best coaching institute for iit-jee booklet (work, energy and power)
TRANSCRIPT
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7/29/2019 BEST COACHING INSTITUTE FOR IIT-JEE BOOKLET (WORK, ENERGY AND POWER)
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APEX INSTITUTE - 62, Nitikhand-III, Indirapuram, Ghaziabad Ph.No.-+91-9910817866, 0120-4901457,Webs i te : w w w.apex i it . co .in /
WORK, ENERGY AND POWER
WORK DONE BY A CONSTANT FORCE
Consider a particle, that undergoes a displacement r along a straight line while upon acted by a constant force
F that makes an angle with r as shown in figure.
The work W done by a constant force F acting on a particle is equal to the product of the component of F
along the direction the displacement of particle and the magnitude of the displacment )r( .Mathematically,
r.cosFW = (from Figure)
= rF. (Applying = cosabb.ar
)
Work is a scalar quantity. Its unit is Joule.
(i) Force does not work if point of application of force does not move ( r = 0)(ii) Work done by a force is zero if displacement is perpendicular to the force ( = 900)(iii) If the angle between the force and the displacement is acute ( < 900), we say that work done by the
force is positive.
(iv) If the angle between the force and the displacement is obtuse ( > 900), we say that work done bythe force is negative.
(v) Work done depends on the frame of reference. With the change of the frame of reference inertial
force does not change while displacement may change.
WORK DONE BY A VARIABLE FORCE
Consider a particle being displaced along the x-axis under the action of a varying force, as shown in the figure.
The particle is displaced in the direction of increasing x from x = xito x = x
f. In such a situation, we cannot use
W = (F cos )r to calculate the work done by the force because this relationship applies only when F is constantin magnitude and direction. However, if we imagine that the particle undergoes a very small displacement x,shown in the figure (1), then the x component of the force, F
x, is approximately constant over this interval, and
we can express the work done by the force for this small displacement as
W1
= Fxx
r
F cos
F
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APEX INSTITUTE - 62, Nitikhand-III, Indirapuram, Ghaziabad Ph.No.-+91-9910817866, 0120-4901457,Webs i te : w w w.apex i it . co .in /
Fx
Area = A = Fxx
Fx
xxfx
xi
(1)
Fx
xxfxi
(2)
Thus is just the area of the shaded rectangle in the figure (1). If we imagine that the Fx
versus x curve is divided
into a large number of such intervals, then the total work done for the displacement from xito x
fis approxi-
mately equal to the sum of a large number of such terms.
W = f
i
x
x
xxF
If the displacements are allowed to approach zero, then the number of terms in the sum increases without limit
but the value of the sum approaches a definite value equal to the area under the curve bounded by Fx
and the x-
axis.
dxFxF
f
i
f
i
x
x
x
x
x
xx =0lim
This definite integral is numerically equal to the area under the Fx
versus x curve between xiand x
f. There-
fore, we can express the work done by Fx
for the displacement of the object from xito x
fas
W = dxF
f
i
x
x
x
when F
x= F cos is constant.
If more than one force acts on a particle, the total work done is just the work done by the resultant force. For
systems that do not act as particles, work must be found for each force separately. If we express the resultant
force in the x-direction as Fx, then the net work done as the particle moves from x
ito x
fis
Wnet
= ( )dxFf
i
x
x
x
Illustration:
A chain of mass m = 0.80 kg and length l = 1.5 m rests on a rough - surfaced table so that one of its
ends hangs over the edge. The chain starts sliding off the table all by itself provided the over hanging
part equals n = 1/3 of the chain length. What will be the total work performed by the friction forces
acting on the chain by the moment it slides completely off the table ?
Solution:
Slipping occurs when the weight of hanging part is just sufficient to overcome the frictional force
exerted by the table. Let be the coefficient of friction between chain and table.
