Chapter 10 Work, Energy and Power Part 2 – Physics free study material by TEACHING CARE online tuition and coaching classes
Elastic Potential Energy.
(1) Restoring force and spring constant : When a spring is stretched or compressed from its normal position (x = 0) by a small distance x, then a restoring force is produced in the spring to bring it to the normal position.
According to Hooke’s law this restoring force is proportional to the displacement x and its direction is always opposite to the displacement.
i.e.
or
F µ –x
F = – k x
…..(i)
where k is called spring constant. If x = 1, F = k (Numerically)
or k = F
Hence spring constant is numerically equal to force required to produce unit displacement (compression or extension) in the spring. If required force is more, then spring is said to be more stiff and viceversa.
Actually k is a measure of the stiffness/softness of the spring.
Dimension : As k = F \ [k] = [F] = [MLT 2 ] = [MT ^{2} ]
x [x] L
Units : S.I. unit Newton/metre, C.G.S unit Dyne/cm.
Note : @Dimension of force constant is similar to surface tension.
 Expression for elastic potential energy : When a spring is stretched or compressed from its normal
position (x = 0), work has to be done by external force against restoring force.
F ext
= F restoring = kx
Let the spring is further stretched through the distance dx, then work done
dW = F ext . dx = F_{ext} . dx cos 0^{o} = kx dx
[As cos 0^{o} = 1]
Therefore total work done to stretch the spring through a distance x from its mean position is given by
x x é x 2 ù ^{x} 1
W = ò dW = ò kx dx = k ê ú = kx ^{2}
0 0 ë 2 û 0 2
This work done is stored as the potential energy of the stretched spring.
\ Elastic potential energy U = 1 kx ^{2}
2
U = 1 Fx éAs k = F ù
2 êë x úû
U = F 2 éAs x = F ù
2k
\ Elastic potential energy U = 1 kx ^{2} = 1 Fx = F 2
ëê k úû
2 2 2k
Note : @If spring is stretched from initial position
x_{1} to final position
x _{2} then work done
= Increment in elastic potential energy = 1 k(x ^{2} – x ^{2} )
2 2 1
(3) Energy graph for a spring : If the mass attached with spring performs simple harmonic motion about its mean position then its potential energy at any position (x) can be given by
U = 1 kx ^{2}
2
So for the extreme position
1
….(i)
2
U = 2 ka
[As x = ± a for extreme]
This is maximum potential energy or the total energy of mass.
\ Total energy
E = 1 ka ^{2}
2
….(ii)
[Because velocity of mass = 0 at extreme \ K = 1 mv^{2} = 0 ]
2
Now kinetic energy at any position
K = E – U = 1 k a ^{2} – 1 k x ^{2}
K = 1 k(a ^{2} – x ^{2} )
2
2 2
….(iii)
From the above formula we can check that
U = 1 ka ^{2}
max 2
K = 1 ka ^{2}
max 2
[At extreme x = ± a] and
[At mean x = 0] and
U min = 0
K min = 0
[At mean x = 0]
[At extreme x = ± a]
E = 1 ka ^{2} =
2
constant (at all positions)
It mean kinetic energy changes parabolically w.r.t. position but total energy remain always constant irrespective to position of the mass
Problem 35. A long spring is stretched by 2 cm, its potential energy is U. If the spring is stretched by 10 cm, the potential energy stored in it will be [CPMT 1976, 86, 96; MP PMT 2002; CBSE PMT 2003]
(a) U / 25 (b) U / 5 (c) 5 U (d) 25 U
Solution : (d) Elastic potential energy of a spring U = 1 kx ^{2}
2
\ U µ x ^{2}
U æ x ö 2 U æ 10 cm ö 2
So ^{ } ^{2} = ç^{ } ^{2} ÷
Þ ^{ } ^{2} = ç ÷ Þ U 2 = 25 U
U1 è x1 ø
U è 2 cm ø
Problem 36. A spring of spring constant 5 ´ 10^{3} N / m is stretched initially by 5 cm from the unstretched position. Then the work required to stretch it further by another 5 cm is [AIEEE 2003] (a) 6.25 Nm (b) 12.50 Nm (c) 18.75 Nm (d) 25.00 Nm
Solution : (c) Work done to stretch the spring from x_{1} to x _{2}
W = 1 k(x ^{2} – x ^{2} ) = 1 5 ´ 10^{3} [(10 ´ 10 ^{2} )^{2} – (5 ´ 10 ^{2} )^{2} ] = 1 ´ 5 ´ 10^{3} ´ 75 ´ 10 ^{4}
= 18.75 N.m .
2 ^{2} ^{1} 2 2
Problem 37. Two springs of spring constants 1500 N / m
and
3000 N / m
respectively are stretched with the same force.
They will have potential energy in the ratio [MP PET/PMT 1998; Pb. PMT 2002]
(a) 4 : 1 (b) 1 : 4 (c) 2 : 1 (d) 1 : 2
Solution : (c) Potential energy of spring U = F 2
2k
Þ U1 U 2
= k 2
k1
= 3000 = 2 : 1
1500
[If F = constant]
Problem 38. A body is attached to the lower end of a vertical spiral spring and it is gradually lowered to its equilibrium position. This stretches the spring by a length x. If the same body attached to the same spring is allowed to fall suddenly, what would be the maximum stretching in this case
(a) x (b) 2x (c) 3x (d) x/2 Solution : (b) When spring is gradually lowered to it’s equilibrium position
kx = mg \ x = mg .
k
When spring is allowed to fall suddenly it oscillates about it’s mean position
Let y is the amplitude of vibration then at lower extreme, by the conservation of energy
Þ 1 ky ^{2} = mgy Þ y = 2mg = 2x.
2 k
Problem 39. Two equal masses are attached to the two ends of a spring of spring constant k. The masses are pulled out symmetrically to stretch the spring by a length x over its natural length. The work done by the spring on each mass is
(a)
1 kx ^{2}
2
(b)
 1kx ^{2}
2
(c)
1 kx ^{2}
4
(d)
 1kx ^{2}
4
Solution : (d) If the spring is stretched by length x, then work done by two equal masses =
1 kx ^{2}
2
So work done by each mass on the spring =
Electrical Potential Energy.
1 kx ^{2}
4
\ Work done by spring on each mass = – 1 kx ^{2} .
4
It is the energy associated with state of separation between charged particles that interact via electric force. For two point charge q_{1} and q_{2} , separated by distance r.
U = 1 . q_{1}q_{2}
4pe _{0} r
While for a point charge q at a point in an electric field where the potential is V U = qV
As charge can be positive or negative, electric potential energy can be positive or negative.
Problem 40. A proton has a positive charge. If two protons are brought near to one another, the potential energy of the system will
 Increase (b) Decrease
 Remain the same (d) Equal to the kinetic energy
Solution : (a) As the force is repulsive in nature between two protons. Therefore potential energy of the system increases.
Problem 41. Two protons are situated at a distance of 100 fermi from each other. The potential energy of this system will be in eV
(a) 1.44 (b)
1.44 ´ 10^{3}
(c)
1.44 ´ 10 ^{2}
(d)
1.44 ´ 10 ^{4}
Solution : (d)
U = 1 q_{1}q_{2} = 9 ´ 10^{9} ´ (1.6 ´ 10 ^{19} )^{2}
= 2.304 ´ 10 ^{–}^{15} J = 2.304 ´ 10 15 eV = 1.44 ´ 10^{4} eV
4pe _{0} r
100 ´ 10 ^{15}
1.6 ´ 10 ^{19}
Problem 42.
80 Hg 208
nucleus is bombarded by a particles with velocity 10^{7} m/s. If the a particle is approaching the Hg
nucleus headon then the distance of closest approach will be
(a)
1.115 ´ 10^{13} m
(b)
11.15 ´ 10^{13} m
(c)
111.5 ´ 10^{13} m
 Zero
Solution : (a) When a particle moves towards the mercury nucleus its kinetic energy gets converted in potential energy of
the system. At the distance of closest approach
1 mv ^{2} =
2
1
4pe _{0}
q1q2
r
Þ 1 ´ (1.6 ´ 10 ^{27} )(10^{7} )^{2} = 9 ´ 10^{9} (2.e)(80 e) Þ r = 1.115 ´ 10 ^{13} m.
2 r
Problem 43. A charged particle A moves directly towards another charged particle B. For the momentum is P and the total energy is E
 P and E are conserved if both A and B are free to move
 (a) is true only if A and B have similar charges
 If B is fixed, E is conserved but not P
 If B is fixed, neither E nor P is conserved
(A + B) system, the total
Solution : (a, c) If A and B are free to move, no external forces are acting and hence P and E both are conserved but when B
is fixed (with the help of an external force) then E is conserved but P is not conserved.
Gravitational Potential Energy.
It is the usual form of potential energy and is the energy associated with the state of separation between two bodies that interact via gravitational force.
For two particles of masses m_{1} and m_{2} separated by a distance r
Gravitational potential energy U = – Gm1m2
r
(1) If a body of mass m at height h relative to surface of earth then
Gravitational potential energy U = mgh
1 + h
R
Where R = radius of earth, g = acceleration due to gravity at the surface of the earth.
 If h << R then above formula reduces to U = mgh.
(3) If V is the gravitational potential at a point, the potential energy of a particle of mass m at that point will be
U = mV
(4) Energy height graph : When a body projected vertically upward from the ground level with some initial velocity then it possess kinetic energy but its potential energy is zero.
As the body moves upward its potential energy increases due to increase in height but kinetic energy decreases (due to decrease in velocity). At maximum height its kinetic energy becomes zero and potential energy maximum but through out the complete motion total energy remains constant as shown in the figure.
Problem 44. The work done in pulling up a block of wood weighing 2kN for a length of 10 m on a smooth plane inclined at an angle of 15 ^{o} with the horizontal is (sin 15^{o} = 0.259) [AFMC 1999] (a) 4.36 k J (b) 5.17 k J (c) 8.91 k J (d) 9.82 k J
Solution : (b) Work done = mg ´ h
= 2 ´ 10^{3} ´ l sinq
= 2 ´ 10^{3} ´ 10 ´ sin 15 ^{o} = 5176 J = 5.17 kJ
Problem 45. Two identical cylindrical vessels with their bases at same level each contains a liquid of density d. The height of the liquid in one vessel is h_{1} and that in the other vessel is h_{2} . The area of either vases is A. The work done by gravity in equalizing the levels when the two vessels are connected, is [SCRA 1996]
(a)
(h1 – h2 ) gd
(b)
(h_{1} – h_{2} ) gAd
(c)
1 (h1 – h2 )^{2} gAd
2
(d)
1 (h1 – h2 )^{2} gAd
4
Solution : (d) Potential energy of liquid column is given by mg h = Vdg h = Ahdg h = 1 Adgh^{2}
2 2 2 2
Initial potential energy = 1 Adgh^{2} + 1 Adgh^{2}
2 1 2 2
Final potential energy =
1 Adgh^{2} + 1 Adh^{2}g = Adgh^{2}
2 2
Work done by gravity = change in potential energy
W = é 1 Adgh^{2} + 1 Adgh^{2} ù – Adgh^{2}
êë 2
é h2
1 2
h2 ù
2 úû
æ h_{1} + h_{2} ö ^{2}
h1 + h2
= Adg ê^{ } ^{1} + ^{ } ^{2} ú – Adgç
÷ [As h = ]
êë 2
2 úû
è 2 ø 2





