Chapter 9 Work, Energy and Power Part 1 – Physics free study material by TEACHING CARE online tuition and coaching classes
Chapter 9 Work, Energy and Power Part 1 – Physics free study material by TEACHING CARE online tuition and coaching classes
Introduction.
The terms ‘work’, ‘energy’ and ‘power’ are frequently used in everyday language. A farmer clearing weeds in his field is said to be working hard. A woman carrying water from a well to her house is said to be working. In a drought affected region she may be required to carry it over large distances. If she can do so, she is said to have a large stamina or energy. Energy is thus the capacity to do work. The term power is usually associated with speed. In karate, a powerful punch is one delivered at great speed. In physics we shall define these terms very precisely. We shall find that there is at best a loose correlation between the physical definitions and the physiological pictures these terms generate in our minds.
Work is said to be done when a force applied on the body displaces the body through a certain distance in the direction of force.
Work Done by a Constant Force.
Let a constant force F be applied on the body such that it makes an angle q with the horizontal and body is displaced through a distance s
By resolving force F into two components :
 F cosq in the direction of displacement of the
 F sinq in the perpendicular direction of displacement of the
Since body is being displaced in the direction of F cos q , therefore work
done by the force in displacing the body through a distance s is given by
W = (F cosq ) s = Fs cosq or W = F.s
Thus work done by a force is equal to the scalar or dot product of the force and the displacement of the body.
If a number of force
F 1 , F 2 , F 3……. Fn
are acting on a body and it shifts from position vector
r _{1} to position
vector
r _{2} then W = (F 1 + F 2 + F 3 + ….F n ).( r _{2} – r _{1} )
Nature of Work Done.
Problem 1. A body of mass 5 kg is placed at the origin, and can move only on the xaxis. A force of 10 N is acting on it in
a direction making an angle of 60 ^{o} with the xaxis and displaces it along the xaxis by 4 metres. The work
done by the force is [MP PET 2003]

(a) 2.5 J (b) 7.25 J (c) 40 J (d) 20 J
Solution : (d) Work done =
r
F.s = F s cosq = 10 ´ 4 ´ cos 60 = 20 J
Problem 2. A force
F = (5ˆi + 3ˆj) N is applied over a particle which displaces it from its origin to the point
r = (2ˆi – 1ˆj)
metres. The work done on the particle is [MP PMT 1995; RPET 2003]
(a) –7 J (b) +13 J (c) +7 J (d) +11 J
Solution : (c) Work done =
r r

F.r = (5i + 3 j).(2i – j) = 10 – 3 = +7 J
Problem 3. A horizontal force of 5 N is required to maintain a velocity of 2 m/s for a block of 10 kg mass sliding over a rough surface. The work done by this force in one minute is
(a) 600 J (b) 60 J (c) 6 J (d) 6000 J
Solution : (a) Work done = Force ´ displacement = F ´ s = F ´ v ´ t = 5 ´ 2 ´ 60 = 600 J.
Problem 4. A box of mass 1 kg is pulled on a horizontal plane of length 1 m by a force of 8 N then it is raised vertically to a height of 2m, the net work done is
(a) 28 J (b) 8 J (c) 18 J (d) None of above
Solution : (a) Work done to displace it horizontally = F ´ s = 8 ´ 1 = 8 J
Work done to raise it vertically F ´ s = mgh = 1 ´ 10 ´ 2 = 20 J
\ Net work done = 8 +20 = 28 J
Problem 5. A 10 kg satellite completes one revolution around the earth at a height of 100 km in 108 minutes. The work done by the gravitational force of earth will be
(a)
108 ´ 100 ´ 10 J
(b)
108 ´ 10 J
100
(c)
100 ´ 10 J
108
(d) Zero
Solution : (d) Work done by centripetal force in circular motion is always equal to zero.
Work Done by a Variable Force.
When the magnitude and direction of a force varies with position, the work done by such a force for an
infinitesimal displacement is given by dW = F. ds
The total work done in going from A to B as shown in the figure is


W = òA F. ds = òA (F cosq )ds
In terms of rectangular component
F = Fxˆi + Fy ˆj + Fz kˆ
ˆ ˆ ˆ
ds = dxi + dyj + dzk
^{B} ˆ ˆ ˆ ˆ ˆ ˆ
\ W = òA (Fxi + Fy j + Fzk).(dxi + dyj + dzk)
or W =
xB F dx +
yB F dy +
zB F dz

òx x
òy y
òz z


Problem 6. A position dependent force
r

F = (7 – 2x + 3x ) N
acts on a small abject of mass 2 kg to displace it from
 = 0
to x = 5m . The work done in joule is [CBSE PMT 1994]
(a) 70 J (b) 270 J (c) 35 J (d) 135 J
Solution : (d) Work done = òx2 F dx = ò5 (7 – 2x + 3x ^{2} )dx = [7x – x ^{2} + x ^{3} ]^{5} = 35 – 25 + 125 = 135 J
0
x1 0
Problem 7. A particle moves under the effect of a force F = Cx from x = 0 to x = x_{1}. The work done in the process is
[CPMT 1982]
(a)
Cx ^{2}
(b)
 Cx ^{2}
(c)
Cx (d) Zero
1 2 1
x x é
1

2 ù x1
Solution : (b) Work done = ò 2 F dx = ò 1 Cx dx = C ê x ú
= 1 C x ^{2}
x1 0
ëê 2
úû 0 2
Problem 8. The vessels A and B of equal volume and weight are immersed in water to a depth h. The vessel A has an opening at the bottom through which water can enter. If the work done in immersing A and B are WA and WB respectively, then
(a)
WA = WB
(b)
WA < WB
(c)
WA > WB
(d)
WA > = < WB
Solution : (b) When the vessels are immersed in water, work has to be done against upthrust force but due to opening at the bottom in vessel A, upthrust force goes on decreasing. So work done will be less in this case.
Problem 9. Work done in time t on a body of mass m which is accelerated from rest to a speed v in time t_{1} as a function of time t is given by
1 v v
1 æ mv ö 2
1 v^{2}
(a)
m t ^{2}
 t1
(b)
m t ^{2}
t1
(c)
ç ÷ t ^{2}
2 è t1 ø
(d)
m t ^{2}



1
æ 1 ö 1
1 æ v ö 2 é v ù
Solution : (d) Work done = F.s = ma.ç a t ^{2} ÷ =
ma ^{2} t ^{2} = mç ÷ t ^{2} êAs acceleration (a) = givenú
è 2 ø 2
2 è t1 ø ë
t1 û
Dimension and Units of Work.
Dimension : As work = Force ´ displacement
\ [W] = [Force] ´ [Displacement]
= [MLT ^{2} ] ´[L] = [ML^{2}T ^{2} ]
Units : The units of work are of two types
Joule [S.I.]: Work done is said to be one Joule, when 1 Newton
force displaces the body through 1 meter in its own direction.
From W = F.s
1 Joule = 1 Newton ´ 1 metre
Erg [C.G.S.] : Work done is said to be one erg when 1 dyne
force displaces the body through 1 cm in its own direction.
From W = F s
1 Erg = 1Dyne ´ 1cm
Relation between Joule and erg
1 Joule = 1 N ´ 1 m = 10^{5} dyne ´ 10^{2} cm
= 10^{7} dyne ´ cm = 10^{7} Erg
kg–m [S.I.]: 1 Kgm of work is done when a force of 1kgwt. displaces the body through 1m in its own direction.
From W = F s
1 kgm = 1 kgwt ´ 1 metre
= 9.81 N ´ 1 metre = 9.81 Joule
gm–cm [C.G.S.] : 1 gmcm of work is done when a force of 1gmwt displaces the body through 1cm in its own direction.
From W = F s
1 gmcm = 1gmwt ´ 1cm. = 981 dyne ´ 1cm
= 981 erg
Work Done Calculation by Force Displacement Graph.
Let a body, whose initial position is
 , is acted upon by a variable force (whose magnitude is changing
continuously) and consequently the body acquires its final position
 f .
Let F be the average value of variable force within the interval dx from position x to (x + dx) i.e. for small displacement dx. The work done will be the area of the shaded strip of width dx. The work done on the body in
displacing it from position
dW = F dx
 to x f
will be equal to the sum of areas of all the such strips

\W = x f
xi
dW = x f

xi
F dx

\W = x f (Area of strip of width dx)
xi
\W = Area under curve Between xi and x f
i.e. Area under force displacement curve with proper algebraic sign represents work done by the force.
Problem 10. A 10 kg mass moves along xaxis. Its acceleration as a function of its position is shown in the figure. What is
the total work done on the mass by the force as the mass moves from x = 0 to x = 8 cm [AMU (Med.) 2000]
(a) 8 ´ 10 ^{–}^{2} J
(b) 16 ´ 10 ^{–}^{2} J
(c) 4 ´ 10 ^{–}^{4} J
(d) 1.6 ´ 10 ^{–}^{3} J
Solution : (a) Work done on the mass = mass ´ covered area between the graph and displacement axis on at graph.
= 10 ´ 1 (8 ´ 10 ^{–}^{2} ) ´ 20 ´ 10 ^{–}^{2} = 8 ´ 10 ^{–}^{2} J.
2
Problem 11. The relationship between force and position is shown in the figure given (in one dimensional case). The work done by the force in displacing a body from x = 1 cm to x = 5 cm is [CPMT 1976]
 20 ergs
 60 ergs
 70 ergs
 700 ergs
Solution : (a) Work done = Covered area on forcedisplacement graph = 1 ´ 10 + 1 ´ 20 – 1 ´ 20 + 1 ´ 10 = 20 erg.
Problem 12. The graph between the resistive force F acting on a body and the distance covered by the body is shown in the figure. The mass of the body is 25 kg and initial velocity is 2 m/s. When the distance covered by the body is 5m , its kinetic energy would be
 50 J
 40 J
 20 J
 10 J
Solution : (d) Initial kinetic energy of the body = 1 mu ^{2} = 1 ´ 25 ´ (2)^{2} = 50 J
2 2
Final kinetic energy = Initial energy – work done against resistive force (Area between graph and displacement axis)
= 50 – 1 ´ 4 ´ 20 = 50 – 40 = 10 J.
2
Work Done in Conservative and NonConservative Field .
 In conservative field work done by the force (line integral of the force e. òF.dl
path followed between any two points.
WA®B = WA®B = WA®B
) is independent of the
Path I
or ò F.dl
Path I
Path II
= ò F.dl
Path II
Path III
= ò F.dl
Path III
 In conservative field work done by the force (line integral of the force e. òF.dl
) over a closed path/loop is zero.
WA® B + WB® A = 0
or ò F.dl = 0
Conservative force : The forces of these type of fields are known as conservative forces.
Example : Electrostatic forces, gravitational forces, elastic forces, magnetic forces etc and all the central forces are conservative in nature.
If a body of man m lifted to height h from the ground level by different path as shown in the figure
Work done through different paths
WI
WII
WIII
= F. s = mg ´ h = mgh
= F. s = mg sinq ´ l = mg sinq ´ h = mgh
sinq
= mgh1 + 0 + mgh2 + 0 + mgh3 + 0 + mgh4 = mg(h1 + h2 + h3 + h4 ) = mgh
WIV
= ò F. ds
= mgh
It is clear that WI
= WII
= WIII
= WIV
= mgh .
Further if the body is brought back to its initial position A, similar amount of work (energy) is released from the system it means WAB = mgh
and
WBA = –mgh .
Hence the net work done against gravity over a round strip is zero.
WNet
= WAB + WBA
= mgh + (mgh) = 0
i.e. the gravitational force is conservative in nature.
Nonconservative forces : A force is said to be nonconservative if work done by or against the force in moving a body from one position to another, depends on the path followed between these two positions and for complete cycle this work done can never be a zero.
Example: Frictional force, Viscous force, Airdrag etc.
If a body is moved from position A to another position B on a rough table, work done against frictional force shall depends on the length of the path between A and B and not only on the position A and B.
WAB = mmgs
Further if the body is brought back to its initial position A, work has to be done against the frictional force, which always opposes the motion. Hence the net work done against the friction
over a round trip is not zero.
WBA = mmgs.
\WNet
= WAB + WBA = mmgs + mmgs = 2mmgs ¹ 0.
i.e. the friction is a nonconservative force.
Problem 13. If W_{1}, W_{2}
and W3
represent the work done in moving a particle from A to B along three different paths 1, 2
and 3 respectively (as shown) in the gravitational field of a point mass m, find the correct relation
 W1> W2 > W3
 W1= W2 = W3
 W1< W2 < W3
 W2> W1 > W3
Solution : (b) As gravitational field is conservative in nature. So work done in moving a particle from A to B does not depends upon the path followed by the body. It always remains same.
Problem 14. A particle of mass 0.01 kg travels along a curve with velocity given by 4ˆi + 16kˆ ms^{–}^{1}. After some time, its
velocity becomes 8ˆi + 20ˆj ms ^{–}^{1}
this interval of time is
due to the action of a conservative force. The work done on particle during
(a) 0.32 J (b) 6.9 J (c) 9.6 J (d) 0.96 J
Solution : (d)
v1 =
= and v 2 = =
Work done = Increase in kinetic energy = 1 m[v ^{2} – v ^{2} ] = 1 ´ 0.01[464 – 272] = 0.96 J .
2 2 1 2
Work Depends on Frame of Reference.
With change of frame of reference (inertial) force does not change while displacement may change. So the work done by a force will be different in different frames.
Examples : (1) If a porter with a suitcase on his head moves up a staircase, work done by the upward lifting force relative to him will be zero (as displacement relative to him is zero) while relative to a person on the ground will be mgh.
(2) If a person is pushing a box inside a moving train, the work done in
the frame of train will
F.s
while in the frame of earth will be
F.(s +
s _{0} ) where
s _{0} is the displacement of the train relative to the ground.
Energy.
The energy of a body is defined as its capacity for doing work.
 Since energy of a body is the total quantity of work done therefore it is a scalar
 Dimension: [ML^{2}T ^{2} ] it is same as that of work or
 Units : Joule [S.I.], erg [C.G.S.]
Practical units : electron volt (eV), Kilowatt hour (KWh), Calories (Cal) Relation between different units: 1 Joule = 107 erg
1 eV = 1.6 ´ 10 19 Joule
1 KWh = 3.6 ´ 106 Joule
1 Calorie = 4.18 Joule
 Mass energy equivalence : Einstein’s special theory of relativity shows that material particle itself is a form of The relation between the mass of a particle m and its equivalent energy is given as
E = mc 2
If m = 1amu = 1.67 ´ 10 ^{27} kg
then
where c = velocity of light in vacuum.
E = 931 MeV = 1.5 ´ 10 10 Joule .
If m = 1kg
then
E = 9 ´ 1016 Joule
Examples : (i) Annihilation of matter when an electron
(e – )
and a positron
(e + )
combine with each other, they
annihilate or destroy each other. The masses of electron and positron are converted into energy. This energy is released in the form of g rays.
e – + e + ® g + g
Each g photon has energy = 0.51 MeV.
Here two g photons are emitted instead of one g photon to conserve the linear momentum.
 Pair production : This process is the reverse of annihilation of In this case, a photon
(g )
having
energy equal to 1.02 MeV interacts with a nucleus and give rise to electron converted into matter.
(e – ) and positron (e + ) . This energy is
 Nuclear bomb : When the nucleus is split up due to mass defect (The difference in the mass of nucleons and the nucleus) energy is released in the form of g radiations and
 Various forms of energy
(i) Mechanical energy (Kinetic and Potential)  (ii) Chemical energy  (iii) Electrical energy 
(iv) Magnetic energy  (v) Nuclear energy  (vi) Sound energy 
(vii) Light energy  (viii) Heat energy 
 Transformation of energy : Conversion of energy from one form to another is possible through various devices and
Problem 15. A particle of mass ‘m’ and charge ‘q’ is accelerated through a potential difference of ‘V’ volt. Its energy is
[UPSEAT 2001]
 qV (b) mqV (c)
æ q ö
ç m ÷
(d) q