Weight of hanging part = (weight of horizontal part)
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APEX INSTITUTE - 62, Nitikhand-III, Indirapuram, Ghaziabad Ph.No.-+91-9910817866, 0120-4901457,Webs i te : w w w.apex i it . co .in /
nmg = (1 n) mg
=n1
n
...(i)
Let x be the length of the hanging part at some time instant.
frictional force f(x) = (normal reaction)
=l
l mg)x(
The work done by the frictional force if the hanging part increases to (x + dx) is :
dW = f (x) dx
W = dW = l
l l
l
n
mg)x(dx
W = l
mgl
l
l
n
2
2
xx
W = mg
)n1(2
)n1( 2l
l
Substituting the value of from (i), we get :
W = 2
mg)n1(n l= 1.3 J.
WORK DONE BY A SPRING
Consider the situation shown in figure. One end of a spring is attached to a fixed vertical support and the other
end to a block which can move on a horizontal table. Let x = 0 denote the position of the block when the spring
is in its natural length. We shall calculate the work done on the block by the spring-force as the block movesfrom x = 0 to x = x
1.
x = 0
A
x = x1
A
The force on the block is k times the elongation of the spring. But the elongation changes as the block moves
and so does the force. We cannot take F out of the integration F rd . We have to write the work done duringa small interval in which the block moves from x to x + dx. The force in this interval is kx and the displacement
is dx. The force and displacement are opposite in direction.
So, F. rd = -F dx = - kx dxduring this interval. The total work done as the block is placed from x = 0 to x
1is
W =2
1
0
2
02
1
2
111
kxkxkxdx
xx
=
=
If the block moves from x = x1
to x = x2, the limits of integration are x
1and x
2and the work done is
W =
22
2
12
1
2
1kxkx
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Note that if the block is displaced from x1
and x2
and brought back to x = x1, the work done by the spring-
force is zero. The work done during the return journey is negative of the work during the onward journey.
The net work done by the spring-force in a round trip is zero.
Illustration:
A block of mass m has a velocity v0
when it just touches a spring. The block moves through a distance l before
it stops after. The spring constant is k. What is the work done on it by the spring force?
Solution :
The net force acting on the block by the spring is equal to Fspring
= kx
Work done by the spring = -Fspring
.ds
= -2
.2
0
kldxkx
l =
WORK DEPENDS ON THE FRAME OF REFERENCE
Work done by a force is given as
S.FW =
Since S displacement of point of application is frame dependent phenomena work is frame
dependent quantity, same force may do different work in different frames
Illustration :
Over a horizontal plank a small block of mass m is lying at rest. Now plank is moved withconstant acceleration a such that there is no relative motion between block and plank. Find the work
done by friction of plank on block in first t seconds.
(a) in ground frame
(b) in plank frame.
Solution:
(a) In ground frame
Friction force acting on the block (f) = ma
Displacement in first t second
2at2
1S =
Work done = F.S. = f
2
at2
1
=222 tma
2
1at
2
1.ma =
(b) In plank frame f = ma
But displacement s = 0
(because there is no relative motion between plank and block)
m
v0 k
ma
fa
f
F (=ma)P
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W = F.S.= f . s
= ma . 0
= 0
Note: While calculating net work done on a body in accelera ted frame work done by pseudo
force need to be taken into account .
WORK-ENERGY THEOREM
If the work done by the net force on a particle can be calculated for a given displacement, the change in the
particle's speed will be easy to evaluate.
As shown in the figure a particle of mass m moving to the right under the action of a constant net force F.
Because the force is constant, we know Newton's second law that the particle will move with a constant
acceleration a. If the particle is displaced a distance s, the net work done by the force F is
Wnet
= Fs = (ma)s
We found that the following relationships are valid when a particle moves at constant acceleration
m
vi vf
s
F
s = ;)(2
1tvv fi + a = t
vvif
where viis the speed at t = 0 and v
fis the speed at time t. Substituting these expressions
Wnet
= m tvvt
vvfi
if)(
2
1+
Wnet
=22
2
1
2
1if mvmv
Wnet
= Kf- K
i= K . . . . (i)
That is, the work done by the constant force Fnet
in displacing a particle equals the change in kinetic energy
of the particle.
Equation (i) is an important result known as the work-energy theorem. For convenience, it was derived
under the assumption that the net force acting on the particle was constant.