é h^{2} h^{2} æ h^{2} + h^{2} + 2h_{1}h_{2} öù Adg _{2}
= Adg ê ^{1} + ^{ } ^{2} – ç 1 2 ÷ú = (h1 – h2 )
êë 2 ç 4 ÷
Problem 46. If g is the acceleration due to gravity on the earth’s surface, the gain in the potential energy of an abject of mass m raised from the surface of earth to a height equal to the radius of the earth R, is [IITJEE1983]
(a)
1 mgR
2
(b)
2mgR
(c) mgR (d)
1 mgR
4
Solution : (a) Work done = gain in potential energy =
mgh
1 + h / R
= mgR
1 + R / R
= 1 mgR
2
[As h = R (given)]
Problem 47. The work done in raising a mass of 15 gm from the ground to a table of 1m height is
(a) 15 J (b) 152 J (c) 1500 J (d) 0.15 J
Solution : (d) W = mgh = 15 ´ 10 ^{–}^{3} ´ 10 ´ 1 = 0.15 J.
Problem 48. A body is falling under gravity. When it loses a gravitational potential energy by U, its speed is v. The mass of the body shall be
 2Uv
 U
2v
2U (d)
v 2
U
2v ^{2}
Solution : (c) Loss in potential energy = gain in kinetic energy Þ U = 1 mv ^{2} \ m = 2U .
2 v 2
Problem 49. A liquid of density d is pumped by a pump P from situation (i) to situation (ii) as shown in the diagram. If the crosssection of each of the vessels is a, then the work done in pumping (neglecting friction effects) is
 2dgh
 dgha
 2dgh^{2}a
 dgh^{2}a
Solution : (d) Potential energy of liquid column in first situation = Vdg h + Vdg h
= Vdgh = ahdgh
= dgh^{2}a
2 2
[As centre of mass of liquid column lies at height h ]
2

Potential energy of the liquid column in second situation = Vdg æ 2h ö = (A ´ 2h)dgh = 2dgh^{2}a
2
è ø
Work done pumping = Change in potential energy = 2dgh^{2}a – dgh^{2}a = dgh^{2}a .
Problem 50. The mass of a bucket containing water is 10 kg. What is the work done in pulling up the bucket from a well of depth 10 m if water is pouring out at a uniform rate from a hole in it and there is loss of 2kg of water from it while it reaches the top (g = 10 m / sec ^{2} )
(a) 1000 J (b) 800 J (c) 900 J (d) 500 J
Solution : (c) Gravitational force on bucket at starting position = mg = 10 ´ 10 = 100 N
Gravitational force on bucket at final position = 8 ´ 10 = 80 N
So the average force through out the vertical motion = 100 + 80 = 90 N
2
\ Work done = Force ´ displacement = 90 ´ 10 = 900 J.
Problem 51. A rod of mass m and length l is lying on a horizontal table. The work done in making it stand on one end will be
(a) mgl (b)
mgl
2
(c)
mgl (d) 2mgl
4
Solution : (b) When the rod is lying on a horizontal table, its potential energy = 0
But when we make its stand vertical its centre of mass rises upto high
l . So it’s potential energy = mgl
\ Work done = charge in potential energy = mg l
2
2 2
 0 = mgl .
2
Problem 52. A metre stick, of mass 400 g, is pivoted at one end displaced through an angle potential energy is
60 ^{o} . The increase in its
 1 J
 10 J
 100 J
(d) 1000 J
Solution : (a) Centre of mass of a stick lies at the mid point and when the stick is displaced through an angle 60^{o} it rises upto height ‘h’ from the initial position.
From the figure h = l
2
 l cosq 2
= l (1 – cosq ) 2
Hence the increment in potential energy of the stick = mgh
= mg l (1 – cosq ) = 0.4 ´ 10 ´ 1 (1 – cos 60 ^{o} ) = 1 J
2 2
Problem 53. Once a choice is made regarding zero potential energy reference state, the changes in potential energy
 Are same
 Are different
 Depend strictly on the choice of the zero of potential energy
 Become indeterminate
Solution : (a) Potential energy is a relative term but the difference in potential energy is absolute term. If reference level is fixed once then change in potential energy are same always.
Work Done in Pulling the Chain Against Gravity.
A chain of length L and mass M is held on a frictionless table with (1/n)^{th} of its length hanging over the edge.
Let
m = M =
L
mass per unit length of the chain and y is the length of the
chain hanging over the edge. So the mass of the chain of length y will be ym and the force acting on it due to gravity will be mgy.
The work done in pulling the dy length of the chain on the table.
dW = F(– dy) [As y is decreasing]
i.e. dW = mgy (– dy)
So the work done in pulling the hanging portion on the table.

0 é y 2 ù ^{0}
W = – L / n mgy dy = mg ê 2 ú
= mg L^{2}
2n^{2}
ë û L / n
\ W = MgL
2n^{2}
[As m = M/L]
Alternative method :
If point mass m is pulled through a height h then work done W = mgh
Similarly for a chain we can consider its centre of mass at the middle point of the hanging part i.e. at a height
of L/(2n) from the lower end and mass of the hanging part of chain = M
n
So work done to raise the centre of mass of the chain on the table is given by
W = M ´ g ´ L
n 2n
or W = MgL
2n^{2}
[As W = mgh]
Velocity of Chain While Leaving the Table.
Taking surface of table as a reference level (zero potential energy)
Potential energy of chain when 1/n^{th} length hanging from the edge = – MgL
2n^{2}
Potential energy of chain when it leaves the table = – MgL
2
Kinetic energy of chain = loss in potential energy
Þ 1 Mv2 = MgL – MgL
2 2 2n^{2}
Þ 1 Mv 2 = MgL é1 – 1 ù
2
\ Velocity of chain v =
2 êë
n2 úû
Problem 54. A uniform chain of length L and mass M is lying on a smooth table and one third of its length is hanging vertically down over the edge of the table. If g is acceleration due to gravity, the work required to pull the hanging part on to the table is [IITJEE 1985; MNR 1990; MP PMT 1994, 97, 2000; JIMPER 2000; AIEEE 2002]
(a)
MgL
(b)
MgL (c)
3
MgL (d)
9
MgL
18
Solution : (d) As 1/3 part of the chain is hanging from the edge of the table. So by substituting n = 3 in standard expression
W = MgL = MgL = MgL .
2n ^{2} 2(3)^{2} 18
Problem 55. A chain is placed on a frictionless table with one fourth of it hanging over the edge. If the length of the chain is 2m and its mass is 4kg, the energy need to be spent to pull it back to the table is
(a) 32 J (b) 16 J (c) 10 J (d) 2.5 J
Solution : (d)
W = MgL = 4 ´ 10 ´ 2 = 2.5 J.
2n^{2} 2 ´ (4)^{2}
Problem 56. A uniform chain of length 2m is held on a smooth horizontal table so that half of it hangs over the edge. If it is released from rest, the velocity with which it leaves the table will be nearest to
(a) 2 m/s (b) 4 m/s (c) 6 m/s (d) 8 m/s
Solution : (b) v = =
= = 3.87 ≃ 4 m/s (approx.)
Law of Conservation of Energy.
(1) Law of conservation of energy
For a body or an isolated system by workenergy theorem we have K
r




– K = ò
…..(i)
But according to definition of potential energy in a conservative field U
r

– U = ò
…..(ii)
So from equation (i) and (ii) we have
K2 – K1 = (U2 – U1 )
_{2} _{1} F. dr
or K2 + U2 = K1 + U1
i.e. K + U = constant.
For an isolated system or body in presence of conservative forces the sum of kinetic and potential energies at any point remains constant throughout the motion. It does not depends upon time. This is known as the law of conservation of mechanical energy.
D(K + U) = DE = 0
\ DK + DU = 0
[As E is constant in a conservative field]
i.e. if the kinetic energy of the body increases its potential energy will decrease by an equal amount and viceversa.
(2) Law of conservation of total energy : If some nonconservative force like friction is also acting on the particle, the mechanical energy is no more constant. It changes by the amount of work done by the frictional force.
D(K + U) = DE = Wf
[where Wf
is the work done against friction]
The lost energy is transformed into heat and the heat energy developed is exactly equal to loss in mechanical energy. We can, therefore, write DE + Q = 0 [where Q is the heat produced]
This shows that if the forces are conservative and nonconservative both, it is not the mechanical energy alone
which is conserved, but it is the total energy, may be heat, light, sound or mechanical etc., which is conserved.
In other words : “Energy may be transformed from one kind to another but it cannot be created or destroyed.
The total energy in an isolated system is constant“. This is the law of conservation of energy.
Problem 57. Two stones each of mass 5kg fall on a wheel from a height of 10m. The wheel stirs 2kg water. The rise in temperature of water would be [RPET 1997]
(a) 2.6° C (b) 1.2° C (c) 0.32° C (d) 0.12° C
Solution : (d) For the given condition potential energy of the two masses will convert into heat and temperature of water will
increase W = JQ Þ 2m ´ g ´ h = J(m_{w} S Dt) Þ
\ Dt = 1000 = 0.119 ^{o} C = 0.12^{o} C .
8.4 ´ 10^{3}
2 ´ 5 ´ 10 ´ 10 = 4.2(2 ´ 10^{3} ´ Dt)
Problem 58. A boy is sitting on a swing at a maximum height of 5m above the ground. When the swing passes through the mean position which is 2m above the ground its velocity is approximately [MP PET 1990]
 6 m/s (b) 9.8 m/s (c) 6.26 m/s (d) None of these
Solution : (a) By the conservation of energy Total energy at point A = Total energy at point B
Þ mgh1 = mgh2 + 1 mv ^{2}
2
Þ 9.8 ´ 5 = 9.8 ´ 2 + 1 v ^{2}
2
Þ v ^{2} = 58.8 \ v = 7.6 m / s
Problem 59. A block of mass M slides along the sides of a bowl as shown in the figure. The walls of the bowl are frictionless and the base has coefficient of friction 0.2. If the block is released from the top of the side, which is 1.5 m high, where will the block come to rest ? Given that the length of the base is 15 m
 1 m from P
 Mid point
 2 m from P
 At Q
Solution : (b) Potential energy of block at starting point = Kinetic energy at point P = Work done against friction in traveling a distance s from point P.
\ mgh = m mgs Þ s = h = 1.5 = 7.5 m
m 0.2
i.e. block come to rest at the mid point between P and Q.
Problem 60. If we throw a body upwards with velocity of initial value ? Take g = 10 m /s ^{2}
4 ms ^{–}^{1}
at what height its kinetic energy reduces to half of the
 4m (b) 2 m (c) 1 m (d) None of these
Solution : (d) We know kinetic energy K = 1 mv ^{2}
2
\ v µ
When kinetic energy of the body reduces to half its velocity becomes v = u =
= 2 2 m/s
From the equation v ^{2} = u ^{2} – 2gh Þ (2
2)^{2} = (4)^{2} – 2 ´ 10 h
\ h = 16 – 8 = 0.4 m .
20
Problem 61. A 2kg block is dropped from a height of 0.4 m on a spring of force constant compression of the spring is
K = 1960 Nm^{–}^{1} . The maximum
(a) 0.1 m (b) 0.2 m (c) 0.3 m (d) 0.4 m
Solution : (a) When a block is dropped from a height, its potential energy gets converted into kinetic energy and finally spring get compressed due to this energy.
\ Gravitational potential energy of block = Elastic potential energy of spring
Þ mgh = 1 Kx ^{2}
2
Þ x = =
= 0.09 m ~– 0.1m .
Problem 62. A block of mass 2kg is released from A on the track that is one quadrant of a circle of radius 1m. It slides down
the track and reaches B with a speed of done against the force of friction is
4 ms ^{–}^{1}
and finally stops at C at a distance of 3m from B. The work
 10 J
 20 J
 2 J
 6 J
Solution : (b) Block possess potential energy at point A = mgh = 2 ´ 10 ´ 1 = 20 J
Finally block stops at point C. So its total energy goes against friction i.e. work done against friction is 20 J.
Problem 63. A stone projected vertically upwards from the ground reaches a maximum height h. When it is at a height
3h , the ratio of its kinetic and potential energies is
4
(a) 3 : 4 (b) 1 : 3 (c) 4 : 3 (d) 3 : 1
Solution : (b) At the maximum height, Total energy = Potential energy = mgh
At the height
3h , Potential energy = mg 3h = 3 mgh
4 4 4
and Kinetic energy = Total energy – Potential energy = mgh – 3 mgh = 1 mgh
4 4
Power.
\ Kinetic energy Potential energy
= 1 .
3
Power of a body is defined as the rate at which the body can do the work.
Average power (P
) = DW = W
 Dt t
Instantaneous power (P ) = dW
r
= F. ds
[As dW = r ]
inst. dt dt
P = r
[As

r = ds ]
inst
F.v
v dt
i.e. power is equal to the scalar product of force with velocity.
 Dimension :
\
[P] = [F][v] = [MLT ^{2} ][LT ^{1} ]
[P] = [ML^{2}T ^{3} ]
 Units : Watt or Joule/sec [S.I.]
Erg/sec [C.G.S.]
Practical units : Kilowatt (kW), Mega watt (MW) and Horse power (hp) Relations between different units : 1watt = 1 Joule / sec = 10^{7} erg / sec
1hp = 746 Watt
1 MW = 10^{6} Watt
1kW = 10^{3} Watt
(3) If work done by the two bodies is same then power µ 1
time
i.e. the body which perform the given work in lesser time possess more power and viceversa.
 As power = work/time, any unit of power multiplied by a unit of time gives unit of work (or energy) and not power, e. Kilowatthour or wattday are units of work or energy.
1 KWh = 10^{3} J ´ (60 ´ 60sec) = 3.6 ´ 10^{6} Joule sec
(5)
The slope of work time curve gives the instantaneous power. As P = dW/dt = tanq
 Area under power time curve gives the work done as
\ W = ò P dt
\ W = Area under P–t curve
P = dW
dt
Position and Velocity of an Automobile w.r.t Time.
An automobile of mass m accelerates, starting from rest, while the engine supplies constant power P, its position and velocity changes w.r.t time.
(1) Velocity : As Fv = P = constant
i.e.
m dv v = P
éAs F = mdv ù
dt
or ò vdv = ò
P dt m
ëê dt úû
By integrating both sides we get v 2 = P t + C
2 m ^{1}
As initially the body is at rest i.e. v = 0 at t = 0, so C_{1} = 0
æ 2Pt ö^{1/} ^{2}
\ v = ç ÷
è m ø
æ 2Pt ö^{1/} ^{2}
(2) Position : From the above expression v = ç ÷
è m ø
ds æ 2Pt ö^{1/} ^{2}
é ds ù
or dt = ç m ÷ êAs v = dt ú
è ø ë û
æ 2Pt ö^{1/} ^{2}


 òds =ò ç m ÷ dt
æ 2P ö^{1} ^{/} ^{2} 2
By integrating both sides we get s = ç ÷ . t ^{3} ^{/} ^{2} + C_{2}
è m ø 3
Now as at t = 0, s = 0, so C_{2} = 0
æ 8P ö^{1/} ^{2}
\ s = ç
÷ t 3 / 2
è 9m ø
Problem 64. A car of mass ‘m’ is driven with acceleration ‘a’ along a straight level road against a constant external resistive force ‘R’. When the velocity of the car is ‘v’, the rate at which the engine of the car is doing work will be
[MP PMT/PET 1998; JIMPER 2000]
 Rv (b) mav (c)
(R + ma)v
(d)
(ma – R)v
Solution : (c) The engine has to do work against resistive force R as well as car is moving with acceleration a.
Power = Force ´ velocity = (R+ma)v.
Problem 65. A windpowered generator converts wind energy into electrical energy. Assume that the generator converts a fixed fraction of the wind energy intercepted by its blades into electrical energy. For wind speed v, the electrical power output will be proportional to [IITJEE 2000]
(a) v (b) v ^{2}
 v ^{3}
 v ^{4}
Solution : (c) Force = v dm = v d (V ´ r) = vr d [A ´ l] = vr A dl
= r Av ^{2}
dt dt dt dt
Power = F ´ v =
r Av ^{2} ´ v = r Av ^{3}
\ P µ v ^{3} .
Problem 66. A pump motor is used to deliver water at a certain rate from a given pipe. To obtain twice as much water from the same pipe in the same time, power of the motor has to be increased to [JIPMER 2002]
(a) 16 times (b) 4 times (c) 8 times (d) 2 times
Solution : (d)
P = work done = mgh
\ P µ m
time t
i.e. To obtain twice water from the same pipe in the same time, the power of motor has to be increased to 2 times.
Problem 67. A force applied by an engine of a train of mass
2.05 ´ 10^{6} kg
changes its velocity from 5 m/s to 25 m/s in 5
minutes. The power of the engine is [EAMCET 2001]
(a) 1.025 MW (b) 2.05 MW (c) 5MW (d) 5 MW
1 m(v ^{2} – v ^{2} ) 1 ´ 2.05 ´ 10^{6} ´ [25^{2} – 5^{2} ]
Solution : (b)
Power = Work done = Increase in kinetic energy = 2
2 1 = 2
time
time t
5 ´ 60
= 2.05 ´ 10^{6} watt = 2.05 MW
Problem 68. From a water fall, water is falling at the rate of 100 kg/s on the blades of turbine. If the height of the fall is 100m then the power delivered to the turbine is approximately equal to [BHU 1997]
(a) 100kW (b) 10 kW (c) 1kW (d) 1000 kW
Solution : (a)
Power = Work done = mgh = 100 ´ 10 ´ 100 = 10^{5} watt = 100 kW éAs m = 100 kg (given)ù
t t êë t sec úû
Problem 69. A particle moves with a velocity
r

v = 5i – 3 j + 6k ms
under the influence of a constant force

r = 10ˆi + 10ˆj + 20kˆ N. The instantaneous power applied to the particle is
(a) 200 Js^{–1} (b) 40 Js^{–1} (c) 140 Js^{–1} (d) 170 Js^{–1}
Solution : (c)
r r

P = F.v = (10i + 10 j + 20k).(5i – 3 j + 6k) = 50 – 30 + 120 = 140 J–s
Problem 70. A car of mass 1250 kg experience a resistance of 750 N when it moves at 30ms^{–1}. If the engine can develop 30kW at this speed, the maximum acceleration that the engine can produce is
(a)
0.8ms ^{–}^{2}
(b)
0.2ms ^{–}^{2}
(c)
 ms ^{–}^{1}
(d)
 ms ^{–}^{2}
Solution : (b) Power = Force ´ velocity = (Resistive force + Accelerating force) ´ velocity
Þ 30 ´ 10^{3} = (750 + ma) ´ 30 Þ ma = 1000 – 750
Þ a =
250
1250
= 0.2 ms ^{–}^{2} .
Problem 71. A bus weighing 100 quintals moves on a rough road with a constant speed of 72km/h. The friction of the road is 9% of its weight and that of air is 1% of its weight. What is the power of the engine. Take g = 10m/s^{2}
(a) 50 kW (b) 100 kW (c) 150 kW (d) 200 kW
Solution : (d) Weight of a bus = mass ´ g = 100 ´ 100 kg ´ 10 m/s ^{2} = 10^{5} N
Total friction force = 10% of weight = 10^{4} N
\ Power = Force ´ velocity = 10^{4} N ´ 72 km / h = 10^{4} ´ 20 watt = 2 ´ 10^{5} watt = 200 kW .
Problem 72. Two men with weights in the ratio 5 : 3 run up a staircase in times in the ratio 11 : 9. The ratio of power of first to that of second is
(a) 15 11
(b)
11 (c)
15
11 (d) 9
9 11
Solution : (a) Power (P) =
mgh or
P µ m Þ
P1 = m1 t 2
= æ 5 ö æ 9 ö = 45 = 15
(g and h are constants)
t t P m t
ç 3 ÷ ç 11 ÷ 33 11
2 2 1
è ø è ø
Problem 73. A dam is situated at a height of 550 metre above sea level and supplies water to a power house which is at a height of 50 metre above sea level. 2000 kg of water passes through the turbines per second. The maximum electrical power output of the power house if the whole system were 80% efficient is
(a) 8 MW (b) 10 MW (c) 12.5 MW (d) 16 MW
Solution : (a)
Power = work done = mgDh = 2000 ´ 10 ´ (550 – 50) = 10 MW
time t 1
But the system is 80% efficient \ Power output = 10 ´ 80% = 8 MW.
Problem 74. A constant force F is applied on a body. The power (P) generated is related to the time elapsed (t) as
(a)
P µ t ^{2}
(b)
P µ t
(c)
P µ (d)
P µ t 3 / 2
Solution : (b)
F = mdv
dt
\ F dt = mdv Þ
v = F t
m
Now P = F ´ v = F ´ F t = F 2 t
m m
Collision.
If force and mass are constants then P µ t.
Collision is an isolated event in which a strong force acts between two or more bodies for a short time as a result of which the energy and momentum of the interacting particle change.
In collision particles may or may not come in real touch e.g. in collision between two billiard balls or a ball and bat there is physical contact while in collision of alpha particle by a nucleus (i.e. Rutherford scattering experiment) there is no physical contact.
(1) Stages of collision : There are three distinct identifiable stages in collision, namely, before, during and after. In the before and after stage the interaction forces are zero.
Between these two stages, the interaction forces are very large and often the dominating forces governing the motion of bodies.
The magnitude of the interacting force is often unknown,
therefore, Newton’s second law cannot be used, the law of conservation of momentum is useful in relating the initial and final velocities.
 Momentum and energy conservation in collision :
(i) Momentum conservation : In a collision the effect of external forces such as gravity or friction are not taken into account as due to small duration of collision (Dt) average impulsive force responsible for collision is much larger than external force acting on the system and since this impulsive force is ‘Internal’ therefore the total momentum of system always remains conserved.
 Energy conservation : In a collision ‘total energy’ is also always Here total energy includes all forms of energy such as mechanical energy, internal energy, excitation energy, radiant energy or even mass energy.
These laws are the fundamental laws of physics and applicable for any type of collision but this is not true for conservation of kinetic energy.
 Types of collision : (i) On the basis of conservation of kinetic
Perfectly elastic collision  Inelastic collision  Perfectly inelastic collision 
If in a collision, kinetic energy after collision is equal to kinetic energy before collision, the collision is said to be perfectly elastic.  If in a collision kinetic energy after collision is not equal to kinetic energy before collision, the collision is said to inelastic.  If in a collision two bodies stick together or move with same velocity after the collision, the collision is said to be perfectly inelastic. 
Coefficient of restitution e = 1  Coefficient of restitution 0 < e < 1  Coefficient of restitution e = 0 
(KE)final = (KE)initial 
Here kinetic energy appears in other forms. In some cases (KE)_{final} < (KE)_{initial} such as when initial KE is converted into internal energy of the product (as heat, elastic or excitation) while in other cases (KE)_{final} > (KE)_{initial} such as when internal energy stored in the colliding particles is released  The term ‘perfectly inelastic’ does not necessarily mean that all the initial kinetic energy is lost, it implies that the loss in kinetic energy is as large as it can be. (Consistent with momentum conservation). 
Examples : (1) Collision between atomic particles
(2) Bouncing of ball with same velocity after the collision with earth. 
Examples : (1) Collision between two billiard balls.
(2) Collision between two automobile on a road. 
Example : Collision between a bullet and a block of wood into which it is fired. When the bullet remains embeded in the block. 
In fact all majority of collision belong to this category. 
(ii) On the basis of the direction of colliding bodies
Perfectly Elastic Head on Collision.
Let two bodies of masses m_{1}
and m2
moving with initial velocities u_{1}
and u2
in the same direction and they
collide such that after collision their final velocities are v_{1} and v_{2} respectively.
According to law of conservation of momentum
m1u1 + m2u2 = m1v1 + m2v2
……(i)
Þ m1 (u1 – v1 ) = m2 (v2 – u2 )
……(ii)
According to law of conservation of kinetic energy
1 m u^{2} + 1 m u^{2} = 1 m v ^{2} + 1 m v ^{2}
…..(iii)
2 1 1 2 2 2 2 1 1 2 2 2



Þ m1 (u^{2} – v ^{2} ) = m
(v ^{2} – u^{2} )
..…(iv)


Dividing equation (iv) by equation (ii)
v1 + u1 = v2 + u2
…..(v)
Þ u1 – u2 = v2 – v1
…..(vi)
Relative velocity of approach = Relative velocity of separation
Further from equation (v) we get v2 = v1 + u1 – u2
Substituting this value of v_{2} in equation (i) and rearranging we get
æ m1 – m2 ö 2m2u2
v1 = ç m + m
÷ u1 + m + m
……(vii)
è 1 2 ø 1 2
æ m2 – m1 ö 2m1u1
Similarly we get v2 = ç m + m
÷ u2 + m + m
……(viii)
è 1 2 ø 1 2
 Special cases of head on elastic collision
æ m1 – m2 ö
2m2
æ m2 – m1 ö
2m1u1
Since v1 = ç m + m
÷ u1 + m + m u2
and
v2 = ç m + m
÷ u2 + m + m
è 1 2 ø 1 2 è 1 2 ø 1 2
Substituting m1 = m2 we get
v1 = u2
and
v2 = u1
It means when two bodies of equal masses undergo head on elastic collision, their velocities get interchanged.
Example : Collision of two billiard balls

Before collision After collision
u_{1} = 50m/s
u_{2} = 20m/s
v_{1} = 20 m/s
v_{2} = 50 m/s
(ii) If massive projectile collides with a light target i.e. m_{1} >> m_{2}
æ m1 – m2 ö
2m2u2
æ m2 – m1 ö
2m1u1
Since v1 = ç m + m
÷ u1 + m + m
and
v2 = ç m + m
÷ u2 + m + m
è 1 2 ø 1 2 è 1 2 ø 1 2

Substituting m_{2} = 0 , we get
v1 = u1 and v2 = 2u1 – u2
Example : Collision of a truck with a cyclist
v_{1} = 120 km/hr
km/hr v_{2} = 230 km/hr
Before collision After collision
(iii) If light projectile collides with a very heavy target i.e. m_{1} << m_{2}
æ m1 – m2 ö
2m2u2
æ m2 – m1 ö
2m1u1
Since v1 = ç m + m
÷ u1 + m + m
and
v2 = ç m + m
÷ u2 + m + m
è 1 2 ø 1 2 è 1 2 ø 1 2
Substituting m_{1} = 0 , we get
v1 = – u1 + 2u2 and v2 = u2
Example : Collision of a ball with a massive wall.
u_{1} = 30 m/s m_{1} = 50gm
u_{2} = 2 m/s
v_{1} = – 26 m/s
v_{2} = 2 m/s
m_{2} = 100 kg
Before collision
After collision
(2) Kinetic energy transfer during head on elastic collision
Kinetic energy of projectile before collision K
= 1 m u^{2}
i 2 1 1
Kinetic energy of projectile after collision K
= 1 m v ^{2}
f 2 1 1
Kinetic energy transferred from projectile to target DK = decrease in kinetic energy in projectile
DK = 1 m u^{2} – 1 m v ^{2} = 1 m (u^{2} – v ^{2} )
2 1 1 2 1 1 2 1 1 1
1 m (u^{2} – v ^{2} ) 2
DK 2 1 1 1
æ v1 ö
Fractional decrease in kinetic energy K = 1
= 1 – ç ÷
u
…..(i)
m u^{2}
è 1 ø
2 1 1
æ m1 – m2 ö
2m2u2
We can substitute the value of v1 from the equation v1 = ç m + m
÷ u1 + m + m
æ m1 – m2 ö
è 1 2 ø 1 2
If the target is at rest i.e. u_{2} = 0 then v1 = ç m + m ÷u1
è 1 2 ø
DK æ m – m ö 2
From equation (i)
= 1 – ç^{ } ^{1} ^{2} ÷
K è m1 + m2 ø
…..(ii)
or DK =
K
or DK =
K
4m1m2 (m1 + m2 )^{2}
4m1m2
(m1 – m2 )^{2} + 4m1m2
…..(iii)
…..(iv)
Note : @ Greater the difference in masses less will be transfer of kinetic energy and vice versa
@ Transfer of kinetic energy will be maximum when the difference in masses is minimum
i.e.
m1 – m2 = 0
or m1 = m2
then
DK = 1 = 100%
K
So the transfer of kinetic energy in head on elastic collision (when target is at rest) is maximum when the masses of particles are equal i.e. mass ratio is 1 and the transfer of kinetic energy is 100%.
@ If m = nm
then from equation (iii) we get
DK = 4n
2 1 K
(1 + n)^{2}
@ Kinetic energy retained by the projectile
æ DK ö
= 1 –
kinetic energy transferred by
projectile
ç ÷

è øRetained
æ DK ö
é æ m – m ö2 ù
æ m – m ö 2
Þ ç ÷
= 1 – ê1 – ç ^{1} ^{2} ÷ ú = ç^{ } ^{1} ^{2} ÷
è K øRetained
ëê è m1 + m2 ø úû
è m1 + m2 ø
(3) Velocity, momentum and kinetic energy of stationary target after head on elastic collision
æ m2 – m1 ö 2m1u1
(i) Velocity of target : We know v2 = ç m + m
÷ u2 + m + m
2m u 2u
è 1 2 ø 1 2
m
Þ v2 = 1 1 = 1
[As u_{2} = 0 and Let ^{2} = n ]
m1 + m2 1 + m2 / m1 m1
\ v2
= 2u1 1 + n
(ii) Momentum of target :
P = m v
= 2nm1u1
éAs m
= m n and v
= 2u_{1} ù
\ P2
2 2 2
= 2m1u1 1 + (1 / n)
1 + n
ëê 2 1
2 1 + n úû
1 1 æ 2u ö ^{2}
2 m u^{2}n
(iii) Kinetic energy of target :
K2 =
m v ^{2} = nm_{1} ç ^{1} ÷
= 1 1
2 2 2 2
è 1 + n ø
(1 + n)^{2}
= 4(K1 )n
é As K = 1 m u^{2} ù
(1 – n)^{2} + 4n
ëê 1 2
1 1 úû
(iv)
Relation between masses for maximum velocity, momentum and kinetic energy
Problem 75. n small balls each of mass m impinge elastically each second on a surface with velocity u. The force experienced by the surface will be [MP PMT/PET 1998; RPET 2001; BHU 2001; MP PMT 2003]
 mnu (b) 2 mnu (c) 4 mnu (d)
1 mnu
2
Solution : (b) As the ball rebounds with same velocity therefore change in velocity = 2u and the mass colliding with the surface per second = nm
Force experienced by the surface
F = m dv
dt
\ F = 2 mnu.