mV
è ø
Solution : (a) Energy of charged particle = charge ´ potential difference = qV
Problem 16. An ice cream has a marked value of 700 kcal. How many kilowatt hour of energy will it deliver to the body as it is digested [AMU (Med.) 2000]
(a) 0.81 kWh (b) 0.90 kWh (c) 1.11 kWh (d) 0.71 kWh
Solution : (a) 700 k cal = 700 ´ 10^{3} ´ 4.2 J = 700 ´ 103 ´ 4.2 = 0.81kWh
3.6 ´ 10^{6}
[As 3.6 ´ 10^{6} J = 1kWh ]
Problem 17. A metallic wire of length L metres extends by l metres when stretched by suspending a weight Mg to it. The mechanical energy stored in the wire is
(a)
2Mgl
(b) Mgl (c)
Mgl (d)
2
Mgl
4
Solution : (c) Elastic potential energy stored in wire U = 1 Fx = Mgl .
2 2
Kinetic Energy.
The energy possessed by a body by virtue of its motion is called kinetic energy.
Examples : (i) Flowing water possesses kinetic energy which is used to run the water mills.
 Moving vehicle possesses kinetic
 Moving air (e. wind) possesses kinetic energy which is used to run wind mills.
 The hammer possesses kinetic energy which is used to drive the nails in
 A bullet fired from the gun has kinetic energy and due to this energy the bullet penetrates into a
(1) Expression for kinetic energy : Let
m = mass of the body, u = Initial velocity of the body (= 0)
F = Force acting on the body, a = Acceleration of the body
s = Distance travelled by the body, v = Final velocity of the body From v 2 = u2 + 2as
Þ v 2 = 0 + 2as
\s = v 2
2a
Since the displacement of the body is in the direction of the applied force, then work done by the force is
W = F ´ s = ma ´ v 2
2a
Þ W = 1 mv2
2
This work done appears as the kinetic energy of the body KE = W = 1 mv2
2
 Calculus method : Let a body is initially at rest and force F is applied on the body to displace it through
ds along its own direction then small work done
dW = F.ds = F ds
Þ dW = ma ds
Þ dW = m dv ds
[As F = ma]
é As a = dv ù
dt
Þ dW = mdv. ds dt
ëê dt úû
Þ dW = mvdv
…….(i)
é As
ds = vù
ëê dt úû
Therefore work done on the body in order to increase its velocity from zero to v is given by
v v év 2 ù ^{v} 1
W = ò mvdv = mò vdv = mê ú = mv^{2}
0 0 ë 2 û 0 2
This work done appears as the kinetic energy of the body KE = 1 mv^{2} .
2
In vector form
As m and v .v
KE = 1 m(v .v ) 2
are always positive, kinetic energy is always positive scalar i.e. kinetic energy can never be negative.
 Kinetic energy depends on frame of reference : The kinetic energy of a person of mass m, sitting in a
train moving with speed v, is zero in the frame of train but
1 mv ^{2} in the frame of the earth.
2
(4) Kinetic energy according to relativity : As we know
E = 1 mv^{2} .
2
But this formula is valid only for (v << c) If v is comparable to c (speed of light in free space = then according to Einstein theory of relativity
3 ´ 10^{8} m / s )
E = mc ^{2}
 mc ^{2}
 Workenergy theorem: From equation (i) dW = mvdv .
Work done on the body in order to increase its velocity from u to v is given by
v v év 2 ù ^{v}
W = ò mvdv = mòu vdv = m ê 2 ú
u
Þ W = 1 m[v ^{2} – u^{2} ] 2
ë û _{u}
Work done = change in kinetic energy
W = DE
This is work energy theorem, it states that work done by a force acting on a body is equal to the change produced in the kinetic energy of the body.
This theorem is valid for a system in presence of all types of forces (external or internal, conservative or non conservative).
If kinetic energy of the body increases, work is positive i.e. body moves in the direction of the force (or field) and if kinetic energy decreases work will be negative and object will move opposite to the force (or field).
Examples : (i) In case of vertical motion of body under gravity when the body is projected up, force of gravity is opposite to motion and so kinetic energy of the body decreases and when it falls down, force of gravity is in the direction of motion so kinetic energy increases.
(ii) When a body moves on a rough horizontal surface, as force of friction acts opposite to motion, kinetic energy will decrease and the decrease in kinetic energy is equal to the work done against friction.
(6) Relation of kinetic energy with linear momentum: As we know
E = 1 mv^{2} = 1 é P ù v ^{2}
[As
P = mv ]
2
\ E = 1 Pv
2
2 êë v úû
or E = P 2 éAs v = P ù
2m
So we can say that kinetic energy
êë
E = 1 mv2 = 1 Pv = p 2
m úû
2 2 2m
and Momentum P = 2E = .
v
From above relation it is clear that a body can not have kinetic energy without having momentum and viceversa.
(7)
Various graphs of kinetic energy
Problem 18. Consider the following two statements
 Linear momentum of a system of particles is zero
 Kinetic energy of a system of particles is zero
Then [AIEEE 2003]
(a) 1 implies 2 and 2 implies 1 (b) 1 does not imply 2 and 2 does not imply 1
(c) 1 implies 2 but 2 does not imply 1 (d) 1 does not imply 2 but 2 implies 1
Solution : (d) Momentum is a vector quantity whereas kinetic energy is a scalar quantity. If the kinetic energy of a system is zero then linear momentum definitely will be zero but if the momentum of a system is zero then kinetic energy may or may not be zero.
Problem 19. A running man has half the kinetic energy of that of a boy of half of his mass. The man speeds up by 1 m/s so as to have same K.E. as that of boy. The original speed of the man will be [Pb. PMT 2001]
(a)
2 m / s
(b)
( – 1)m / s
(c)
1 m / s
(d)
1 m / s
( – 1)
Solution : (c) Let m = mass of the boy, M = mass of the man, v = velocity of the boy and V = velocity of the man
Initial kinetic energy of man = 1 MV ^{2} = 1 é 1 mv ^{2} ù = 1 é 1 æ M ö v ^{2} ù éAs m = M givenù
ê ú ê ç ÷ ú ê ú
Þ V ^{2} = v 2
4
Þ V = v
2
2
…..(i)
2 ë 2
û 2 ë 2 è 2 ø û
ë 2 û
When the man speeds up by 1 m/s ,
1 M(V + 1)^{2} = 1 mv ^{2} = 1 æ M ö v ^{2} Þ (V + 1)^{2} = v 2
ç ÷
2 2 2 2 2
Þ V + 1 = v
è ø
…..(ii)
From (i) and (ii) we get speed of the man V = m / s .
Problem 20. A body of mass 10 kg at rest is acted upon simultaneously by two forces 4N and 3N at right angles to each other. The kinetic energy of the body at the end of 10 sec is [Kerala (Engg.) 2001]
(a) 100 J (b) 300 J (c) 50 J (d) 125 J
Solution : (d) As the forces are working at right angle to each other therefore net force on the body F = = 5 N
Kinetic energy of the body = work done = F ´ s = F ´ 1 a t ^{2} = F ´ 1 æ F ö t ^{2} = 5 ´ 1 æ 5 ö(10)^{2} = 125 J.
2 2 ç m ÷ ç ÷
è ø 2 è 10 ø
Problem 21. If the momentum of a body increases by 0.01%, its kinetic energy will increase by [MP PET 2001]
(a) 0.01% (b) 0.02 % (c) 0.04 % (d) 0.08 %
Solution : (b) Kinetic energy
E = P 2
2m
\ E µ P ^{2}
Percentage increase in kinetic energy = 2(% increase in momentum) [If change is very small]
= 2(0.01%) = 0.02%.
Problem 22. If the momentum of a body is increased by 100 %, then the percentage increase in the kinetic energy is
[NCERT 1990; BHU 1999; Pb. PMT 1999; CPMT 1999, 2000; CBSE PMT 2001]
(a) 150 % (b) 200 % (c) 225 % (d) 300 %
P ^{2} E æ P ö 2 æ 2P ö ^{2}
Solution : (d) E =
Þ 2 = ç 2 ÷ = ç
÷ = 4
2m E_{1}
è P1 ø
è P ø
E2 = 4 E1 = E1 + 3E1 = E1 + 300 % of E1 .
Problem 23. A body of mass 5 kg is moving with a momentum of 10 kgm/s. A force of 0.2 N acts on it in the direction of motion of the body for 10 seconds. The increase in its kinetic energy is [MP PET 1999]
(a) 2.8 J (b) 3.2 J (c) 3.8 J (d) 4.4 J
Solution : (d) Change in momentum = P2 – P1 = F ´ t Þ P2 = P1 + F ´ t = 10 + 0.2 ´ 10 = 12kg–m / s
Increase in kinetic energy E =
1 [P ^{2} – P ^{2} ] = 1 [(12)^{2} – (10)^{2} ] =
1 [144 – 100] = 44 = 4.4 J.
2m ^{2} ^{1} 2m
2 ´ 5 10
Problem 24. Two masses of 1g and 9g are moving with equal kinetic energies. The ratio of the magnitudes of their respective linear momenta is [CBSE PMT 1993; CPMT 1995]
(a) 1 : 9 (b) 9 : 1 (c) 1 : 3 (d) 3 : 1
Solution : (c) P =
\ P µ if E = constant . So
P1 =
P2
= = 1 .
3
Problem 25. A body of mass 2 kg is thrown upward with an energy 490 J. The height at which its kinetic energy would become half of its initial kinetic energy will be [ g = 9.8 m / s ^{2} ]
(a) 35 m (b) 25 m (c) 12.5 m (d) 10 m
Solution : (c) If the kinetic energy would become half, then Potential energy =
Þ mgh = 1 [490] Þ 2 ´ 9.8 ´ h = 1 [490] Þ h = 12.5 m
1 (Initial kinetic energy)
2
2 2
Problem 26. A 300 g mass has a velocity of (3ˆi + 4ˆj) m/sec at a certain instant. What is its kinetic energy (a) 1.35 J (b) 2.4 J (c) 3.75 J (d) 7.35 J
Solution : (c)
r = (3ˆi + 4ˆj)
\ v =
= 5 m/s . So kinetic energy = 1 mv ^{2} = 1 ´ 0.3 ´ (5)^{2} = 3.75 J
v 2 2
Stopping of Vehicle by Retarding Force.
If a vehicle moves with some initial velocity and due to some retarding force it stops after covering some distance after some time.
 Stopping distance : Let m = Mass of vehicle, v
= Velocity, P = Momentum, E = Kinetic energy
F = Stopping force, x =
Stopping distance, t = Stopping time
Then, in this process stopping force does work on the vehicle and destroy the motion.
By the work energy theorem
W = DK = 1 mv2
2
Þ Stopping force (F) ´ Distance (x) = Kinetic energy (E)
Þ Stopping distance (x) = Kinetic energy(E)
Stopping force(F)
Þ x = mv2
2F
…..(i)
 Stopping time : By the impulsemomentum theorem
F ´ t = DP Þ F ´ t = P
\ t = P
F
or t = mv
F
…..(ii)
 Comparison of stopping distance and time for two vehicles : Two vehicles of masses m_{1} and m_{2}
are moving with velocities v_{1} and v_{2} respectively. When they are stopped by the same retarding force (F).
The ratio of their stopping distances
x1 = E1
x 2 E2
m v^{2}
= 1 1
m v^{2}
and the ratio of their stopping time
t1 = P1 t 2 P2
2 2
= m1v1
m2v2
t1 = P1 = 2m_{1} E_{1} =  
m1  
t 2 P2 2m_{2} E_{2}  m2 
Note : @If vehicle is stopped by friction then
1 mv^{2} 1 mv^{2} v 2
Stopping distance
x = 2 = 2 =
[As a = mg]
F ma 2mg
Stopping time
t = mv = mv = v
F mm g m g
Problem 27. Two carts on horizontal straight rails are pushed apart by an explosion of a powder charge Q placed between the carts. Suppose the coefficients of friction between the carts and rails are identical. If the 200 kg cart travels a distance of 36 metres and stops, the distance covered by the cart weighing 300 kg is [CPMT 1989]
 32 metres
 24 metres
 16 metres
 12 metres
Solution : (c) Kinetic energy of cart will goes against friction. \ E = P 2
2m
= m mg ´ s
Þ s =
P 2
2m gm ^{2}
As the two carts pushed apart by an explosion therefore they possess same linear momentum and coefficient of friction is same for both carts (given). Therefore the distance covered by the cart before coming to rest is given by
1 s æ m ö 2 æ 200 ö ^{2} 4 4
s µ \ ^{ } ^{2} = ç 1 ÷ = ç ÷ = Þ S_{2} =
 36 = 16 metres .
m2 s1
è m2 ø
è 300 ø 9 9
Problem 28. An unloaded bus and a loaded bus are both moving with the same kinetic energy. The mass of the latter is
twice that of the former. Brakes are applied to both, so as to exert equal retarding force. If distance covered by the two buses respectively before coming to a stop, then
x1 and
x _{2} be the
(a)
x1 = x 2
(b)
2x1 = x 2
(c)
4 x1 = x2
(d)
8 x1 = x 2
Solution : (a) If the vehicle stops by retarding force then the ratio of stopping distance
x1 =
x 2
E1 .
E2
But in the given problem kinetic energy of bus and car are given same i.e. E_{1} = E_{2}. \ x_{1} = x_{2} .
Problem 29. A bus can be stopped by applying a retarding force F when it is moving with a speed v on a level road. The distance covered by it before coming to rest is s. If the load of the bus increases by 50 % because of passengers, for the same speed and same retarding force, the distance covered by the bus to come to rest shall be
(a) 1.5 s (b) 2 s (c) 1 s (d) 2.5 s
Solution : (a) Retarding force (F) ´ distance covered (x) = Kinetic energy æ 1 mv ^{2} ö
If v and F are constants then x µ m \
x 2 = m2 x1 m1
ç
è
= 1.5 m = 1.5
m
÷

ø
Þ x_{2}
= 1.5 s.
Problem 30. A vehicle is moving on a rough horizontal road with velocity v. The stopping distance will be directly proportional to
(a) (b) v (c) v ^{2} (d) v ^{3}
Solution : (c) As s = v 2
2a
\ s µ v ^{2} .
Potential Energy.
Potential energy is defined only for conservative forces. In the space occupied by conservative forces every point is associated with certain energy which is called the energy of position or potential energy. Potential energy generally are of three types : Elastic potential energy, Electric potential energy and Gravitational potential energy etc.
 Change in potential energy : Change in potential energy between any two points is defined in the terms of the work done by the associated conservative force in displacing the particle between these two points without any change in kinetic
_{r} r
U – U = ò 2
r = –W
……(i)
2 1 r1
F.dr
We can define a unique value of potential energy only by assigning some arbitrary value to a fixed point called the reference point. Whenever and wherever possible, we take the reference point at infinite and assume potential
energy to be zero there, i.e. if take r1 = ¥
and r2 = r
then from equation (i)

_{r} r

U = ò
r = –W
In case of conservative force (field) potential energy is equal to negative of work done in shifting the body from reference position to given position.
This is why in shifting a particle in a conservative field (say gravitational or electric), if the particle moves opposite to the field, work done by the field will be negative and so change in potential energy will be positive i.e. potential energy will increase. When the particle moves in the direction of field, work will be positive and change in potential energy will be negative i.e. potential energy will decrease.

 Three dimensional formula for potential energy: For only conservative fields F equals the negative gradient (Ñ) of the potential
So F = ÑU
( Ñ read as Del operator or Nabla operator and
r = d ˆi + d ˆj + d kˆ )
Þ r é dU ˆ dU ˆ dU ˆù
dx dy dz


F = ê dx i + dy j + dz kú
where
dU = Partial derivative of U w.r.t. x (keeping y and z constant)
dx
dU = Partial derivative of U w.r.t. y (keeping x and z constant)
dy
dU = Partial derivative of U w.r.t. z (keeping x and y constant)
dz
 Potential energy curve : A graph plotted between the potential energy of a particle and its displacement from the centre of force is called potential energy
Figure shows a graph of potential energy function U(x) for one dimensional motion.
As we know that negative gradient of the potential energy gives force.
dU
\ – dx = F
(4) Nature of force :
 Attractive force : On increasing x, if U increases
dU = positive
dx
then F is negative in direction i.e. force is attractive in nature. In graph this is represented in region BC.
 Repulsive force : On increasing x, if U decreases
dU = negative
dx
then F is positive in direction i.e. force is repulsive in nature. In graph this is represented in region AB.
 Zero force : On increasing x, if U does not changes
dU = 0
dx
then F is zero i.e. no force works on the particle. Point B, C and D represents the point of zero force or these points can be termed as position of equilibrium.
 Types of equilibrium : If net force acting on a particle is zero, it is said to be in
For equilibrium
dU = 0 , but the equilibrium of particle can be of three types :
dx
Stable  Unstable  Neutral 
When a particle is displaced slightly from a position, then a force acting on it brings it back to the initial position, it is said to be in stable equilibrium position.
Potential energy is minimum. F = – dU = 0 dx d 2U = positive dx ^{2} i.e. rate of change of dU is positive. dx Example :
A marble placed at the bottom of a hemispherical bowl.  When a particle is displaced slightly from a position, then a force acting on it tries to displace the particle further away from the equilibrium position, it is said to be in unstable equilibrium. Potential energy is maximum. F = – dU = 0 dx d 2U = negative dx ^{2} i.e. rate of change of dU is negative. dx Example :
A marble balanced on top of a hemispherical bowl.  When a particle is slightly displaced from a position then it does not experience any force acting on it and continues to be in equilibrium in the displaced position, it is said to be in neutral equilibrium. Potential energy is constant. F = – dU = 0 dx d ^{2}U = 0 dx ^{2} i.e. rate of change of dU is zero. dx Example : A marble placed on horizontal table. 
Problem 31. A particle which is constrained to move along the xaxis, is subjected to a force in the same direction which varies
with the distance x of the particle from the origin as
F(x) = –kx + ax ^{3} . Here k and a are positive constants. For
x ³ 0 , the functional from of the potential energy U(x) of the particle is
U(x)
(a)
(b)
U(x)
(c)
U(x)
(d)
U(x)
Solution : (d)
2 ax 4

F = – dU Þ dU = –F.dx Þ U = ò x –kx + ax ^{3} ) dx Þ U = kx –
dx
\ We get U = 0
0 2 4
at x = 0 and x =
Also we get U = negative for x >
From the given function we can see that F = 0 at x = 0 i.e. slope of U–x graph is zero at x = 0.
Problem 32. The potential energy of a body is given by
A – Bx ^{2}
(where x is the displacement). The magnitude of force
acting on the particle is [BHU 2002]
(a) Constant (b) Proportional to x
Solution : (b)
 Proportional to x ^{2}
F = –dU = – d (A – Bx ^{2} ) = 2Bx
\ F µ x .
 Inversely proportional to x
dx dx
Problem 33. The potential energy of a system is represented in the first figure. The force acting on the system will be represented by
U(x)
(a)
F(x)
(b)
O
F(x)
a
(c)
x

F(x)
(d) F(x)
Solution : (c) As slope of problem graph is positive and constant upto distance a then it becomes zero. Therefore from
F = – dU
dx
we can say that upto distance a force will be constant (negative) and suddenly it becomes zero.
Problem 34. A particle moves in a potential region given by U = 8 x ^{2} – 4 x + 400
 J. Its state of equilibrium will be
(a)
x = 25 m
(b)
x = 0.25 m
(c)
x = 0.025 m
(d)
x = 2.5 m
Solution : (b)
F = – dU = –
dx
d (8 x ^{2} – 4 x + 400)
dx
For the equilibrium condition F = – dU = 0 Þ 16x – 4 = 0 Þ x = 4 / 16 \ x = 0.25 m .
dx
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