Now, we shall show that the work-energy theorem is valid even when the force is varying. If the resultant
force acting on a body in the x direction is Fx, then Newton's second law states that
Fx
= ma. Thus, we express the net work done as
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Wnet
= ( ) =f
i
f
i
x
x
x
x
x dxmadxF
Because the resultant force varies with x, the acceleration and speed also depend on x. We can now use the
following chain rule to evaluate Wnet
.
a =dx
dvv
dt
dx
dx
dv
dt
dv ==
Wnet
=22
2
1
2
1if mvmv
Illustration:
A bullet leaving the muzzle of a rifle barrel with a velocity v penetrates a plank and loses one fifth of
its velocity. It then strikes second plank, which it just penetrates through. Find the ratio of the thickness
of the planks supposing average resistance to the penetration is same in both the cases.
Solution:Let R = resistance force offered by the planks,
t1
= thickness of first plank,
t2
= thickness of second plank.
For first p lank :
Loss in KE = work against resistance
2
1mv2
2
1m (
5
4v)2 = Rt
1
21 mv2 25
9 = Rt1
.................(i)
For second plank
2
1m (
5
4v)2 0 = Rt
2
2
1mv2
25
16= Rt
2.................(ii)
dividing (I) & (II) 21
t
t
= 16
9.
POTENTIAL ENERGY AND CONSERVATION OF MECHANICAL ENERGY
We define the change in potential energy of a system corresponding to a conservative internal force as
Uf- U
i= -W = -
f
i
rdF.
where W is the work done by the internal force on the system as the system passes from the initial configu-
ration i to the final configuration f.
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Note: We don't (or can't) define potential energy corresponding to a nonconservative internal force.
Suppose only conservative internal forces operate between the parts of the system and the potential energy U
is defined corresponding to these forces. There are either no external forces or the work done by the them is
zero. We have
Uf- U
i= -W = -(K
f- K
i)
or Uf+ Kf= Ui + Ki . . . . . (i)
The sum of the kinetic energy and the potential energy is called the total mechanical energy. We see from
equation (i) that the total mechanical energy of a system remains constant if the internal forces are conser-
vative and the external forces do not work. This is called the principle of conservation of mechanical energy.
The total mechanical energy K + U is not constant if nonconservative forces, such as friction, act between
the parts of the system. We can't apply the principle of conservation of energy in presence of nonconservation
forces. The work-energy theorem is still valid even in the presence of nonconservative forces.
Note: that only a change in potential energy is defined above. We are free to choose the zero potential
energy in any configuration just as we are free to choose the origin in space anywhere we like.
Brain Teaser: One persons says that the potiential energy of a particular book kept in an almirah is 20 J and
the other says it is 30 J. Is one of them necessarily wrong?
If nonconservative internal forces operate within the system, or external forces do work on the system, the
mechanical energy changes as the configuration changes. According to the work-energy theorem, the work
done by all the forces equals the change in the kinetic energy. Thus,
Wc
+ Wnc
+ Wext
= Kf- K
i
where the three terms on the left denote the work done by the conservative internal forces, nonconservative
internal forces and the external forces.
As Wc
= -(Uf- U
i),
We get Wnc + Wext = (Kf+ Uf) - (Ki + Ui)= E
f- E
i. . . . . (ii)
Where E = K + U is the total mechanical energy.
If the internal forces are conservative but external forces also act on the system and they do work, Wnc
= 0
and from (ii)
Wext
= Ef- E
i. . . . .(iii)
The work done by the external forces equals the change in the mechanical energy of the system.
Let us summarise the concepts developed so far in this chapter.
(1) Work done on a particle is equal to the change in its kinetic energy.
(2) Work done on a system by all the (external and internal) forces is equal to the change in its kinetic
energy.
(3) A force is called conservative if the work done by it during a round trip of a system is always zero. The
force of gravitation, Coulomb force, force by a spring etc. are conservative. If the work done by it
during a round trip is not zero, the force is nonconservative. Friction is an example of nonconservative
force.
(4) The change in the potential energy of a system corresponding to conservative internal forces is equal
to negative of the work done by these forces.
(5) If no external forces act (or the work done by them is zero) and the internal forces are conserva-
tive, the mechanical energy of the system remains constant. This is known as the principle of
conservation of mechanical energy.
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(6) If some of the internal forces are nonconservative, the mechanical energy of the system is not constant.
(7) If the internal forces are conservative, the work done by the external forces is equal to the change in
mechanical energy.
Brain Teaser: When you lift a box from the floor and put it on an almirah the potential energy of the box
increases, but there is no change in its kinetic energy. Is it a violation of conservation of
energy?
Illustration:
In the figure shown stiffness is k and mass of the block is m. The pulley is fixed. Initially
the block m is held such that, the elongation in the spring is zero and then released from
rest. Find the maximum elongation in the spring
Neglect the mass of the spring, pulley and that of the string.
Solution:
Let the maximum elongation in the spring be x, when the block is at position 2.
The displacement of the block m is also x. If E1
and E2
are the energies of the system
when the block is at position 1 and 2 respectively. ThenE1
= U1g
+ U1s
+ T1
where U1g
= gravitational P.E. with respect to surface S.
U1S
= P.E. stored in the spring.
T1
= initial K.E. of the block.
E1
= mgh1
+ 0 + 0 = mgh1
. . . . (i)
and E2
= U2g
+ U2s
+ T2
= mgh2
+ 02
1 2 +kx . . . . . (ii)
From conservation of energy E1
= E2
mgh1 = mgh2 +2
21 kx
mgx)hh(mgkx2
121
2 ==
x = 2mg/k
Illustration:
A block is placed on the top of a plane inclined at 37 with horizontal. The length of the plane is 5 m.
The block slides down the plane and reaches the bottom.
(a) Find the speed of the block at the bottom if the inclined plane is smooth.
(b) Find the speed of the block at the bottom if the coefficient of friction is 0.25
Solution:
Let h be the height of inclined plane
h = 5 sin 37 = 3 m(a) As the block slides down the inclined plane, it loses GPE and
gains KE.
Loss in GPE = gain in KE
mg (loss in height) = KEf KE
i
m
1
m
2h2
h1
S
m m
R
s
C
A
B
mgmg cos37
37
h37
v
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mgh =2
1mv2 0
v = gh2 = 38.92 = 7.67 m/s.
Note : 1. Loss in energy = initial energy - final energy
2. gain in energy = final energy - initial energy
(b) As the block comes down, it loses GPE. It gains KE and does work against friction.
loss in GPE = gain in KE + work done against friction
mgh = (1/2 mv2 0) + ( mg cos 37) s 3mg = 1/2 mv2 + (0.25) mg 4/5 5
v = g4 = 6.26 m/s
Illustration:A 1.0 kg block collides with horizontal light spring of force constant 2 N/m. The block compresses the
spring 4 m from the rest position. Assuming that the coefficient of kinetic friction between the block
and the horizontal surface is 0.25, what was the speed of the block at the instant of collision ?
Solution:
When the block compresses the spring, let x m be the amount of compression, i.e. x = 4m.
Let v = velocity of the block when it collides with the spring.
Loss in KE of the block = (gain in elastic potential energy of the spring) + (work done against friction)
21 mv2 0 =
21 kx2 + mg x
2
1mv2 =
2
1(2) (4)2 + 0.25 1 9.8 4
v2 = 51.6 v = 6.51 = 7.18 m/s
Illustration:
A pump is required to lift 1000 kg of water per minutes from a well 20 m deep and eject it at a rate of
20 m/s.
(a) How much work is done in lifting water ?
(b) How much work is done in giving it a KE ?
Solution:
(a) Work done in lifting water = gain in PE (potential energy)
work = 1000 g 20 = 1.96 105 J per minute
(b) Work done (per minute) in giving it KE = 1/2 mv2
= 1/2 (1000) (20)2 = 2 105 J per minute
R
s
C
A
B
mgmg cos37
37
h37
mgcos37
v
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Brain Teaser: A "Violation" of the Law of the Conservation of Energy
The following argument seems to prove a violation of the law of the energy conservation.
Suppose that a resting hand-cart of mass m is hit by and retains a projectile of the same mass.
The projectile had been flying horizontally before the collision at a velocity v in the same
direction as the hand-cart. As a result of this impact the hand-cart and projectile will together
set in motion at a initial velocity which can be found from the law of the conservation of
momentum
221
v
m
mvv == .
Hence, the kinetic energy of the hand-cart and projectile together is
42
22
2
2
1
mv
vm
W =
= ,
while before the collision the projectile had a kinetic energy of
2
2
mvW= .
i.e., twice as larger. Thus, after the collision half the energy has vanished altogether.
Can you say where it has gone?
TYPES OF POTENTIAL ENERGY
1. Elastic
We have seen that the work done by the spring force (of course conservative for an ideal spring) is
- 221 kx when the spring is stretched or compressed by an amount x from its unstretched or natural position.
Thus, U = -W = -
2
2
1kx
or U =2
2
1kx (k = spring constant)
Note that elastic potential energy is always positive.
2. Gravitational
The gravitational potential energy of two particles of masses m1 and m2 separated by a distance r is given by
U =r
mmG 21
Here, G = universal gravitation constant
= 6.67 10-11 22
kg
mN
If a body of mass m is raised to a height 'h' from the surface of earth, the change in potential energy of the
system (earth + body) comes out to be:
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U =
+
R
h
mgh
1 (R = radius of earth)
or U mgh if h
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power as t approaches zero.
i.e., P =dt
dW
t
W
t=
0lim
where we have represented the infinitesimal value of the work done by dW (even though it is not a change
and therefore not a differential).
P = vFdt
sdF
dt
dW.. ==
where we have used the fact thatdt
sdv = .
This SI unit of power is joule per second (J/s), also called watt(W)' (after James Watt);
1W = 1 J/s = 1 kg.m2/s3.
CONSERVATIVE AND NON-CONSERVATIVE FORCES
If the work done along a closed path is zero, the force is said to be conservative otherwise it is called non-
conservative. Under conservative forces, the work done depends only the initial and final positions and it ispath independent.Conservative forces are non-dissipative and we store work (or energy) whereas the non-conservative forces are
dissipative where in we do not store work. Examples of non-conservative force are friction, viscous force etc.
Three Types of Equilibrium
As we have studied in the chapter of 'Laws of motion' a body is said to be translatory equilibrium if net force
acting on the body is zero, i.e.,
0=netFIf the forces are conservative
F =drdU
and for equilibrium F = 0
So, ,0=dr
dUor 0=
dr
dU
i.e., at equilibrium position slope of U- r graph is zero or the potential energy is optimum (maximum or mini-mum or constant). Equilibrium are of three types, i.e. the situation where F = 0 and dU/dr = 0 can be obtained
under three conditions. These are stable equilibrium, unstable equilibrium and neutral equilibrium.
If 02
2
>dr
Udit is stable equilibrium
02
2
Rg2
Therefore, if ,gR5ugR2 R if u2 > 2gR. Thus, the particle, will
leave the circle when h > R or 900 < < 1800. This situation is shown in the figure.
gR5ugR2
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Illustration:
A heavy particle hanging from a fixed point by a light inextensible string of length l is projected horizontally
with speed ).(gl Find the speed of the particle and the inclination of the string to the vertical at the
instant of the motion when the tension in the string is equal to the weight of the particle.
Solution:Let tension in the string becomes equal to the weight of the particle when particle reaches the point B and
deflection of the string from vertical is . Resolving mg along the string and perpendicular to the string, weget net radial force on the particle at B i.e.
FR
= T - mg cos . . . . (i)If v be the speed of the particle at B, then
FR
=l
mv2
. . . . (ii)
From (i) and (ii), we get
T - mg cos = lmv 2
. . . . (iii)
Since at B, T = mg
mg(1 - cos ) =l
mv 2
v2 = gl(1 - cos ) . . . . .(iv)
Conserving the energy of the particle at point A and B, we have
22
02
1)cos1(
2
1mvmglmv +=
where v0
= gl and v = )cos1( gl
gl = 2gl(1 - cos) + gl(1 - cos) cos = 2/3 . . . . .(v)Putting the value of cos in equation (iv) we get
v =3
gl
A BODY MOVING INSIDE A HOLLOW TUBE
The same discussion holds good for this case, but instead of tension in the string we
have the normal reaction of the surface. If N is the normal reaction at the lowest
point, then
N - mg =r
mv 21; N = m
+ g
r
v 21
At the highest point of the circle,
N + mg =r
mv 22
N = m
g
r
v 22
A
BT
mg cos
mg sin
O
v1
O
N
mg
mg
N
O
v2
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The condition v1 rg5
All other equations (can be) similarly obtained by replacing tension T by reaction N.
BODY MOVING ON A SPHERICAL SURFACE
The small body of mass m is placed on the top of a smooth sphere of radius r.
If the body slides down the surface, at what point does it fly off the surface?Consider the point C where the mass is, at a certain instant. The forces are the
normal reaction R and the weight mg. The radial component of the weight is mg
cos acting towards the center. The centripetal force is
mg cos - R =r
mv 2
where v is the velocity of the body at O.
R = m
r
vg
2
cos . . . . . (i)
The body flies off the surface at the point where R becomes zero.
i.e., g cos =r
v 2
; cos =rg
v2
. . . . (ii)
To find v, we use conservation of energy
i.e., )(2
1 2 BNmgmv =
= mg(OB - ON) = mgr(1 - cos )v2 = 2rg(1 - cos )
2(1 - cos ) = rgv
2
. . . .(iii)
From equation (ii) and (iii) we get
cos = 2 - 2 cos ; 3 cos = 2
cos =3
2; = cos-1
3
2. . . . (iv)
This gives the angle at which the body goes off the surface.
The height from the ground of that point = AN = r(1 + cos)
= r r
3
5
3
21 =
+
Illustration:
A point mass m starts from rest and slides down the surface of a frictionless
solid sphere of radius R as shown in the figure. At what angle will this body
break off the surface of sphere? Find the velocity with which it will break off.
Solution:
Applying COE, we have at the point A and B,
N
A
mg
C
RB
O
m
RO
R
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m
R
O
R
N
m
A
B
v
we have mgR (1 cos ) = 2mv2
1. . (i)
Force equation gives, mg cos N = mv2/R . . (ii)N = 0 for break off.
v = cosgR . . . . (iii)
Putting it in (i)
We get cos = 2/3
Putting this in (iii) we get
v = gR3
2.
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SOLVED OBJECTIVE
Example 1. A block of mass M is pulled a distance x along a horizontal table. The work done by
the weight is
(a) 0 (b) Mg x
(c)2
mgl(d) none of these.
Solution: (a) 0E.P = (because of no vertical movement) work done = 0.E.P =
Example 2. The potential energy of a particle varies with position x according to the relation
x27x2)x(U 4 = the point2
3x = is point of
(a) Unstable equilibrium (b) Stable equilibrium
(c) neutral equilibrium (d) none of these.
Solution : (b) 027x8dx
dUF 3 === (for equilibrium)
02
324x24
dx
Ud2
2
2
2
>
==
stable equilibrium.
Example 3. A conservative force F is acting on a body and body moves from Point A to B and
then B to A then work done by the force is
(a) 0W < (b) 0W >(c) 0W = (d) 0W .
Solution : (c) When conservative force applied, work done by the force do not depend on path.
Example 4. It is kinetic energy of a body increases by 21%, the momentum of a body increase by
(a) 10 (b) 11
(c) 9 (d) 12.
Solution : (a) 11 mK2P =
22 mK2P =
12 K21.1K =
22 K)21.1(m2P =
1.121.1P
P
1
2 ==
% change = 10%
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Example 5. If a compressed spring is dissolved in acid
(a) Total energy of molecules of acid increase
(b) The energy of the acid remains constant
(c) The energy of the acid decrease
(d) none of the above.
Solution : (a) Compressed spring has some potential energy. When this spring dissolved in acid
P.E. of spring transfer to molecules of acid.
Example 6. There are two identical mass less springs A and B of spring constants AK and BK
respectively and BA KK > . Then(a) If they are compressed to same distance, then work done on A > work done on B
(b) If they are compressed by same force work done on A < work done on B
(c) If they are compressed by same force work done on A> work done on B
(d) Both a and b are correct.
Solution : (d) Case I:
Compressed by same distance2
AA xK2
1W =
2BB xK
2
1W =
Q BA KK >
BA WW > .
Case II:
Compressed by same force
BBAA xKxKF ==
A
AK
Fx =
B
BK
Fx =
2
A
A2AAA
K
FK
2
1xK
2
1W
==
A
2
K
F
2
1=
B
2
BK
F
2
1W =
AB WW > .
Example 7. Force acting on a particle moving in a straight line varies with the velocity of the
particle as KVF = . Where K is constant. The work done by this force in time t is(a) KVt (b) K2V2t2
(c) K2Vt (d) KV2t
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Solution : (d) KVF =P = KV.V = KV2
W = Pt
tKV2=
Example 8. A particle is projected with a velocity and making an angle with horizontal. Thenpower of the gravitational force(a) varies linearly with time
(b) is constant throughout
(c) is negative for complete path
(d) none of the above.
Solution : (a) jvivv yx +=r
gtcosvvy =
jmgF =
V.FP =
)jviv(.)jmg( yx +=
ymgv=
)gtcosv(mg =
)cosmgvtmg(P 2 = .
tP
Example 9. A block of mass M slides down a rough inclined plane of inclination with horizontal,with initial velocity v and from a height h. At the bottom point. Its velocity becomes
zero. Then the work done by the friction force in stopping the block.
(a)
+ mghmv
2
1 2(b) mghmv
2
1 2
(c) 22mv
2
1mgh (d) mghmv2 .
Solution: (a) fg WWK +=
f2
wmghmv2
1
+=
Work done by the friction force =
+ mghmv
2
1 2
Example 10. A ball is released from the top of a tower. The ratio of work done by force of gravity
in first, third and fifth second of the motion of ball is
(a) 1 : 3 : 5 (b) 5 : 3 : 1
(c) 1 : 5 : 9 (d) 9 : 5 : 1.
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Solution : (c) Work done by the force = E.Pdistance travelled in nth second
)1n2(g2
1uS
n+=
here u = 0
1.g2
1S
1=
S3
= 5.g2
1
9.g2
1S
5=
321 W;W;W 331 mgS;mgS;mgS== 1 : 5 : 9.
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SOLVED EXAMPLES (SUBJECTIVE)
Example 1. An object of mass 5 kg falls from rest through a vertical distance of 20 m and attains a
velocity of 10 m/s. How much work is done by the resistance of the air on the object?
(g = 10 m/s2).
Solution: Work done by all forces = change in K.E...EKWW gravityair =+
2
2
1mvmghWair =+
mghmvWair
= 22
1
20105101052
1=
JWair 750=
Example 2. A particle of mass m moves along a straight line on smooth horizontal plane, acted
upon by a force delivering a constant power P. If the initial velocity of the particle is
zero, then find its displacement as a function of time t.
Solution: Work done in time t = change in K.E.
2
2
1. mvtP =
2/12t
m
Pv = 2/1t
m
P2
dt
dx=
=t
0
2/1
x
0
dttm
P2dx
3tm
P2
3
2x =
Example 3. A block is projected horizontally on a rough hori-
zontal floor. The coefficient of friction between the
block and the floor is . The block strikes a lightspring of stiffness k with a velocity v
0. Find the maxi-
mum compression of the spring.
Solution: Since the block slides and the spring is com-
pressed through a distance x the net retarding
force acting on it
= F = (kx + N) = ( mg + kx) Work done by net force for the displace-ment x, W = F. dx
mk
v
mx
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W = x
dxF0
KE = +x
dxkxmg
0
)(
+=
2
kxmgxmv
2
10
220
x2 + 02 2
0= v
k
mx
k
mg
x =2
4422
0
2
222
k
mv
k
gm
k
mg+
x =
+ 11
2
0
g
v
m
k
k
mg
Example 4. A small block is projected with a speed v0on a horizontal
track which turns into a semi circle (vertical) of radius R.
Find the minimum value of V0
so that the body will hit
the point A after leaving the track at its highest point.
The arrangement is shown in the figure, given that the
straight part is rough and the curved part is smooth. The
coefficient of friction is .
Solution: Let the block escape the point at C with a velocity v horizontally. Since it hits the
initial spot A after falling through a height 2R we can write (2R) = (1/2)gt2
where t = time of its fall.
t = 2 gR/
the distance AB = 2v gR/
d = 2v gR/ . . . . . (i)
work energy theorem is applied to the motion of the body from A to B leads
KE = Wf
mgdmvmv = 212
0
2
1
2
1
v0
= gdv 221
+ . . . .(ii)
Energy conservation between B and C yields
)2(2
1
2
1 221 Rmgmvmv =
v1
= gRv 42 + . . . .(iii)
N
N
kx
mg
R
Ad
B
v1
v
C
Rv0
smoothRough surface A
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when the bead escapes C, its minimum speed v can be given as
mgR
mv=
2
( the normal contact force = 0)
v = gR . . . . (iv)
By using (iii) and (iv) we obtain
v1
= gR5 . . . . . (v)
using (i) and (iv) we obtain
d = ( ) Rg
RgR 22 =
. . . . .(vi)
Putting the values of v1
and d in (ii) we obtain
v0
= )2(25 RggR +
v0 = gR)45( +
Example 5. Two smooth balls of mass m1
and m2
connected by a light inextensible string are at the
opposite points of horizontal diameter of a smooth semi cylindrical surface of radius
R. If m1
is released, find its speed at any angular distance moved m2.
Solution: Let the ball m2
move through an angle , the mass m will fall through a distanceh
1= R.
The ball m2
rises through a height h2
as, h2
= R sinThe change in gravitational potential energy of m
1is
PE1 = -m1gh1 = = -m1 gR(since m1
loses its potential energy as it falls down).
The change in gravitational potential energy of m2
is
PE2
= m2gh
2= m
2gRsin
(Since m2
gains potential energy as it rises up)
The total change in gravitational potential energy= PE = PE
1+ PE
2
PE = -m1gR + m
2gR sin = gR(m
2sin - m
1) . . . . .(i)
= KE =2
)(
2
1
2
12
212
2
2
1
vmmvmvm
+=+ . . . . .(ii)
where v = speed of m1
and m2
at the positions as shown in the figure.
From the principle of conservation of energy we obtain,
KE + PE = 0 . . . . . (iii)Using (i), (ii) and (iii), we obtain,
0)sin()(2
121
2
21 =+ mmgRvmm
v =)(
)sin(2
21
21
mm
mmgR
+
.
R
m2
m2
m1 h2
v
m1
v h1
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Example 6: In Figure (a) and (b) AC, DG and GF are fixed inclined planes. BC = EF = x and AB =
DE = y. A small block of mass m is released from rest from the point A. Its slides
down AC and reaches C with a speed Cv . The same blocked is released from rest from
the point D; it slides down DGF and reaches the point F with speed Fv . The coefficient
of kinetic friction between the block and the surfaces AC and DGF is . Calculate Cvand
F
v .
R
s
C
A
B
(a)
mg
mg cos
x
y
(b)
x
y
x1 x2
s2
s1
G
D
E F
Solution:(a) ME at A = mgy + 0 and if Cv is the velocity at C,
So ME at C = 0 + (1/2) 2Cmv
Loss in ME = mgy (1/2) 2Cmv .
This loss in ME is equal to work done against friction i.e.,
scosmgmv)2/1(mgy 2C =
mxmgymv)2/1( 2C = g [as s/xcos = ]
or )xy(g2vc = ...(i)
(b) In this situation,
22112F sfsfmv)2/1(mgy +=
or21
2F scosmgscosmgmgymv)2/1( =
or )s/x(cosas][xxy[gv)2/1( 11212F == and )]s/x(cos 22=
or CF v)xy(g2v == [as xxx 21 =+ ] ...(ii)
Example 7. In the figure shown a massless spring of stiffness k and natural
length 0l is rigidly attached to a block of mass m and is invertical position. A wooden ball of mass m is released fromrest to fall under gravity. Having fallen a height h the ball strikesthe spring and gets stuck up in the spring at the top. What
should be the minimum value of h so that the lower block willjust lose contact with the ground later on? Find also the
corresponding maximum compression in the spring. Assume
thatk
mg4l0 >> . