolie carries a load while moving on a ground, the displacement is in the horizontal direction while the force of gravity acts vertically downward. So the work done by the force of gravity is zero.

Solution 13.

S.I unit of work is Joule.
C.G.S unit of work is erg.
Relation between joule and erg :
1joule = 1N x 1m
But 1N = 105dyne
And 1m = 100 cm = 10cm
Hence, 1 joule = 105dyne x 102cm
= 107dyne x cm = 107erg
Thus, 1 Joule = 107 erg

Solution 14.

S. I unit of work is Joule.
1 joule of work is said to be done when a force of 1 newton displaces a body through a distance of 1 metre in its own direction.
Previous

Solution 15.

Relation between joule and erg :
1 joule = 1N x 1m
But 1N = 105dyne
And 1m = 100 cm= 10cm
Hence, 1 joule = 105dyne x 102cm
= 107dyne x cm = 107erg
Thus, 1 Joule = 107 erg

Solution 16.

Let a body of mass m fall down through a vertical height h either directly or through an inclined plane e.g. a hill, slope or staircase. The force of gravity on the body is F = mg acting vertically downwards and the displacement in the direction of force (i.e., vertical) is S=h. Therefore the work done by the force of gravity is
W = FS = mgh

Solution 17.

Let a boy of mass m climb up through a vertical height h either through staircase of using a lift. The force of gravity on the boy is F=mg acting vertically downwards and the displacement in the direction opposite to force (i.e., vertical) is S=-h. Therefore the work done by the force of gravity on the boy is
W= FS =-mgh
or,the work W=mgh is done by the boy against the force of gravity.

Solution 18.

The energy of a body is its capacity to do work. Its S.I unit is Joule (J).

Solution 19.

eV measures the energy of atomic particles.
1eV= 1.6 x 10-19J

Solution 20.

1 J = 0.24 calorie

Solution 21.

Calorie measures heat energy.
1calorie = 4.18 J

Solution 22.

1kWh is the energy spent (or work done) by a source of power 1kW in 1 h.
1kWh = 3.6 x 106J

Solution 23.

The rate of doing work is called power. The S.I. unit of power is watt (W).

Solution 24.

Power spent by a source depends on two factors:
(i) The amount of work done by the source, and
(ii) The time taken by the source to do the said work.
Example: If a coolie A takes 1 minute to lift a load to the roof of a bus, while another coolie B takes 2 minutes to lift the same load to the roof of the same bus, the work done by both the coolies is the same, but the power spent by the coolie A is twice the power spent by the coolie B because the coolie A does work at a faster rate.

Solution 25.

WorkPower
1. Work done by a force is equal to the product of force and the displacement in the direction of force.1. Power of a source is the rate of doing work by it.
2. Work done does not depend on time.2. Power spent depends on the time in which work is done.
3. S.I unit of work is joule (J).3. S.I unit of power is watt (W).

Solution 26.

EnergyPower
1. Energy of a body is its capacity to do work.1. Power of a source is the energy spent by it in 1s.
2. Energy spent does not depend on time.2. Power spent depends on the time in which energy is spent.
3. S.I unit of energy is joule (J).3. S.I unit of power is watt (W).

Solution 27.

S.I unit of power is watt (W).
If 1 joule of work is done in 1 second, the power spent is said to be 1 watt.

Solution 28.

Horse power is another unit of power, largely used in mechanical engineering. It is related to the S.I unit watt as :
1 H.P =746 W

Solution 29.

Watt (W) is the unit of power, while watt hour (Wh) is the unit of work, since power x time = work.

Solution 30.

a. Energy is measured in kWh
b. Power is measure in kW
c. Energy is measured in Wh
d. Energy is meaused in eV
Concept insight: Energy has bigger units like kWh (kilowatt hour) and Wh (watt hour). Similarly bigger unit of power is kW (kilo watt).
The energy of atomic particles is very small, and hence, it is measured in eV (electron volt).

Solution 1 (MCQ).

746 W

Solution 2 (MCQ).

The unit kWh is the unit of energy.

Numericals

Solution 1.
Selina Concise Physics Class 10 ICSE Solutions Work, Energy and Power 1

Solution 2.
Selina Concise Physics Class 10 ICSE Solutions Work, Energy and Power 2

Solution 3.
Selina Concise Physics Class 10 ICSE Solutions Work, Energy and Power 3

Solution 4.
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Solution 5.
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Solution 6.
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Solution 7.
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Solution 8.
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Solution 9.
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Solution 10.
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Solution 12.
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Solution 13.

(i) The work done by persons A and B is independent of time. Hence both A and B will do the same amount of work. Hence,
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(ii) Power developed by the person A and B is calculated as follows:
A takes 20 s to climb the stairs while B takes 15 s, to do the same. Hence B does work at a much faster rate than A; more power is spent by B.
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Solution 15.
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Exercise 2(B)

Solution 1.

Two forms of mechanical energy are:

  1. Kinetic energy
  2. Potential energy

Solution 2.

Elastic potential energy is possessed by wound up watch spring.

Solution 3.

(a) Kinetic energy (K)
(b) Potential energy (U)
(c) Kinetic energy (K)
(d) Potential energy (U)
(e) Kinetic energy (K)
(f) Potential energy (U)

Solution 4.

Potential energy: The energy possessed by a body by virtue of its specific position (or changed configuration) is called the potential energy.
Different forms of P.E. are as listed below:

  1. Gravitational potential energy: The potential energy possessed by a body due to its position relative to the centre of Earth is called its gravitational potential energy.
    Example: A stone at a height has gravitational potential energy due to its raised height.
  2. Elastic potential energy: The potential energy possessed by a body in the deformed state due to change in its configuration is called its elastic potential energy.
    Example: A compressed spring has elastic potential energy due to its compressed state.

Solution 5.

Potential energy is possessed by the body even when it is not in motion. For example: a stone at a height has the gravitational potential energy due to its raised position.

Solution 6.

Gravitational potential energy is the potential energy possessed by a body due to its position relative to the centre of earth.
For a body placed at a height above the ground, the gravitational potential energy is measured by the amount of work done in lifting it up to that height against the force of gravity.
Let a body of mass m be lifted from the ground to a vertical height h. The least upward force F required to lift the body (without acceleration) must be equal to the force of gravity (=mg) on the body acting vertically downwards. The work done W on the body in lifting it to a height h is
W= force of gravity (mg) x displacement (h)
= mgh
This work is stored in the body when it is at a height h in the form of its gravitational potential energy.
Gravitational potential energy U= mgh

Solution 7.

The work done W on the body in lifting it to a height h is
W= force of gravity (mg) x displacement (h)
=mgh
This work is stored in the body when it is at a height h in the form of its gravitational potential energy.
Gravitational potential energy U= mgh

Solution 8.

A body in motion is said to possess the kinetic energy. The energy possessed by a body by virtue of its state of motion is called the kinetic energy.

Solution 9.

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Solution 10.

According to the work-energy theorem, the work done by a force on a moving body is equal to the increase in its kinetic energy.

Solution 11.

Body of mass m is moving with a uniform velocity u. A force is applied on the body due to which its velocity changes from u to v and produces an acceleration a in moving a distance S.Then,
Work done by the force= force x displacement
W = F x S———(i)
Selina Concise Physics Class 10 ICSE Solutions Work, Energy and Power 16

Solution 12.

Selina Concise Physics Class 10 ICSE Solutions Work, Energy and Power 17
Both the masses have same momentum p. The kinetic energy, K is inversely proportional to mass of the body.
Hence light mass body has more kinetic energy because smaller the mass, larger is the kinetic energy.

Solution 13.

Kinetic energy is related to momentum and mass as
p = √2mK
As the kinetic energy of both bodies are same, momentum is directly proportional to square root of mass.
Now, mass of body B is greater than that of body A.
Hence, body B will have more momentum than body A.

Solution 15.
The three forms of kinetic energy are:

  1. Translational kinetic energy- example: a freely falling body
  2. Rotational kinetic energy-example: A spinning top.
  3. Vibrational kinetic energy-example: atoms in a solid vibrating about their mean position.

Solution 16.

Potential energy (U)

Kinetic energy (K)
    1. The energy possessed by a body by virtue of its specific position or changed configuration is called potential energy.

    1.The energy possessed by a body by virtue of its state of motion is called the kinetic energy.

     2.Two forms of potential energy are gravitational potential energy and elastic potential energy.

2. Forms of kinetic energy are translational, rotational and vibrational kinetic energy.
     3.Example: A wound up watch spring has potential energy.

3. For example: a moving car has kinetic energy.

Solution 17.

(a) Motion.
(b) Position.

Solution 18.

When the string of a bow is pulled, some work is done which is stored in the deformed state of the bow in the form of its elastic potential energy. On releasing the string to shoot an arrow, the potential energy of the bow changes into the kinetic energy of the arrow which makes it move.

Solution 19.

The compressed spring has elastic potential energy due to its compressed state. When it is released, the potential energy of the spring changes into kinetic energy which does work on the ball if placed on it and changes into kinetic energy of the ball due to which it flies away.
Selina Concise Physics Class 10 ICSE Solutions Work, Energy and Power 18

Solution 20.

When water falls from a height, the potential energy stored in water at a height changes into the kinetic energy of water during the fall. On striking the ground, a part of the kinetic energy of water changes into the heat energy due to which the temperature of water rises.

Solution 21.

Yes, when force is normal to displacement, no transfer of energy takes place.

Solution 22.

Kinetic energy.

Solution 23.

The six different forms of energy are:

  1. Solar energy
  2. Heat energy
  3. Light energy
  4. Chemical or fuel energy
  5. Hydro energy
  6. Nuclear energy

Solution 24.

(a) Potential energy of wound up spring converts into kinetic energy.
(b) Chemical energy of petrol or diesel converts into mechanical energy (kinetic energy)
(c) Kinetic energy to potential energy
(d) Light energy changes into chemical energy
(e) Electrical energy changes into chemical energy
(f) Chemical energy changes into heat energy
(g) Chemical energy changes into heat and light energy
(h) Chemical energy changes into heat, light and sound energy

Solution 25.

(a) Electrical energy into sound energy
(b) Heat energy into mechanical energy
(c) Sound energy into electrical energy
(d) Electrical energy to mechanical energy
(e) Electrical energy into light energy
(f) Chemical energy to heat energy
(g) Light energy into electrical energy
(h) Chemical energy into heat energy
(i) Chemical energy into electrical energy
(j) Chemical energy to mechanical energy
(k) Electrical energy into heat energy
(l) Light energy into electrical energy
(m) Electrical energy into magnetic energy.

Solution 27.

No. This is because, whenever there is conversion of energy from one form to another apart of the energy is dissipated in the form of heat which is lost to surroundings.

Solution 1 (MCQ).

Potential energy
Hint: P.E. is the energy possessed by a body by virtue of its position.

Solution 2 (MCQ).

Chemical to electrical
Hint: When current is drawn from an electric cell, the chemical energy stored in it changes into electrical energy.

Numericals

Solution 1.
Selina Concise Physics Class 10 ICSE Solutions Work, Energy and Power 19

Solution 2.

Mass , m=1kg
Height, h=5m
Gravitational potential energy = mgh
=1 x 10 x 5=50J

Solution 3.

Gravitational potential energy=14700 J
Force of gravity = mg= 150 x 9.8N/kg= 1470N
Gravitational potential energy= mgh
14700 =1470 x h
h=10m

Solution 4.
Selina Concise Physics Class 10 ICSE Solutions Work, Energy and Power 20

Solution 5.

Mass =0.5 kg
Energy= 1 J
Gravitational potential energy= mgh
1=0.5 x10 x h
1=5h
Height, h= 0.2 m

Solution 6.

Force of gravity on boy=mg= 25 x 10 =250N
Increase in gravitational potential energy= Mg (h2-h1)
= 250 x (9-3)
=250 x6=1500 J

Solution 7.

Mass of water, m= 50kg
Height, h=15m
Gravitational potential energy= mgh
=50 x10 x 15
=7500 J

Solution 8.

Mass of man=50kg
Height of ladder, h2=10m

(i) Work done by man =mgh2
=50 x 9.8 x10= 4900J

(ii) increase in his potential energy:
Height, h2= 10m
Reference point is ground, h1=0m
Gravitational potential energy= Mg (h2-h1)
= 50 x9.8x (10-0)
= 50 x 9.8 x10= 4900J

Solution 9.

F=150N

(a) Work done by the force in moving the block 5m along the slope =Force x displacement in the direction of force
=150 x 5=750 J.

(b) The potential energy gained by the block U =mgh where h =3m
=200 x 3=600 J
he potential energy gained by the block

(c) The difference i.e., 150 J energy is used in doing work against friction between the block and the slope, which will appear as heat energy.

Solution 10.
Selina Concise Physics Class 10 ICSE Solutions Work, Energy and Power 21

Solution 11.

If the speed is halved (keeping the mass same), the kinetic energy decreases, it becomes one-fourth (since kinetic energy is proportional to the square of velocity).

Solution 13.
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Solution 14.
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Solution 15.
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Solution 16.
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Solution 17.
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Solution 18.
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Solution 19.
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Solution 20.
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Solution 21.
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Solution 22.
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Solution 23.
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Exercise 2(C)

Solution 1.

According to the law of conservation of energy, energy can neither be created nor can it be destroyed. It only changes from one form to another.

Solution 2.

According to the law of conservation of mechanical energy, whenever there is an interchange between the potential energy and kinetic energy, the total mechanical energy (i.e., the sum of kinetic energy K and potential energy U) remains constant i.e., K + U = constant when there are no frictional forces.

Mechanical energy is conserved only when there are no frictional forces for a given system (i.e. between body and air). Thus, conservation of mechanical energy is strictly valid only in vacuum, where friction due to air is absent.

Solution 3.

Motion of a simple pendulum and motion of a freely falling body.

Solution 4.

Kinetic energy of the body changes to potential energy when it is thrown vertically upwards and its velocity becomes zero.

Solution 5.

(a) Potential energy
(b) Potential energy and kinetic energy
(c) Kinetic energy

Solution 6.

Let a body of mass m be falling freely under gravity from a height h above the ground (i.e., from position A). Let us now calculate the sum of kinetic energy K and potential energy U at various positions, say at A (at height h above the ground), at B (when it has fallen through a distance x) and at C (on the ground).
Selina Concise Physics Class 10 ICSE Solutions Work, Energy and Power 33

(i) At the position A (at height h above the ground):
Initial velocity of body= 0 (since body is at rest at A)
Hence, kinetic energy K =0
Potential energy U = mgh
Hence total energy = K + U= 0 + mgh =mgh—–(i)

(ii) At the position B (when it has fallen a distance x):
Selina Concise Physics Class 10 ICSE Solutions Work, Energy and Power 34
Thus from equation (i), (ii) and (iii), we note that the total mechanical energy i.e., the sum of kinetic energy and potential energy always remain constant at each point of motion and is equal to initial potential energy at height h.

Solution 7.

When the bob swings from A to B, the kinetic energy decreases and the potential energy becomes maximum at B where it is momentarily at rest.
Selina Concise Physics Class 10 ICSE Solutions Work, Energy and Power 35
From B to A, the potential energy again changes into the kinetic energy and the process gets repeated again and again.
Thus while swinging, the bob has only the potential energy at the extreme position B or C and only the kinetic energy at the resting position A. At an intermediate position (between A and B or between A and C), the bob has both the kinetic energy and potential energy, and the sum of both the energies (i.e., the total mechanical energy) remains constant throughout the swing.

Solution 8.

(a) At position A, pendulum has maximum kinetic energy and its potential energy is zero at its resting position. Hence, K=mgh and U= 0.
(b) At B, kinetic energy decreases and potential energy increases. Hence, K= 0 and U=mgh
(c) At C also, kinetic energy K= 0 and potential energy U=mgh.

Solution 9.

a) Extreme position: Potential energy
b) Mean position: Kinetic energy
c) Between mean and extreme: Both kinetic energy and potential energy

Solution 10.

The gradual decrease of useful energy due to friction etc. is called the degradation of energy.
Examples:

  1. When we cook food over a fire, the major part of heat energy from the fuel is radiated out in the atmosphere. This radiated energy is of no use to us.
  2. When electrical appliances are run by electricity, the major part of electrical energy is wasted in the form of heat energy.

Solution 1 (MCQ).

Potential energy of the ball at the highest point is mgh.
Hint: At the highest point, the ball momentarily comes to rest and thus its kinetic energy becomes zero.

Solution 2 (MCQ).

The sum of its kinetic and potential energy remains constant throughout the motion.
Hint: In accordance with law of conservation of mechanical energy, whenever there is an interchange between the potential energy and kinetic energy, the total mechanical energy remains constant.

Numericals

Solution 1.
Selina Concise Physics Class 10 ICSE Solutions Work, Energy and Power 36

Solution 2.
Selina Concise Physics Class 10 ICSE Solutions Work, Energy and Power 37

Solution 3.

(a)Potential energy of the ball =mgh
=2 x 10 x 5=100J
(b)Kinetic energy of the ball just before hitting the ground = Initial potential energy= mgh=2x10x5=100J
(c)Mechanical energy converts into heat and sound energy.

Solution 4.
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Solution 5.
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Solution 6.
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More Resources for Selina Concise Class 10 ICSE Solutions

How Do You Find The Kinetic Energy

What is Kinetic Energy

Kinetic Energy: The energy of a body due to its motion is called kinetic energy. In other words. The ability of a body to do work by virtue of its motion is called its kinetic energy.

Expression for Kinetic Energy: The kinetic energy of a body is measured in terms of the amount of work done by an opposing force that brings the body to rest from its present state of motion.
Kinetic-Energy
Suppose a body of mass m is moving with a velocity v and is brought to rest by an opposing force F.
Now retarding force is given by
F = ma          …(1)
Now using the equation of motion,
v2 – u2 = 2as, we get
02 = v2 – 2as
∴s=\( \frac{{{v}^{2}}}{2a} \) ….. (2)
Kinetic energy of the body = work done by the retarding force
or Kinetic energy = force × displacement
= F . s …(3)
Substituting the value of F from equation (1) and the value of s from equation (2) in equation (3), we get
\( K.E.=ma\times \frac{{{v}^{2}}}{2a} \)
\( K.E.=\frac{1}{2}m{{v}^{2}} \)
Thus, a body of mass m and moving with a velocity v has the capacity of doing work equal to \(\frac { 1 }{ 2 }\) mv2 before it stops.

Kinetic Energy Example Problems With Solutions

Example 1:    A bullet of mass 100 gm is fired with a velocity 50 m/s from a gun. Calculate the kinetic energy of the bullet.
Solution:    Kinetic energy is given by
\( K.E.=\frac{1}{2}m{{v}^{2}} \)
Here m = 100 gm = 0.1 kg; v = 500 m/s
K.E. = \(\frac { 1 }{ 2 }\) × 0.1 × (50)2
= \(\frac { 1 }{ 2 }\) × 0.1 × 50 × 50 = 125 J

Example 2:    A 4 kg body is dropped from the top of a building of height 2.5 m. With what velocity will it strike the ground ? What is its kinetic energy when it strikes the ground ?
(Takes g = 9.8 m/s2)
Solution:    Velocity of the body with which it strikes the ground can be calculated by using the equation, v2 = u2 + 2gh
Here u = 0; g = 9.8 m/s2 ; h = 2.5 m
Substituting these values, we get
v2 = 02 + 2 × 9.8 × 2.5 = 49
or v = 7 m/s
Thus, the speed of the body with which it strikes the ground = 7 m/s.

Example 3:   Calculate the velocity of 4 kg mass with kinetic energy of 128 J.
Solution:    The formula for kinetic energy is given by
\( K.E.=\frac{1}{2}m{{v}^{2}} \)
Here K.E. = 128 J; m = 4 kg
128 = \(\frac { 1 }{ 2 }\) × 4 × v2
v2 = 64
v = 8 m/s

Example 4:    Which would have a greater effect on the kinetic energy of an object, doubling the mass or doubling the velocity ?
Solution:    (i) The kinetic energy of a body is directly proportional to its “mass” (m). So, if we double the mass (so that it becomes 2m), then the kinetic energy will also get doubled.
(ii) On the other hand, kinetic energy of a body is directly proportional to the “square of its velocity” (v2). So, if we double the velocity (so that it becomes 2v), then the kinetic energy will become four times. This is because : (2v)2 = 4v2.
It is clear from the above discussion that doubling the velocity has a greater effect on the kinetic energy of an object.

Example 5:    Two bodies of equal masses move with uniform velocity v and 3v respectively. Find the ratio of their kinetic energies.
Solution:    In this problem, the masses of the bodies are equal, so let the mass of each body be m. We will now write down the expression for the kinetic energies of both the bodies separately.
(i) Mass of first body = m
Velocity of first body = v
So, K.E. of first body = \(\frac { 1 }{ 2 }\) mv2         ….. (1)
(ii) Mass of second body = m
Velocity of second body = 3v
So, K.E. of second body = \(\frac { 1 }{ 2 }\) m (3v)2
= \(\frac { 1 }{ 2 }\) m × 9 v2
= \(\frac { 9 }{ 2 }\) mv2                 ….. (2)
Now, to find out the ratio of kinetic energies of the two bodies, we should divide equation (1) by equation (2), so that
\( \frac{\text{K}\text{.E}\text{.}\,\text{of}\,\text{first}\,\text{body}}{\text{K}\text{.E}\text{.}\,\text{of}\,\text{sec}\,\text{ondbody}\,}=\frac{\frac{\text{1}}{\text{2}}\text{m}{{\text{v}}^{\text{2}}}}{\frac{\text{9}}{\text{2}}\text{m}{{\text{v}}^{\text{2}}}} \)
\( \frac{\text{K}\text{.E}\text{.}\,\text{of}\,\text{first}\,\text{body}}{\text{K}\text{.E}\text{.}\,\text{of}\,\text{sec}\,\text{ond}\,\text{body}}=\frac{\text{1}}{\text{9}}\text{ }….\text{ (3)} \)
Thus, the ratio of the kinetic energies is 1 : 9. We can also write down the equation (3) as follows:
K.E. of second body = 9 × K.E. of first body
That is, the kinetic energy of second body is 9 times the kinetic energy of the first body. It is clear from this example that when the velocity (or speed) of a body is “tripled” (from v to 3v), then its kinetic energy becomes “nine times”.

Example 6.  What is the kinetic energy of a 1200 kg car travelling at a velocity of 25 m s-1?
Solution:
How Do You Find The Kinetic Energy

Example 7. Table shows the masses and velocities of two boys, Bek Sar and Kek Cik respectively.
How Do You Find The Kinetic Energy 1
Which of the boys has greater kinetic energy?
Solution:
How Do You Find The Kinetic Energy 2

How to Calculate the Energy Used

What is Energy

Definition: Energy is the ability to do work. The amount of energy possessed by a body is equal to the amount of work it can do when its energy is released. Thus, energy is defined as the capacity of doing work. Energy is a scalar quantity and it exists in various forms.

Units of energy: The units of energy are the same as that of work. In SI system, the unit of energy is joule (J). In CGS system, the unit of energy is erg.

  • 1 Joule = 107 ergs
  • Other units of energy in common use are watt-hour and kilowatt hour.
  • 1 watt-hour = 1 watt × 1 hour
    = 1 watt × 60 × 60 sec
    = 3600 J
  • 1 kilowatt-hour (kWh) = 3.6 × 106 Joule
  • Heat energy is usually measured in calorie or kilocalorie such that
    1 calorie = 4.18 J
  • A very small unit of energy is electron volt(eV).
    1 eV = 1.6 × 10-19 J

The energy possessed by a body due to its state of rest or state of motion is called mechanical energy.
Mechanical energy is of two types
(A) Kinetic Energy            (B) Potential Energy.

Principle of Conservation of Energy

  1. Principle of conservation of energy states that energy cannot be created or destroyed but can be changed from one form to another.
  2. The total energy of the universe is constant. The total energy of an isolated system is constant.
  3. Figure shows the transformation of energy from one form to another.
    Principle of Conservation of EnergyPrinciple of Conservation of Energy 1

Principle of Conservation of Energy Experiment

Aim: To study the principle of conservation of energy.
Materials: Ticker tape, a polystyrene sheet, string, a 300 g slotted mass, a pulley, cellophane tape
Apparatus: Ticker timer, a.c. power supply, two retort stands with clamps, trolley, electronic balance, plane
Method:

  1. A friction-compensated plane is arranged as shown in Figure.
  2. The mass of the trolley, m1 is measured with an electronic balance.
  3. A slotted mass of mass, m2 = 300 g is tied to one end of a non-elastic string.
  4. The other end of the string is tied to one end of the trolley.
  5. The string is placed over the pulley and held at 0.5 m above the polystyrene sheet.
  6. The ticker timer is switched on and the slotted mass is released so that it falls downward, pulling the trolley down the runway.
  7. The ticker timer is analysed to determine the final velocity of the trolley.
    Principle of Conservation of Energy 2

Results:
Principle of Conservation of Energy 3
Discussion:

  1. The plane is friction-compensated to minimise energy loss due to friction.
  2. When the slotted mass drops, it loses gravitational potential energy. The trolley and slotted mass gain kinetic energy.
  3. Ideally the gravitational potential energy loss, Ep equals to the kinetic energy gain, Ek. This is in accordance to the principle of conservation of energy.
  4. However, the experimental results shows that Ek is slightly less than Ep. This is because of unavoidable energy loss due to friction of the trolley as well as air friction.

 

How do you find Power in Physics

What is Power

PowerFigure shows two electric motors, A and B respectively. Each motor lifts an identical load from the floor.
Motor A can lift the load more quickly than motor B. Hence, motor A can do the same amount of work in a shorter time.
Motor A is said to be more powerful than motor B.
Definition: Power is defined as the rate of doing work.
\( \text{Power}=\frac{\text{Work}\,\text{done}}{\text{Time}\,\text{taken}} \)
\( \text{P}=\frac{\text{W}}{\text{t}} \)
In other words, power is the work done per unit time, power is a scalar quantity.
Since W = F.S therefore
\( \text{P}=\frac{\text{W}}{\text{t}}=\frac{\text{FS}}{\text{t}}=F\times V=\text{force}\times \text{velocity} \)
Unit of power: The S.I. unit of power is watt and it is the rate of doing work at 1 joule per second.
\(1\text{ watt}=\frac{\text{1}\,\text{joule}}{\text{1}\,\text{seconds}}\)
1 kilowatt = 1 kW = 1000 W
1 Horse power = 1 H.P. = 746 W

Activity 1

Aim: To measure the power generated by a student running up the stairs.
Apparatus: Stopwatch, metre rule, weighing machine
Method:
Power 1

  1. The mass, m of a student is measured.
  2. The student is asked to run up the stairs from one floor to the next floor of a building in the school.
  3. The vertical height, h gained by the student in running up to the next floor is measured by multiplying the number of steps by the height of each step.
  4. The time taken, t for the run is recorded by another student using a stopwatch.

Analysis of Data:
Mass of student = m kg
Vertical height = h m
Time taken = t s
Work done = Gain in gravitational potential energy = mgh, where g = 9.8 m s-2
Power generated, P = W/t watt
Discussion:

  1. For a more accurate result, more than one person should time the run and the average time should be calculated.
  2. This activity assumes that all the work done goes to the increase in gravitational potential energy only. However, in reality, some work is also used in overcoming frictional forces.

What is the Efficiency?

  1. Useful energy is the energy that can be used to do a certain work.
  2. Wasted energy is the energy that is lost to the surrounding and cannot be used to do useful work.
  3. Figure shows an efficient device. Most of the input energy is converted into useful energy. Very little of the input energy is wasted.
    Efficiency 1
  4. Figure shows an inefficient device. Most of the input energy is wasted. Only a small portion of the input energy is converted into useful energy.
    Efficiency 2
  5. The efficiency of a device is defined as the percentage of the input energy that is transformed into useful energy.
    Efficiency 5
  6. The efficiency of a device can also be calculated in terms of power.
    Efficiency 6
  7. In the engine of a vehicle as shown in Figure, the energy needed is obtained by burning fuels like petrol and diesel. The energy obtained is converted into kinetic energy of the car. However, a large portion of the energy is wasted.
    Efficiency 7
  8. There are two types of engines that are commonly used in vehicles; the petrol engine and the diesel engine. The efficiency of a diesel engine is higher than that of a petrol engine. However, diesel engines are heavier and more costly to construct. Diesel engines are generally noisier than petrol engines. Diesel engines are preferred by heavy vehicles like lorries, buses, tractors and locomotives.
  9. Electric vehicles are more efficient because less energy is wasted as heat. However, electric vehicles are not common because of the high cost of technology required to build them.

Power and Efficiency Example Problems with Solutions

Example 1. A machine raises a load of 750 N through a height of 15 m in 5s. Calculate:
(i) the work done by the machine.
(ii) the power at which the machine works.
Solution:     (i) Work done is given by W = F.s
Here      F = 750 N; s = 15 m
∴ W = 750 × 15 = 11250 J
= 11.250 kJ
(ii) Now, power of the machine is given by
\( \text{P}=\frac{\text{W}}{\text{t}} \)
Here,    W = 11250 J; t = 5 s
∴ Power P = \(\frac { 11250J }{ 5s }\) = 2250 W = 2.250 kW

Example 2. A weight lifter lifted a load of 100 kg to a height of 3 m in 10 s. Calculate the following:
(i) amount of work done
(ii) power developed by him
Solution:     (i) Work done is given by
W = F . s
Here, F = mg = 100 × 10 = 1000 N
W = 1000 N × 3 m = 3000 joule
(ii) Now, P = \(\frac { W }{ t }\) , where W = 3000 J and t = 10 s
∴ P = \(\frac { 3000J }{ 10s }\) = 300 W

Example 3. A water pump raises 60 liters of water through a height of 20 m in 5 s. Calculate the power of the pump. (Given: g = 10 m/s2, density of water = 1000 kg/m3)
Solution:     Work done, W = F.s             …(1)
Here, F = mg             …(2)
But, Mass = volume × density
Volume = 60 liters = 60 × 10-3 m3
Density = 1000 kg/m3
∴ Mass, m = (60 × 10-3 m3) × (1000 kg/m3) = 60 kg
∴ Equation (2) becomes
F = 60 kg × 10 m/s2 = 600 N
Now, W = F . s = 600 N × 20 m = 12000 J
\( \text{P}=\frac{\text{W}}{\text{t}} \)
= \(\frac { 12000J }{ 5s }\) = 2400 W

Example 4. A woman pulls a bucket of water of total mass 5 kg from a well which is 10 m deep in 10 s. Calculate the power used by her.
Solution:     Given that m = 5 kg; h = 10 m; t = 10 s
g = 10 m/s2
\( \text{P}=\frac{\text{W}}{\text{t}} \)
\(=\frac{\text{mgh}}{\text{t}}=\frac{\text{5 }\!\!\times\!\!\text{ 10 }\!\!\times\!\!\text{ 10}}{\text{10}}=\text{50W}\)

Example 5. A 40 kg boy climbs a vertical height of 3.5 m in 4 s.
Power 2
(a) Calculate the gain in gravitational potential energy of the boy.
(b) Determine the power generated by the boy. [Take g=9.8m s-2]
Solution:
Power 3

Example 6. A crane lifts a heavy bucket to a height of 2.5 m from the ground in 3.5 s.
Efficiency 3
(a) Calculate the power generated by the crane in lifting the bucket if the mass of the bucket is 840 kg.
(b) Explain why the power generated by the crane is actually higher than the value calculated in (a), [g = 9.8 m s-2]
Solution:
Efficiency 4
(b) This is because besides lifting the bucket, work is also done to overcome frictional forces between the cable and the pulley and other parts of the crane.

Example 7. An electric motor has an input power of 120 W. It lifts a 20 kg load to a vertical height of 1.5 m in 5 s. What is the efficiency of the electric motor? [g = 9.8 m s-2]

Efficiency 8

How do you find work in physics?

What is the formula for work?

Definition: In our daily life “work” implies an activity resulting in muscular or mental exertion. However, in physics the term ‘work’ is used in a specific sense involves the displacement of a particle or body under the action of a force. “work is said to be done when the point of application of a force moves.
Work done in moving a body is equal to the product of force exerted on the body and the distance moved by the body in the direction of force.
Work = Force × Distance moved in the direction of force.

The work done by a force on a body depends on two factors
(i) Magnitude of the force, and
(ii) Distance through which the body moves (in the direction of force)

Unit of Work
Work-and-Energy
When a force of 1 newton moves a body through a distance of 1 metre in its own direction, then the work done is known as 1 joule.
Work = Force × Displacement
1 joule = 1 N × 1 m
or 1 J = 1 Nm (In SI unit)

Work Done Analysis

Work done when force and displacement are along same line:
Work done by a force: Work is said to be done by a force if the direction of displacement is the same as the direction of the applied force.
Work done against the force: Work is said to be done against a force if the direction of the displacement is opposite to that of the force.
Work done against gravity: To lift an object, an applied force has to be equal and opposite to the force of gravity acting on the object. If ‘m’ is the mass of the object and ‘h’ is the height through which it is raised, then the upward force
(F) = force of gravity = mg
If ‘W’ stands for work done, then
W = F . h = mg . h
Thus W = mgh
Therefore we can say that, “The amount of work done is equal to the product of weight of the body and the vertical distance through which the body is lifted.

Work done when force and displacement are inclined (Oblique case):
Consider a force ‘F’ acting at angle θ to the direction of displacement ‘s’ as shown in fig.
Work-done
Work done when force is perpendicular to Displacement
θ = 90º
W = F.S × cos 90º = F.S × 0 = 0
Thus no work is done when a force acts at right angle to the displacement.

 

Special Examples

  1. When a bob attached to a string is whirled along a circular horizontal path, the force acting on the bob acts towards the centre of the circle and is called as the centripetal force. Since the bob is always displaced perpendicular to this force, thus no work is done in this case.
  2. Earth revolves around the sun. A satellite moves around the earth. In all these cases, the direction of displacement is always perpendicular to the direction of force (centripetal force) and hence no work is done.
  3. A person walking on a road with a load on his head actually does no work because the weight of the load (force of gravity) acts vertically downwards, while the motion is horizontal that is perpendicular to the direction of force resulting in no work done. Here, one can ask that if no work is done, then why the person gets tired. It is because the person has to do work in moving his muscles or to work against friction and air resistance.

Work Done By A Force Example Problems With Solutions

Example 1:    How much work is done by a force of 10N in moving an object through a distance of 1 m in the direction of the force ?
Solution:    The work done is calculated by using the formula:
W = F × S
Here,      Force F = 10 N
And, Distance, S = 1 m
So, Work done, W = 10 × 1 J
= 10 J
Thus, the work done is 10 joules

Example 2:    Find the work done by a force of 10 N in moving an object through a distance of 2 m.
Solution:     Work done = Force × Distance moved
Here,         Force = 10 N
Distance moved = 2 m
Work done, W = 10 N × 2 m
= 20 Joule = 20 J

Example 3:    Calculate the work done in pushing a cart, through a distance of 100 m against the force of friction equal to 120 N.
Solution:    Force, F = 120 N; Distance, s = 100 m
Using the formula, we have
W = Fs = 120 N × 100 m = 12,000 J

Example 4:    A body of mass 5 kg is displaced through a distance of 4m under an acceleration of 3 m/s2. Calculate the work done.
Sol. Given: mass, m = 5 kg
acceleration, a = 3 m/s2
Force acting on the body is given by
F = ma = 5 × 3 = 15 N
Now, work done is given by
W = Fs = 15 N × 4 m = 60 J

Example 5:    Calculate the work done in raising a bucket full of water and weighing 200 kg through a height of 5 m. (Take g = 9.8 ms-2).
Solution:    Force of gravity
mg = 200 × 9.8 = 1960.0 N
h = 5 m
Work done, W = mgh
or W = 1960 × 5 = 9800 J

Example 6:    A boy pulls a toy cart with a force of 100 N by a string which makes an angle of 60º with the horizontal so as to move the toy cart by a distance horizontally. Calculate the work done.
Solution:    Given F = 100 N, s = 3 m, θ= 60º.
Work done is given by
W = Fs cos θ= 100 × 2 × cos 60º
= 100 × 3 × 1/2 = 150 J (∵ cos 60º = 1/2 )

Example 7:    An engine does 64,000 J of work by exerting a force of 8,000 N. Calculate the displacement in the direction of force.
Solution:    Given W = 64,000 J; F = 8,000 N
Work done is given by W = Fs
or    64000 = 8000 × s
or    s = 8 m

Example 8:    A horse applying a force of 800 N in pulling a cart with a constant speed of 20 m/s. Calculate the power at which horse is working.
Solution:    Power, P is given by force × velocity, i.e.
P = F . v
Here F = 800 N; v = 20 m/s
∴ P = 800 × 20 = 16000 watt
= 16 kW

Example 9:    A boy keeps on his palm a mass of 0.5 kg. He lifts the palm vertically by a distance of 0.5 m. Calculate the amount of work done.
Use g = 9.8 m/s2.
Solution:    Work done, W = F . s
Here, force F of gravity applied to lift the mass, is given by
F = mg
= (0.5 kg) × (9.8 m/s2)
= 4.9 N
and s = 0.5 m
Therefore, W = (4.9) . (0.5m) = 2.45 J.

Example 10:    A truck of mass 2500 kg is stopped by a force of 1000 N. It stops at a distance of 320 m. What is the amount of work done ? Is the work done by the force or against the force?
Solution:    Here the force, F = 1000 N
Displacement, s = 320 m
∴ Work done, W = F . s
= (1000N) . (320 m)
= 320000 J
In this case, the force acts opposite to the direction of displacement. So the work is done against the force.

Example 11:    A car weighing 1200 kg and travelling at a speed of 20 m/s stops at a distance of 40 m retarding uniformly. Calculate the force exerted by the brakes. Also calculate the work done by the brakes.
Solution:    In order to calculate the force applied by the brakes, we first calculate the retardation.
Initial speed, u = 20 m/s; final speed,
v = 0, distance covered, s = 90 m
Using the equation, v2 = u2 + 2as, we get
02 = (20)2 + 2 × a × 40
or   80a = –400
or   a = –5 m/s2
Force exerted by the brakes is given by
F = ma
Here m = 1200 kg; a = – 5 m/s2
∴ F = 1200 × (–5) = – 6000 N
The negative sign shows that it is a retarding force.
Now, the work done by the brakes is given by
W = Fs
Here F = 6000 N; s = 40 m
∴ W = 6000 × 40 J = 240000 J
= 2.4 × 105 J
∴ Work done by the brakes = 2.4 × 105 J

Example 12. A worker uses a horizontal force of 400 N to push a rubbish cart.
work 1
Calculate the work done to push the cart a distance of 180 m on a horizontal surface.
Solution:
Since the direction of the movement of the cart is the same as the direction of the applied force, therefore W = F x s
= 400 x 180 = 72 000 J

Example 13. A boy uses a force of 120 N to pull a crate along a straight corridor. The applied force is at an angle of 30° with the horizontal floor.
work 2
Calculate the work done by him after pulling the crate a distance of 90 m.
Solution:
W = F cos θ x s
= (120 cos 30°) x 90 = 9353 J

Example 14. Jin uses a force of 16 N to mop a floor. The applied force makes an angle of 28° with the floor.
work 3
What is the work done by Jin when he pushes the mop a horizontal distance of 2.5 m?
Solution:
W = F cos θ x s
= (16 cos 28°) x 2.5 = 35.3 J