WORK AND ENERGY

LEARNING OBJECTIVES
At the end of this chapter the students will be able to:
1. Understand the concept of work in terms of the product of a force and displacement in the
direction of the force.
→ →
2. Understand and derive the formula Work = w . d = mgh for work done in a gravitational field
near Earth’s surface.
3. Understand that work can be calculated from area under the force-displacement graph.
4. Relate power to work done.
5. Define power as the product of force and velocity.
6. Quote examples of power from everyday life.
7. Explain the two types of mechanical energy.
8. Understand the work-energy principle.
9. Derive an expression for absolute potential energy.
10. Define escape velocity.
11. Understand that in a resistive medium loss of potential energy of a body is equal to gain in
kinetic energy of the body plus work done by the body against friction.
12. Give examples of conservation of energies from everyday life.
13. Describe some non-conventional sources of energy.

INTRODUCTION:
Work is often thought in terms of physical or mental effort. In Physics, however, the term work
involves two things (i) force (ii) displacement. We shall begin with a simple situation in which work is
done by a constant force.

WORKDONE BY A CONSTANT FORCE:

The force whose value does not change while performing work is called constant force.

Work:
“The dot product of force and displacement is called work”.
(OR)
“The product of magnitude of displacement and component of force along the direction of
displacement is called work”.
155
156 Physics Intermediate Part-I

Mathematical Form:
→ →
Suppose a constant force F displaces a body through d and the angle between force and
displacement is ‘θ’ then by definition of work.
→ →
W = F . d = Fd cos θ …… (1)
Nature: →
F
Work is a scalar quantity.
θ
Dimensions of W = [ML2T−2]
d (F cos θ)
Cases of Work:
We have following cases for work.
θ < 90°°):
Case-I: Maximum or Positive Work (θ
When angle between force and displacement is zero i.e., ‘0’. Then work done will maximum.
→
W = Fd cos (0)° F
W = Fd (Maximum)
→
θ = 90°°):
Case-II: Zero Work (θ d
When the angle between force and displacement is 90°. →
F
W = Fd cos (90°) = 0
→
Examples: d

→
(i) No work is done on the pail when a person holding the pail by force F is moving forward
because angle between force and displacement is 90°.
(ii) When a person pushes a wall with a force work done is zero because d = 0.
→
W = F .0 = 0
θ > 90°°):
Case-III: Negative Work (θ
→ → C.G.S unit of energy is erg.
When angle between force F and displacement d is 180° then 1 erg = 10−7 J
negative work will be done i.e., At atomic level unit of energy is
electron volt.
W = Fd cos (180°) = −Fd 1 eV = 1.6 x 10−19 J

Example:
Work done by frictional force.
Unit:
→ →
Its unit is joule (J) and is defined as: d F

1J = N.m
“Work done is said to be one joule when one newton of force displaces a body through one
metre”.
[Chapter-4] Work and Energy 157
Graph for Workdone by Constant Force:
Graph between constant force along vertical axis and displacement along horizontal axis is
drawn below.
From graph area of rectangle PQRO is equal to:
Area of graph = Length × Width P Q

F cos θ
= dF cos θ F

= Fd cos θ O Displacement R
Area of graph = Work done d

WORKDONE BY A VARIABLE FORCE:

Variable Force:
“The force whose value changes while performing work is called variable force”.
Examples of Workdone by Variable Force:
Workdone by earth on a rocket moving towards it or moving away from it is variable because
force varies with inverse of square of distance from the earth’s surface.
Similarly, force exerted by the spring due to its compression and extension is variable so it
performs variable work.
Explanation:
Suppose a variable force displaces a body from a point ‘a’ to point ‘b’ in xy-plane, as shown in
figure.
Let F1, F2, F3, ……, Fn be variable forces displace a body through Fn θn
∆d1, ∆d2, ……, ∆dn and angles between force and displacement are b
∆dn
θ1, θ2, ……, θn respectively.
Workdone by force F1 is:
F3 θ3
→ → θ
F2 2
∆W1 = F1 . ∆ d1 = F1 ∆d1 cos θ1 ∆d3
θ1 ∆d2
Workdone by force F2 is: F1 ∆d
1
→ → a
∆W2 = F2 . ∆ d2 = F2 ∆d2 cos θ2
Similarly workdone by nth force Fn:
→ →
∆Wn = Fn . ∆ dn = Fn ∆dn cos θn
Total work done is:
Wtotal = ∆W1 + ∆W2 + ∆W3 + …… + ∆Wn
Wtotal = F1 ∆d1 cos θ1 + F2 ∆d2 cos θ2 + F3 ∆d3 cos θ3 + …… + Fn ∆dn cos θn
Total work done:
n
Wtotal = ∑ Fi cos θi ∆di
i=1
158 Physics Intermediate Part-I

Graph for Workdone by Variable Forces: y os θ 1 θi

F 1 c F i cos
os θ n
In order to draw a graph for the work done by variable force. Fn c

F cos θ
We take ‘F cos θ’ along vertical axis and ‘d’ along horizontal axis. The
displacement ‘d’ is subdivided into equivalent displacements ∆d1, ∆d2,
……, ∆dn.
So that workdone by such small value of force when it displaces x
a b
∆d1 ∆d2 ∆d1 ∆dn
the body through ‘∆d’ is given by:
d
n y
Wtotal = Limit ∑ Fi ∆di cos θi

F cos θ
∆d → 0 i = 1

Thus the workdone by variable force in moving a body between

two points is equal to the area under ‘F cos θ’ verses ‘d’ curve between x
two points ‘a’ and ‘b’. 0 a d b

Significance:
If we let each ∆d to approach zero then we obtain an exact result for work done i.e.,
n
Wtotal = Limit ∑ Fi ∆di cos θi
∆d → 0 i = 1

WORKDONE BY GRAVITATIONAL FIELD:

Conservative Field:
“The field in which work done is independent of the path followed by the body is called
conservative field”.
(OR)
“The field in which work done along a closed path is equal to zero is called conservative
field”.
Following are the examples of conservative field.
• Gravitational field
• Electric field

To Show that Gravitational Field is Conservative:

Suppose a body of mass m is displaced from point A to point B in gravitational field along three
possible paths:
Path-I: From A to B.
Path-II: From A to D then from D to B.
Path-III: From A to C then from C to B.
[Chapter-4] Work and Energy 159
Workdone along Path-I:
In order to calculate the workdone along path-I. We divide the whole path into small intervals
along x and y-axis.
C Path III B y ∆xn
yn B
mg x

Path II
Path III I
th
Pa ∆y2 h
h
mg ∆x2
∆y1
Path II ∆x1
A D A
∆x

Workdone along each horizontal interval is zero because θ = 90°. While the workdone along
each vertical interval is negative because θ = 180°. Thus total work can be calculated as:
→ →
WAB = F . d
WAB = mg(∆y1 + ∆y2 …… ∆yn) cos θ
WAB = mg(∆y1 + ∆y2 …… ∆yn) cos (180°) (θ = 180°)
WAB = −mgh (As ∆y1 + ∆y2 …… ∆yn = h)

WAB = −mgh …… (1)

Path-II:
WA → D → B = WA → D + WD → B
= mg (AD) cos 90° + mg (h) cos 180°
= 0 − mgh

WA → D → B = −mgh …… (2)
Path-III:
Workdone along path III is given by:
WA → C → B = WA → C + WC → B
WA → C → B = mg (h) cos (180°) + mg (CB) cos 90°
WA → C → B = mg (h)(−1) + (0)

WA → C → B = −mgh …… (3)
Conclusion:
From eq. (1), (2), (3):
It is clear that:
WAB = WA → D → B = WA → C → B = −mgh
i.e., workdone is independent of path through which the body is displaced. Hence gravitational field is
conservative field.
160 Physics Intermediate Part-I

Workdone Along Closed Path:

Workdone along a close path in gravitational field is zero.
Proof:
C B

mg

A D
mg mg mg
Let us consider workdone along a close path ACBA. As:
WACBA = WAC + WCB + WBA
WACBA = mg(AC) cos (180°) + mg(CB) cos (90°) + mg(∆y1 + ∆y2 + ∆y3 + … + ∆yn) cos (0°)
= mg(h)(−1) + mg(h)(0) + (mg)(h)(1)
= −mgh + 0 + mgh
WACBA = 0
Conclusion:
Hence workdone along a close path in gravitational field is zero.
Non-conservative Force:
Workdone by non-conservative force depends upon the path followed.
Wokdone along closed path is not zero.
Examples of non-conservative force are:
(i) Frictional force
(ii) Air resistance
(iii) Tension in string
(iv) Normal force
(v) Propulsion force of rocket/motor
POWER:
“Rate of doing work is called power”. Conservative Forces
Mathematical Form: Gravitational force
Suppose work W is done in time ‘t’ then power can be written as: Elastic spring force
Electric force
W
P = t Non-Conservative Forces
Frictional force
Average Power: Air resistance
If ∆W is amount of work is done in time ∆t then average power Tension in a string
can be written as: Normal force
∆W Propulsion force of a rocket
Pav = Propulsion force of a motor
∆t
which is “ratio of total workdone to the time in which it has been done”.
[Chapter-4] Work and Energy 161
Instantaneous Power:
If work is done for such small interval of time such that this time approaches to zero. Then the
ratio of work to time is called instantaneous power.
Instantaneous power can be written as:
∆W
Pinst = Lim
∆t → 0 ∆t

Unit:
S.I unit of power is watt which can be defined as:
If one joule of work is done in time of one second, then power obtained will be one watt.
1 watt = Js−1

Kilowatt Hour (KWh):

Energy is said to be one kilowatt hour, if one thousand watt power is maintained for one hour.
It is commercial unit of electrical energy.
1 KWh = 3.6 MJ Area under power-time graph gives
work.
1 kilowatt hour = 1 × 1000 watt × 3600 s
= 3600000 W-s
= 3.6 × 106 J
1 kilowatt hour = 3.6 MJ (∴ 1 h.p = 746 W)

Power and Velocity:

To prove that power is dot product of force and velocity i.e.,
→ →
P = F . v
Proof:
Suppose the propeller of a motor boat causes the water to exert Approximate Powers
→ → Device Power (W)
a constant force F on the boat and it moves with constant velocity v . Jumbo Jet Aircraft 1.3 × 108
The power delivered by the motor to motor boat for short interval of Car at 90 km h −1
1.1 × 105
time is: Electric heater 2 × 103
Colour TV 120
∆W Flash light (two cells) 1.5
P = Lim …… (1) Pocket calculator 7.5 × 10−4
∆t → 0 ∆t

→ →
But ∆W = ∆( F . d ) (d = Distance covered by boat)
→ →
∆W = F . ∆ d
162 Physics Intermediate Part-I

So eq. (1):
→
→ ∆d
∴ P = Lim F . −2 −3
∆t → 0 ∆t [ P ] = [ML T ]

 →
→ ∆ d 
P = F . Lim   …… (2)
∆t → 0  ∆t 

→
∆d →
As Lim = v
∆t → 0 ∆t

So, eq. (2) becomes:

→ →
P = F . v
→ →
So power is the dot product of force F and velocity v .
ENERGY:
“The ability of a body to do work is called energy”.
Types of Energy:
9
Following are the types of energy: It takes about 9 x 10 J to make a
car and the car then uses about
Kinetic Energy: 12
1 x 10 J of energy from petrol in
its life time.
The energy possessed by a body because of its motion is called
kinetic energy.
Mathematical Form:
Suppose a body of mass is moving with the velocity v, then its Approximate Energy Values Source
Source Energy (J)
K.E will be: Burning 1 ton coal 30 × 109
Burning 1 litre petrol 5 × 107
1
K.E = 2 mv2 K.E. of car at 90 km h −1 1 × 106
Running person at 10 km h −1 3 × 102
Fission of one atom of uranium 1.8 × 10−11
Potential Energy: K.E. of a molecule of air 6 × 10−21

Energy possessed by a body because of its position in a force

field is called potential energy.
Gravitational Potential Energy:
Energy of a body because of its position from the surface of earth is called gravitational potential
energy.
Mathematical Form:
Suppose a body of mass ‘m’ is taken to height h from the surface of earth, then the gravitational
potential energy is given by:
G.P.E = mgh
[Chapter-4] Work and Energy 163
Elastic Potential Energy:
The energy stored in a compressed or stretched spring is called elastic potential energy.
1
E = 2 kx2

Work-Energy Principle:
“Workdone on a body or system is equal to change in its energy, which may be kinetic energy,
potential energy or both”.
Mathematical Form:
Mathematically it is given by: All the food you eat in one day has
about the same energy as 1/3 litre
W = K.Ef − K.Ei = ∆K.E of petrol.

or W = P.Ef − P.Ei = ∆P.E

Proof:
→ W = Change in P.E is for
Suppose a force F is applied upon a body and its velocity conservative forces. P.E. is
changes from its initial value ‘vi’ to final value vf when it gets defined only for conservative
forces.
acceleration ‘a’ through distance ‘d’.
vi vf
F F

d
Using 3rd equation of motion:
2 2
2ad = vf − vi
2 2
vf − vi
a = 2d …… (1)

Also from Newton 2nd law of motion:

F = ma
F
a = m …… (2)

Comparing (1) and (2):

2 2
F vf − vi
m = 2d
2 2
2Fd = mvf − mvi
1 2 2
Fd = 2 [mvf − mvi ]

1 2 1 2
Fd = 2 mvf − 2 mvi …… (3)
164 Physics Intermediate Part-I

Now as:
Fd = Workdone = W
1 2
mvf = Final kinetic energy = K.Ef
2
1 2
2 mv i = Initial kinetic energy = K.Ei

So, eq. (3) becomes:

W = K.Ef − K.Ei

W = ∆K.E
Thus workdone is equal to change in energy.
Example of Work Energy Principal: There is more energy reaching
Earth in 10 days of sunlight than
If a body is raised up from the earth surface then workdone in all the fossil fuels on the Earth.
changes into gravitational potential energy.
If a spring is compressed, then:
Workdone = Increase in elastic P.E
Absolute Potential Energy:
“Workdone by gravitational force in displacing a body from gravitational field to a distance at
infinity where its gravitational field does not exist is called absolute potential energy”.
Mathematical Form:
Mathematically absolute gravitational potential energy is given by:
−GMm
U = r
Derivation:
Suppose a body of mass ‘m’ is displaced from point 1 where N
gravitational field exist to point N where gravitational field does not
exist. We divide the distance between point 1 and point N into points 2, 3,
4, ……, N − 1 such that r1, r2, r3, ……, rn are distances of points 1, 2, 3, 4
……, N from centre of earth. Let ∆r is the separation between two steps. It 3
has such a small value such that the force remains constant for each small 2
∆r
1
step. Let r is midpoint of r1 and r2, then
r1 + r2
r = …… (1)
2
r1 r2
Now by Newton law of gravitation:
GMm
F = …… (2)
r2
O
Here, M = Mass of earth Earth

m = Mass of an object
[Chapter-4] Work and Energy 165
Also from figure:
∆r = r2 − r1
r1 + ∆r = r2 …… (a)
So, eq. (1) becomes:
r1 + ∆r + r1
r =
2
2r1 + ∆r
r =
2
Taking square on both sides:
2
2r1 + ∆r
r2
=  
 2 
2
2 4r1 + ∆r2 + 4r1∆r
r = 4
2
2 4r1 + ∆r2 + 4r1∆r
r = …… (3)
4
∆r2 ∆r2
As is very very small so it can be neglected i.e.,
4 4 = 0.
Also ∆r = r2 − r1
So, eq. (3) becomes:
2
2 4r1 + 0 + 4r1(r2 − r1)
r = 4
2 2
2 4r1 + 4r1r2 − 4r1 4r1r2
r = 4 = 4
r2 = r1r2
So, eq. (2) become:
GMm
F = r1r2 …… (b)

Now workdone in displacing the body from point 1 to point 2 is:

→ →
W1→2 = F .∆ r
= F∆r cos 180°
W1→2 = −F∆r …… (4)
Using eq. (a) and (b) in (4):
−GMm
r1r2 (r2 − r1)
W1→2 =

 r2 r1 
W1→2 = −GMm r r − r r 
 1 2 2 1
166 Physics Intermediate Part-I

1 1
W1→2 = −GMm r − r 
 1 2
1 1
Similarly, W2→3 = −GMm r − r 
 2 3
M
M
 1 1
WN−1→N = −GMm  − 
rn−1 rN
Total work will be equal to sum of works from 1 to N steps i.e.,
WTotal = W1→2 + W2→3 + W3→4 + …… + WN−1→N

= −GMm r − r  − GMm r − r  + …… + (−GMm)  − 

1 1 1 1 1 1
 1 2  2 3  N−1 N
r r
1 1 1 1 1 1 1 1
WTotal = −GMm  − r + r − r + r − …… − + −r 
r1 2 2 3 3 rN−1 rN−1 N
1 1 
So, WTotal = −GMm r − r  …… (5)
 1 N
For absolute potential energy we have point N to be at infinity i.e., rN = ∞.
So, eq. (5) becomes:
1 1 More coal has been used since
WTotal = −GMm r − 
 1 ∞ 1945 than was used in the whole
of history before that.
1 
= −GMm r − 0
1 
1
WTotal = −GMm r 
 1
−GMm
WTotal =
r1
This work is stored in the body in the form of absolute potential energy. Therefore, the general
expression for gravitational potential energy is:
−GMm
Absolute potential energy = U = …… (6)
r
On the surface of earth:
r = R
−GMm
So, U = Ug =
R
From eq. (6) we can conclude:
(i) When ‘r’ increases, the gravitational forces does negative work and ‘U’ increases i.e.,
become less negative.
(ii) When ‘r’ decreases, the body falls toward earth, the work is positive and ‘U’ decreases i.e.,
becomes more negative.
(iii) Negative sign shows that gravitational field for a mass ‘m’ is attractive.
[Chapter-4] Work and Energy 167
Escape Velocity:
“The initial velocity of an object with which it goes out of the earth’s gravitational field is
known as escape velocity”.
Mathematical Form:
Escape velocity can be mathematically given by:
vesc = 2gR Some Escape Speeds (kms−1)
Heavenly body Escape speed
Where vesc = Escape velocity Moon 2.4
Mercury 4.3
g = Gravitational acceleration Mars 5.0
Venus 10.4
R = Radius of earth
Earth 11.2
Explanation and Derivation: Uranus 22.4
Neptune 25.4
Suppose a body of mass m is given escape velocity due to its Saturn 37.0
Jupiter 61
initial kinetic energy.
1 2
Initial K.E = 2 mvesc …… (1)

The initial absolute energy on the surface of earth.

−GMm
P.Ei = R
Final gravitational potential energy = P.Ef = 0
Increase in potential energy = P.Ef − P.Ei
−GMm
Increase in P.E = 0− R 
 
GMm
Increase in P.E = R …… (2)

Since K.E is converting into potential energy.

Initial kinetic energy = Increase in potential energy
1 2 GMm
mv esc =
2 R
2 2GM
vesc = R
2GM
vesc = …… (3)
R
From Newton’s gravitational law:
GMm
F = …… (4)
R2
But F = W = mg …… (5)
168 Physics Intermediate Part-I

Comparing (4) and (5):

GMm
mg =
R2
GMm
R2 = g

gR2
G = M …… (6)

Put (6) in eq. (3):

2gR2
vesc = ×M
M×R

vesc = 2gR …… (7)

Result:
This equation shows that escape velocity is independent of mass of object.
Calculation for Escape Velocity:
As, g = 9.8 ms−2
R = 6.4 × 106 m
vesc = 2 × 9.8 × 6.4 × 106
vesc = 11.2 × 103 ms−1

vesc = 11 kms−1

INTERCONVERSION OF POTENTIAL ENERGY AND KINETIC ENERGY:

K.E and P.E are interconvertible in such a way that increase in one form of energy results in
decrease in other form of energy and vice versa.
Proof: m
A
Suppose a body of mass ‘m’ lifted to the height h at position ‘A’
from surface of earth as shown in figure.
x
We release the body and as it falls to the ground ‘C’. Let ‘B’ is the
point between A and C such that AC = h and AB = x and BC = h − x as B
h
shown in figure.
Total Energy at Point A: (h − x)

The total energy at point A is sum of K.E and P.E. C

T.EA = K.EA + P.EA …… (1)

P.EA = mgh
1
Also K.EA = 2 m(0)2

As mass is at rest v = 0.
[Chapter-4] Work and Energy 169
So, K.EA = 0
So, eq. (1) becomes:
T.EA = mgh …… (a)
Total Energy at Point B:
The total energy at point B is sum of its potential and kinetic energy at this point.
T.EB = K.EB + P.E B …… (2)
P.EB = mg(h − x)
Because h − x is the height in this case:
1 2
K.EB = mvB …… (3)
2
Now for vB:
vi = 0, vf = vB, a = g and S = x
So from 3rd equation of motion:
2 2
2aS = vf − vi
2
2gS = vB − (0)2
2
2gx = vB
So, (3) becomes:
1
K.EB = m (2)(gx)
2
K.EB = mgx
So, eq. (2) becomes:
T.EB = mg(h − x) + mgx
= mgh − mgx + mgx
T.EB = mgh …… (b)
Total Energy at Point C:
Total energy at point C is the sum of potential energy and kinetic energy at that point.
T.Ec = K.Ec + P.Ec …… (4)
At point C:
P.Ec = mg (0) (h = 0)
P.Ec = 0 …… (5)
1 2
and K.Ec = 2 mvc …… (6)
170 Physics Intermediate Part-I

For vc:
vi = 0, vf = vc, a = g, S = h
So 3rd equation of motion becomes:
2 2
2aS = vf − vi
2
2gh = vc
So, eq. (6) becomes:
1
K.Ec = 2 (2gh)m

K.E = mgh
So, eq. (4) becomes:
P.Ec = 0 + mgh
T.Ec = mgh …… (c)
Conclusion:
From eq. (a), (b), (c) it is clear that energy at points (A), (B), (C) is same. Thus total amount of
energy is same ‘K.E’ and P.E are interconvertible but total energy is same.
Another Form of Interconversion of Energy (In Absence of Air Friction):
When a body falls. Its kinetic energy increases because its velocity increases. On the other hand
as body falls its potential energy decreases because its height decreases.
It is clear from figure:
Loss in potential energy = Gain in kinetic energy
v1
1 2 1 2
mg(h1 − h2) = 2 mv2 − 2 mv1 h1
m
1 2 2 h2
mg(h1 − h2) = 2 m[v2 − v1] v2

This result holds good only when frictional forces are not considered.
Mathematical Form of Interconversion of Energy when Air Friction Exists:
When air friction exists then the potential energy losses to kinetic energy and work done against
friction i.e.,
Loss in potential energy = Gain in K.E + Work done against friction …… (1)
Loss in P.E = mgh
1
Gain in K.E = 2 mv2

Work done against friction of air = Force × Distance

= f×h
1
So, eq. (1) becomes mgh = 2 mv2 + fh

Where ‘f’ is frictional force and ‘h’ is height through which frictional force exists due to air
resistance.
[Chapter-4] Work and Energy 171

CONSERVATION OF ENERGY:
One form of energy can be converted into another form of Source of Original
energy but the total amount of energy remains constant. energy source
Solar Sun
NON-CONVENTIONAL ENERGY SOURCES: Bio mass Sun
The energy sources which are not common in these days. Fossil fuels Sun
However it is expected, that these sources will contribute, substantially Wind Sun
to energy demand of future. Following are some examples of non- Waves Sun
conventional sources. Hydro electric Sun
Tides Moon
Energy from Tides: Geothermal Earth
Tidal Energy:
Energy obtained from tides is called tidal energy.
Source: Renewable Non-renewable
Source of tidal energy are the tides produced due to Hydroelectric Coal
Wind Natural gas
gravitational pull between earth and moon. The tides in the sea roughly
Tides Oil
raise twice a day. Geothermal* Uranium
Application: Biomass Oil shale
Sunlight Tar sands
Dam can be constructed to store water with valves and tides Ethanol/Methanol**
present in store water. Such plants are named as tidal power plant * Individual fields may run off
whose working is based upon the following steps: ** Renewable when made from bio mass

(a) Water is stored in reservoir and then valve is closed at high tide.
Reservior
Land Dam Turbine
High tide
Ocean

High tide:
Water level equalized
(b) Water starts to flow from basin to ocean during low tide driving turbines.

Low tide
Water is beginning to flow
out of basin to ocean,
driving turbines
(c) Valve is closed and water level is equalized at low level.
Valve is
closed

Fig. Water level equalized

(d) Valve is opened at high tide and water starts to flow back, driving turbines.

Open valve

Fig. Tidal power plant. Turbines

are located inside the dam
Result:
This rushing water drives the turbine and generates electricity.
172 Physics Intermediate Part-I

Energy from Waves:

The tidal moment and the wind blowing across the surface of the ocean produced strong water
waves. Their energy can be utilized to generate electricity. A device used to produce electricity from the
waves is known as “Salter’s Duck”.
Salter’s Duck: Duck Float
Balance Float

It is a device used to produced electricity and it was

invented by “Professor Salter”. It consists of two parts:
(i) Duck float
Waves
(ii) Balance float

Working:
The waves energy make duck float move relative to balance The pull of the Moon does not
float. The relative motion of the duck float is then used to run only pull the sea up and down.
This tidal effect can also distort
electricity generators. the continents pulling land up and
down by as much as 25 cm.
Solar Energy:
Energy obtained from sun is known as solar energy. Sun produces energy at a rate of 4 × 1026
watt. Average temperature of surface of the sun is 1.5 × 107 K.

Energy Received on Earth:

Solar Constant:
“The rate at which radiant solar energy is received at outer layer of the earth’s atmosphere”.
Outside earth’s atmosphere value of solar constant is 1.4 KWm−2.
Intensity of solar energy reaching on the surface of earth is about 1 KWm−2.
This reduction of energy is due to reflection, scattering and absorption by dust particles, water
vapours and other gases. Solar energy can be used as:

Solar Energy as Heat Source (Direct Use):

Solar energy can be direct collected to heat water for this purpose a device is used called collector.
Collector:
A device that collects heat directly from solar energy by using large solar reflectors and thermal
absorber.
Working:
A collector consists of blackened surface covered with the glass. This black surface absorbs heat
directly from solar radiation cold water passes over this surface and is heated upto about 70°C.

Solar Energy as Source of Electricity:

Solar energy can be used by converting it into electricity. The device used for this purpose is
called solar cell. Single solar cell has low voltage so large no of solar cells are arranged in series to form
solar cell panel.
[Chapter-4] Work and Energy 173
Hot water
Cost and Life Time:
D
Solar cells are expensive but they have long life thus reducing Storage
tank
the running cost.
Cold water
Storage: Insulation A
C
Energy produced by solar cells can be stored during day light
Glass covered
in nickel cadmium batteries by connecting them to solar panels. box
These batteries can then provide power to electrical appliances at
Dull black
nights or on cloudy days. B surface

Uses of Solar Cell: (a)

• In satellites having large solar panels which are kept facing sun.
• In communication systems.
• In calculators.
• In wrist watches.
• In remote ground based weather stations.
(b)

Energy from Biomass:

Crop residue, natural vegetation, trees, animal dung, sewage The rapid growth of human population
are sources of biomass. Following are the methods to obtain energy has put a strain on our natural
resources. A sustainable society
from biomass: minimizes waste and maximizes
the benefit from each resource.
(i) Direct Combustion: Minimizing the use of energy is an
other method of conservation we
The process in which energy is obtained from biomass in the can save energy by,
presence of oxygen is called direct combustion. (i) turning off lights and electrical
appliances when not in use
(ii) Fermentation: (ii) using fluorescent bulbs instead
of incandescent bulbs
The process in which biomass is converted into biofuel by (iii) using sunlight in offices,
using enzymes and decomposition through bacterial action in the commercial centers and houses
during daylight hours
absence of oxygen. Biofuel such as ethanol (alcohol) is replacement (iv) taking short hot shawers.
of gasoline.
Direct Combustion Definition Fermentation Definition
• Solid waste is necessary. • Biomass is converted into biofuel and then
it is burnt to get energy.
• It is carried out in the presence of oxygen • It is carried out in the presence of enzymes
but not enzymes. and not oxygen.
• It is direct process of energy supply. • It is indirect method of energy supply.

Digester:
“The rotting of biomass is taken place in a closed tank named as digester”.
Energy from Waste Products:
Waste product like wood waste, crop residue and municipal solid waste are burnt to heat water in
boiler to produce steam. This steam can run a turbine generator.
174 Physics Intermediate Part-I

Geothermal Energy: Gas

The energy obtained from inside the earth in the form of hot
water or steam is called geothermal energy is obtained from following Methane gas
process:
(i) Radioactive Decay: Digester
Manure and
The rocks inside the earth emit different types of radiations. rotting plant
These radiations contributes toward the generation of geothermal
energy.
(ii) Residual Heat of the Earth: Pollution can be reduced if
The interior of the earth (magma) is very hot such that rocks (i) People use mass
above it are in molten form or in partially molten form. These rocks are transportation
(ii) Use geothermal, solar, hydro-
usually with in 10 km inside the earth with temperature of about 200°C electrical and wind energy as
or more. alternative forms of energy.

(iii) Compression of Materials:

The compression of material deep inside the earth also causes generation of heat.
Energy comes out of the interior of the earth either naturally or artificially depends upon the
existing of water on the molten rocks.
Water Existence on Hot Rocks:
In some places beneath the ground, water is contact with a layer of rocks. This layers of rock is
known as aquifer. This water is raised to high temperature and pressure and comes out of earth’s surface
as hot spring geysers or steam vents. They can run turbine of electrical generator.
If water is not existing on molten rocks naturally then it can be pumped down to get steam at the
other end. This steam can run turbines of electrical generators.
Aquifer:
“An aquifer is a layer of rock holding water that allows water to percolate through it with
pressure”.
Geysers:
“It is a hot spring that discharges steam and hot water, intermittently releasing an explosive
column into air”.
[Chapter-4] Work and Energy 175

SOLVED EXAMPLES
EXAMPLE 4.1
A force F acting on an object varies with distance x as shown in the figure. Calculate the
work done by the force as the object moves from x = 0 to x = 6 m.
Data:
Force = F
Distance = x
To Find:
Work done = W = ?

SOLUTION
[Total area under the graph] = (Area of rectangle) + (Area of triangle)
1
= (4 x 5) + 2 (6 – 4) (5)

1
= 20 + 2 (2) (5)

= 20 + 5 = 25 J
Since area under F–d graph is equal to work done
Hence, W = 25 J
Result:
Work done by the force = W = 25 J

EXAMPLE 4.2
A 70 kg man runs up a long flight of stairs in 4.0 s. The vertical height to the stairs is 4.5 m
calculate his power out put in watts.
Data:
Mass of man = m = 70 kg
= t = 4 sec
Height of stairs = h = 4.5 m
To Find:
Power output = P = ?

SOLUTION
Using
W
P = t

Since W = P.E. = mgh

176 Physics Intermediate Part-I

mgh
∴ P =
t
70 x 9.8 x 4.5
=
4
= 7.7 x 102 watt
Result:
Power output = P = 7.7 × 102 watt

EXAMPLE 4.3
A brick of mass 2.0 kg is dropped from a rest position 5.0 m above the ground. What is its
velocity at a height of 3.0 m above the ground?
Data:
Mass of brick = m = 2 kg
rest position = h1 = 5m
At a height = h2 = 3m
To Find:
Velocity = v = ?

SOLUTION
Using Loss of P.E. = Gain in K.E.
1 2 2
mg (h1 – h2) = 2 m (v2 – v1)

∴ v1 = 0 and v2 = v
1
∴ mg (h1 – h2) = 2 m (v2 – 02)

1
g (h1 – h2) = 2 v2

v2 = 2g (h1 – h2)
Taking square root
v = 2 g (h1 – h2)
Putting values
v = 2 x 9.8 (5 – 3)
= 2 x 9.8 x 2
= 39.2
v = 6.26 m/s
= 6.3 m/s
Result:
Velocity at a height 3.0 m = v = 6.3 m/s
[Chapter-4] Work and Energy 177

SHORT QUESTION ANSWERS

4.1 A person holds a bag of groceries while standing still, talking to a friend. A car is stationery
with its engine running. From the stand point of work, how are these two situations similar?
Ans. In both cases workdone is zero.
Reason: As the person and the car, both are rest i.e., displacement is zero. Thus, from the stand
point of work if displacement = 0 then workdone is also zero.
→ →
W = F . d
Since, d = 0 so,
W = F(0)
W = 0
4.2 Calculate the work done in kilo joules in lifting a mass of 10 kg (at a steady velocity)
through a vertical height of 10 m.
Ans. In order to lift an object, a force F equal in magnitude to that of weight is applied which produces
displacement equal to height i.e.,
→
W = |F| = mg
The workdone (in the form of P.E) is given by:
W = P.E = mgh cos θ
W = (10)(9.8)(10)(cos 0°)
d=h F
W = 980 J (cos 0° = 1)
θ = 0O
980
W = 1000 kJ

W = 0.98 kJ
4.3 A force F acts through a distance L. The force is then increased to 3 F, and then acts
through a further distance of 2 L. Draw the work diagram to scale.
Ans. In this case, take force F along y-axis and displacement C D
3F
‘d’ along x-axis. As area under force displacement graph
is equal to the workdone so,
2F 6FL
Total workdone = (Area OABE)
A B
+ (Area CDFE) 1F

W = (F)(L) + (3F)(2L) 1FL

E F
W = FL + 6FL O 1L 2L 3L

W = 7FL units
178 Physics Intermediate Part-I

4.4 In which case is more work done? When a 50 kg bag of books is lifted through 50 cm, or
when a 50 kg crate is pushed through 2 m across the floor with a force of 50 N?
Ans. Case-I: m1 = 50 kg
h = 50 cm = 0.5 m
50 cm
As W1 = mgh
W1 = (50)(9.8)(0.5)
W1 = 245 J
2m
Case-II: d = 2m
F = 50 N
θ = 0°
W2 = Fd cos θ = Fd (cos 0°)
W2 = Fd (cos 0° = 1)
W2 = (50)(2) = 100 J
The workdone in raising 50 kg of bag of books is more than to push a 50 N block through 2m
i.e., W1 > W2.
4.5 An object has 1 J of potential energy. Explain what does it mean?
Ans. An object having 1 J of P.E has the ability to do work of 1 J. An object lifted up at 1 m with 1 N
force when dropped freely vertically has the capacity to do work of 1 J. The work may be in
gravitational field, magnetic or electrical field. Also it may have K.E of 1 J.
4.6 A ball of mass m is held at a height h1 above a table. The table top is at a height h2 above
the floor. One student says that the ball has potential energy mgh1 but another says that it
is mg (h1 + h2). Who is correct?
Ans. Since P.E is always with respect to some reference point. If
table top is considered as reference then P.E of ball of mass h1
m at height h1 is mgh1. But if floor is considered as reference
h1 + h2

point than the P.E of the same ball is mg(h1 + h2) because
height of ball from floor is h1 + h2. So both are correct.
h2

4.7 When a rocket re-enters the atmosphere, its nose cone becomes very hot. Where does this
heat energy come from?
Ans. When a rocket re-enters the atmosphere, its P.E is converted into K.E. Air resistance opposes its
motion. The rocket works against the frictional forces caused by dust particles and air which is
converted into heat. This causes the nose cone to become hot.
4.8 What sort of energy is in the following:
(a) Compressed spring
(b) Water in a high dam
(c) A moving car
[Chapter-4] Work and Energy 179
Ans. (a) Compressed Spring: It has elastic potential energy, stored in spring during compression of
spring. i.e., E = (1/2) kx2.
(b) Water in a High Dam: It has gravitational potential energy, due to its height.i.e E = mgh.
(c) A Moving Car: Because of its motion produced by engine it has kinetic energy. i.e.,
E = (1/2) mv2.
4.9 A girl drops a cup from a certain height, which breaks into pieces. What energy changes
are involved?
Ans. During drop of cup, its gravitational P.E is converted to K.E which then becomes zero by
striking the ground. Some K.E is used to break cup into pieces and thus becomes K.E of pieces
while the remaining is converted into sound and heat energy.
P.E → K.E → Work done + Heat + Sound
If air resistance is considered then
P.E → Work done against friction + K.E → Work + Heat + Sound
4.10 A boy uses a catapult to throw a stone which accidentally smashes a green house window.
List the possible energy changes.
Ans. The elastic P.E stored in the stone during the stretching of string of catapult, firstly changes into
K.E during motion. With smashing, K.E is consumed in the breaking of glass and the rest is
converted into heat and sound energy.
P.E → K.E + Work against friction → Breaking work + Sound + Heat

EXTRA SHORT QUESTIONS

Q. Give one example of work done by a variable force.
Ans. When a body is moved away from the center of gravity. The magnitude of the force continuously
decreases; however, its direction always remains the same.e.g work done by the gravitational
force.
Q. Define conservative force.
Ans. A force is said to be conservative, if the amount of work done in moving an object against that
force depends only on the initial and the final position of the object.e.g electric force,
gravitational force.
Q. A man whose mass is 75 kg walks up 10 steps, each 20 cm high, in 5 s. Find the power he
develops. Take g = 10 ms−2.
Ans. Mass m = 75 kg
10 × 20
Total height =  100  m = 2m
 
t = 5s
Work 75 × 10 × 2
∴ Power = Time = = 300 W
5
180 Physics Intermediate Part-I

Q. Two protons are brought toward each other. Will the potential energy of the system de-
crease or increase?
Ans. The potential energy of the system will increase. Because, we have to do work in order to bring
the two patrons close to each other.
Q. Can K.E. of a body be negative?
1
Ans. No, K.E. = mv2
2
Since m and v2 can never have negative value. So, K.E. can never be negative.
Q. Is work done by frictional force positive?
Ans. No, the work done by frictional force is negative. Because force and displacement are anti-parallel
W = Fd cos (180°) = −Fd
Q. Differentiate between renewable and non-renewable resources of energy.
Ans. Renewable energy resources Non-renewable energy resources
The energy resources which cannot be The energy resources which can be exhausted
exhausted and can be used again are called one day and cannot be used repeatedly are
renewable energy sources. called non-renewable energy sources.
It can be used again and again throughout its It cannot be used again and again but one day
life. it will be exhausted.
It has low carbon emission and hence It has high carbon emission and hence not
environment friendly. environment friendly.
It is present in unlimited quantity. It is present in limited quantity and vanishes
one day.
Cost is low. Cost is high.
Renewable energy resources are pollution free. The non-renewable energy resources are not
pollution free.
Life of resources in infinite. Life of resources is finite and vanishes one
day.
Solar energy, wind energy, tidal energy etc., Coal, petroleum, natural gases are the
are the examples of renewable resources. examples of non-renewable resources.
[Chapter-4] Work and Energy 181

PROBLEMS WITH SOLUTIONS

PROBLEM 4.1
A man pushes a lawn mower with a 40 N force directed at an angle of 20°° downward from
the horizontal. Find the work done by the man as he cuts a strip of grass 20m long.
Data:
F = 40 N
θ = 20°
d = 20 m
Required:
W = ?
Calculations:
By using formula:
→ →
W = F . d
W = Fd cos θ
Putting the values:
W = (40)(20) cos (20)

W = 7.5 × 102 J

Result:
Workdone = 7.5 × 102 Joules

PROBLEM 4.2
A rain drop (m = 3.35 × 10−5 kg) falls vertically at a constant speed under the influence of
the forces of gravity and friction. In falling through 100 m, how much work is done by.
(a) Gravity and (b) Friction
Data:
h = 100 m
m = 3.35 × 10−5 kg
Required:
(a) Workdone by gravitational force = ?
(b) Workdone by frictional force = ?
182 Physics Intermediate Part-I

Calculations:
Workdone by Gravity:
→ →
Wg = F . d
Wg = Fd cos θ
F
θ = 0°
Wg = Fd cos (0°)
Wg = Fd (cos 0° = 1)
Wg = mgh (F = mg)
−5
Wg = (3.35 × 10 )(9.8)(100)
Wg = 0.0328 Joules
Workdone by Frictional Force:
→ →
Wf = F . d
Wf = Fd cos θ
θ = 180°
Wf = Fd cos (180°) (cos 180° = −1) F

Wf = Fd (−1)
Wf = −mgh
Wf = −(3.35 × 10−5)(9.8)(100)
Wf = −0.0328 joules
Result:
(a) Workdone by gravitational force = 0.0328 J
(b) Workdone by frictional force = −0.0328 J
PROBLEM 4.3
Ten bricks, each 6.0 cm thick and mass 1.5 kg, lie flat on a table. How much work is
required to stack them one on the top of another?
Data: 10
−2
h = 6 cm = 6 × 10 m 9
m = 1.5 kg
8
Number of bricks = n = 10
7
Required:
6
Workdone = W = ?
5
Calculations:
4
W = mgh1 + mgh2 + …… + mgh10
3
W = mg(h1 + h2 + …… + h10)
W = mg(0h + 1h + 2h + 3h + 4h + 5h 2
1 0h 1h 9h
+ 6h + 7h + 8h + 9h)
[Chapter-4] Work and Energy 183
W = mgh(45)
W = (45)(1.5)(9.8)(6 × 10−2)

W = 40 J

Result:
Workdone = W = 40 J

PROBLEM 4.4
A car of mass 800 kg travelling at 54 km h−1 is brought to rest in 60 meters. Find the
average retarding force on the car. What has happened to original kinetic energy?
Data:
m = 800 kg
vi = 54 km h−1
54 × 103
=
3600
vi = 15 ms−1
d = 60 m
vf = 0 ms−1
Required:
(a) Average retarding force = F = ?
(b) Kinetic energy = K.E = ?
Calculations:
Using work energy principle.
For F:
1 2 2
Fd = 2 (mvf − mvi )

1 2 2
F = 2d (m)(vf − vi )
 
800
F = 2(60) [(0)2 − (15)2]

F = (6.666)(−(15)2)
F = (6.666)(−225)
F = −1499.85 N

F ≈ −1500 N

Negative sign shows that the force is retarding force.

184 Physics Intermediate Part-I

For Kinetic Energy:

Since the body comes to rest so,
Velocity = 0 m/s
1
K.E = mv2
2
800
K.E =  (0)2
 2 

K.E = 0

Result:
(a) Retarding force F = 1500 N
(b) Kinetic energy K.E = 0 J

PROBLEM 4.5
A 1000 kg automobile at the top of an incline 10 metre high and 100m long is released and
rolls down the hill. What is its speed at the bottom of the incline if the average retarding force due
to friction is 480 N?
Data:
m = 1000 kg
h = 10 m
S = d = 100 m
F = 480 N
Required:
Speed = v = ?
Calculations:
Using:
Loss of P.E = Gain in K.E + Workdone against friction
1
mgh = 2 mv2 + fd

1
mgh − fd = 2 mv2

Putting the values:

1
(1000)(9.8)(10) − (480)(100) = 2 (1000)(v2)

98000 − 48000 = 500 v2

50,000 = 500 v2
[Chapter-4] Work and Energy 185

50,000
v2 =
500

h = 10 m
v2 = 100 d

v = 100

v = 10 ms−1

Result:
Final speed of automobile at bottom of incline = v = 10 ms−1

PROBLEM 4.6
100 m3 of water is pumped from a reservoir into a tank, 10 m higher than the reservoir, in
20 minutes. If density of water 1000 kg m−3, find.
(a) The increase in P.E.
(b) The power delivered by the pump.

Data:
Volume of water = V = 100 m3
Height of tank = h = 10 m
Time taken = t = 20 min. = 20 × 60 = 1200 sec.
Density of water = ρ = 1000 kg/m3

Required:
(a) Increase in P.E = mgh = ?
(b) Power delieved by pump = P = ?

Calculations:
Mass
As Density = Volume

Mass = Density × Volume

m = 100 × 1000
m = 1 × 105 kg
For P.E:
Increase in P.E = mgh
= (1 × 105)(9.8)(10)
P.E = 9.8 × 106 J
186 Physics Intermediate Part-I

For P:
Work
P =
Time
P.E
P = t
9.8 × 106
P = 1200

P = 8166.6
P = 8.16 × 103 watt

P = 8.2 kW

Result:
(a) Increase in P.E = 9.8 × 106 J
(b) Power delivered by pump = P = 8.2 kW

PROBLEM 4.7
A force (thrust) of 400 N is required to overcome road friction and air resistance in
propelling an automobile at 80 km h−1. What power (kW) must the engine develop?
Data:
Force = F = 400 N
Velocity = v = 80 km h−1
80 × 1000
= 3600
v = 22.22 m/s

Required:
Power = P = ?

Calculations:
→ →
As, P = F . v
P = Fv cos θ
As, force and velocity are in same direction. So
θ = 0°
P = Fv cos (0°)
P = Fv
[Chapter-4] Work and Energy 187
Putting the values:
P = 400 × 22.22
P = 8888 watt
P = 8.88 × 103 W

P = 8.9 kW

Result:
Power = P = 8.9 kW

PROBLEM 4.8
How large a force is required to accelerate an electron (m = 9.1 × 10−31 kg) from rest to a
speed of 2.0 × 107 ms−1 through a distance of 5.0 cm?
Data:
Mass of electron = m = 9.1 × 10−31 kg
Initial velocity = vi = 0 m/s
Final velocity = vf = 2 × 107 m/s
Distance = d = 5 cm = 0.05 m
Required:
Force required = F = ?
Calculations:
By work energy principle:
1 2 2
Fd = 2 m(vf − vi )

As, vi = 0
1 2
So, Fd = 2 mvf
2
mvf
F = 2d

(9.1 × 10−31)(2 × 107)2

F = 2(0.05)
3.64 × 10−16
F =
0.1

F = 3.6 × 10−15 N

Result:
The force required is 3.6 × 10−15 N.
188 Physics Intermediate Part-I

PROBLEM 4.9
A diver weighing 750 N dives from a board 10 m above the surface of a pool of water. Use
the conservation of mechanical energy to find his speed at a point 5.0 m above the water surface,
neglecting air friction.
Data:
Weight of the diver = W = 750 N
h1 10m
Height (above water) = h1 = 10 m
h2 5m
Height (above water surface) = h2 = 5m
Required:
Speed at h2 = v = ?
Calculations:
Gain of K.E = Loss of P.E
1 2
2 mv = mg(h1 − h2)

v2 = 2g(h1 − h2)
Putting the values:
v2 = (2)(9.8)(10 − 5)
v2 = 19.6 × 5
v2 = 98
v = 98
v = 9.9 m/s

Result:
Speed at h2 = v = 9.9 m/s

PROBLEM 4.10
A child starts from rest at the top of a slide of height 4.0m, (a) what is his speed at the
bottom if the slide is frictionless? (b) if he reaches the bottom, with a speed of 6 ms−1, what
percentage of his total energy at the top of the slide is lost as a result of friction?
Data:
Initial velocity = vi = 0 m/s
Height = h = 4m
Required:
(a) Speed = v = ?
(If slide is frictionless)
(b) % of total energy lost = ?
(If velocity at bottom v′ = 6 ms−1)
[Chapter-4] Work and Energy 189
Calculations:
For Speed (For Frictionless System):
Gain of K.E = Loss of P.E
1 2
2 mv = mgh

v2 = 2gh
v = 2gh
v = 2(9.8)(4)
v = 78.4
v = 8.8 m/s
For % age Loss of K.E:
Speed = v′ = 6 m/s (For friction)
Speed = v = 8.8 m/s (For frictionless)
Now,
1 1 
2 mv2 − 2 mv′2
  100
% loss of K.E = × 100
1 
2 mv2
 
1
m(v2 − v′2)
2 100
% loss of K.E = 1 × 100
2
2 mv
(v2 − v′2) 100
% loss of K.E = v2 × 100

[(8.8)2 − (6)2] 100

% loss of K.E =
(8.8)2 × 100

% loss of K.E = 53.5%

Result:
(a) v = 8.8 m/s
(b) % loss of K.E = 54%
190 Physics Intermediate Part-I

MULTIPLE CHOICE
CHOICE QUESTIONS
Note: You have four choices for each objective type question as a, b, c and d. Tick the
correct choice.
1. At what angle between force and displacement work done by force is half of its maximum
value?
(a) 30° (b) 45° (c) 60° (d) 90°
2. Dimensions of work are same as that of:
(a) Momentum (b) Torque (c) Power (d) Impulse
3. If θ < 90°° work is said to be:
(a) Negative (b) Positive (c) Zero (d) None of these
4. When force and displacement are in opposite direction, then the work done is taken as:
(a) Negative (b) Positive (c) Zero (d) Infinite
o
5. If θ > 90 work is said to be:
(a) Negative (b) Positive (c) Zero (d) None of these
6. The unit of work is CGS system is:
(a) Joule (b) Newton (c) Dyne (d) Erg
−
7. When a person holding a pail by the force F is moving forward then the work being done
on the pail is:
(a) Maximum (b) Negative (c) Minimum (d) Zero
8. 1 erg =
(a) 1010 J (b) 10−7 J (c) 107 J (d) 10−5 J
9. When rocket moves away from the Earth, then work is done:
(a) Against centripetal force (b) Against gravitational force
(c) Against magnetic force (d) Against electrostatic force
10. Which of the following is non-conservative force?
(a) Friction (b) Air resistance (c) Tension in string (d) All of them
11. Slope of work-time graph is equal to:
(a) Force (b) Velocity (c) Power (d) Energy
12. Kilo watt hour is unit of:
(a) Power (b) Energy (c) Torque (d) Momentum
13. If an agent consumes a power of 1 kilo-watt in one hour, then energy of the source is:
(a) One watt (b) One kilo-watt (c) One kilowatt-hour (d) one kilo joule
14. When a conservative force performs a positive work than P.E will:
(a) Increases (b) Decrease (c) Remains same (d) None of these
[Chapter-4] Work and Energy 191
15. If the work is done at uniform rate than:
(a) Pav = Pins (b) Pav < Pins (c) Pav > Pins (d) P = 0
16. If velocity of a body is doubled and its mass is also doubled then K.E of the body becomes:
(a) Doubles (b) Four times (c) Eight times (d) One half
17. Escape velocity of a body is independent of:
(a) Mass of earth (b) Radius of earth (c) Mass of object (d) Both (a) and (b)
18. Absolute P.E of a body of mass one kg over the surface of the earth is:
GMm GM GMm GRm
(a) − (b) − (c) (d)
R R R M
19. When we raise the body above the surface of the earth its P.E within the gravitational field:
(a) Increases (b) Decreases (c) Become zero (d) Remain same
20. P.E of a body increases this means work done by gravity is:
(a) Positive (b) Negative (c) Zero (d) Infinite
21. If proton and alpha have same momentum. Which of the following is true?
(a) K.Ep > K.Eα (b) K.Ep < K.Eα (c) K.Ep = K.Eα (d) None of these
22. In a resistive medium, the loss of P.E of any body is:
(a) Equal to gain in K.E plus the work done against friction
(b) Equal to loss in K.E plus the work done against friction
(c) Equal to gain in K.E minus the work done against friction
(d) Only equal to gain in K.E
23. An object at rest may have:
(a) Momentum (b) Velocity (c) K.E (d) Potential energy
24. When two protons are brought close to each other than their:
(a) K.E increases (b) K.E and P.E both increase
(c) P.E increases (d) P.E and K.E remain same
25. A brick of mass 2 kg falls from height 10 m. It velocity when its height is 5 m:
(a) 10 ms−1 (b) 5 ms−1 (c) 2 ms−1 (d) 15 ms−1
26. The initial velocity with which an object can reach to an infinite distance from earth is called:
(a) Terminal velocity (b) Orbital velocity (c) Drift velocity (d) Escape velocity
27. All the food a person eat in one day has about the same energy as:
(a) 3 liter of petrol (b) 1.3 liter of petrol (c) 1/3 liter of petrol (d) 2/3 liter of petrol
GMm
28. In the expression Ug = − , the negative sign shows the earth’s gravitational field for
R
mass is:
(a) Zero (b) Attractive (c) Repulsive (d) None of these
192 Physics Intermediate Part-I

29. The escape velocity of the earth is:

(a) Greater than moon (b) Less than moon (c) Equal to the moon (d) None of these
30. The relation between orbital and escape velocity is:
1 1
(a) Vesc = (b) Vesc = 2 Vo (c) Vesc = 2Vo (d) Vesc =
2Vo 2V
31. The ratio of escape velocity to the orbital velocity is:
1
(a) (b) 2 (c) 1 (d) 2
2
32. The total energy of an object falling freely any point during downward motion:
(a) Decreases (b) Increases (c) Remains constant (d) becomes Zero
33. The original source of energy of tides of:
(a) Earth (b) Sun (c) Moon (d) Star
34. Which one is not a nonrenewable energy source?
(a) Wind (b) Oil (c) Tar sands (d) Natural gas
35. On a clear day solar energy reaching the earth’s surface is about:
(a) 1.4 KWm−2 (b) 1 KWm−2 (c) 1000 Wm−2 (d) Both (b) and (c)
36. Bio fuel ethanol is produced by:
(a) Direct combustion (b) Fermentation of biomass
(c) Radioactive decay (d) Compression of material
37. Biomass is a potential source of:
(a) Light energy (b) Non-renewable energy
(c) Renewable energy (d) None of these
38. A hot spring that discharges steam and hot water intermittently releasing an explosive
column in air is called:
(a) Aquifers (b) Mountain (c) Burner (d) Geyser
39. Heat within the earth generated by the:
(a) Compression of material (b) Residual heat of the earth
(c) Radioactive decay (d) All of these
40. The igneous rocks with in the 10 km of the Earth’s surface having the temperature about:
(a) 2000°C (b) 150°C (c) 200°C or more (d) 900°C
41. The function of photo cells is to convert the light energy in to:
(a) Mechanical energy (b) Electrical energy (c) Heat energy (d) Chemical energy
42. Which of the following is not expression of power?
W Mgh
(a) t (b) t (c) Fv cos θ (d) Fv sin θ
[Chapter-4] Work and Energy 193
43. Electric power of colour TV is  watts.
(a) 10 (b) 100 (c) 120 (d) 240
44. 1 kWh =  GJ.
(a) 3.6 × 103 (b) 3.6 × 10–3 (c) 3.6 × 106 (d) 3.6 × 10–6
45. According to work energy principle (vf – vi)(vf + vi) = .
F 2Fd
(a) Fd (b) (c) Fdm (d)
d m
46. Escape velocity of surface of moon is  km/h.
(a) 86 40 (b) 86.4 (c) 8.64 (d) 0.864
47. Value of solar constant is  W/m2.
(a) 1400 (b) 140 (c) 14 (d) 1.4
48.  batteries are used to store electricity produced by solar cells.
(a) Copper (b) Silver (c) Nickel cadmium (d) Uranium
49. Ratio of max power to velocity is .
(a) Force (b) Work (c) Energy (d) Torque
50. A body of mass M is dropped from height “H”, its speed at midpoint of its falls is:
gH
(a) MgH (b) gH (c) gH (d) M
51. Which of the following force cannot do any work when it acts on the body?
(a) Elastic force (b) Centripetal force (c) Frictional force (d) Electric force
52. The power needed to lift a mass of 5 kg to a height in 10 m in 1 sec. is:
(a) 4.9 W (b) 49 W (c) 4.9 kW (d) 0.49 kW
53. The momentum of particle is numerically equal to its K.E. What is the velocity of the particle?
(a) 9 ms−1 (b) 3 ms−1 (c) 2 ms−1 (d) 1 ms−1
54. A body of mass 3 kg lies on the surface of the table 2 m high. It is moved on the surface by 4 m.
The change of P.E. will be:
(a) Zero (b) 9.8 J (c) 19.6 J (d) 329 J
55. Ratio of dimensions of power and K.E is:
(a) 1:1 (b) T:1 (c) 1:T (d) M:T
56. The kinetic energy of a body of mass m is E. Its momentum is:
mE 2E
(a) 2 mE (b) 2 mE (c) (d)
2 m
57. The gravity does no work, when the body moves:
(a) Horizontally (b) Vertically upward
(c) Vertically downward (d) At an angle of 45° with horizontal
194 Physics Intermediate Part-I

58. A body falls freely under gravity. Its velocity is v when it has lost a potential energy of U.
The mass of the body is:
2U U U
(a) (b) (c) (d) v×U
v2 v2 2v2
59. Two bodies with K.E. in the ratio of 4 : 1 are moving with equal linear momentum. The
ratio of their masses is:
(a) 1:2 (b) 1:1 (c) 4:1 (d) 1:4
60. If momentum of a body having K.E (E) is doubled (by keeping mass the same). The new
K.E. is:
(a) E (b) 2E (c) 4E (d) 8E

ANSWERS
Sr. Ans. Sr. Ans. Sr. Ans. Sr. Ans. Sr. Ans.
1. (c) 2. (b) 3. (b) 4. (a) 5. (a)
6. (d) 7. (d) 8. (b) 9. (b) 10. (d)
11. (c) 12. (b) 13. (c) 14. (b) 15. (a)
16. (c) 17. (c) 18. (b) 19. (a) 20. (b)
21. (a) 22. (a) 23. (d) 24. (c) 25. (a)
26. (d) 27. (c) 28. (b) 29. (a) 30. (b)
31. (b) 32. (c) 33. (c) 34. (a) 35. (d)
36. (b) 37. (c) 38. (d) 39. (d) 40. (c)
41. (b) 42. (d) 43. (c) 44. (b) 45. (d)
46. (a) 47. (a) 48. (c) 49. (a) 50. (c)
51. (b) 52. (d) 53. (c) 54. (a) 55. (c)
56. (a) 57. (a) 58. (a) 59. (d) 60. (c)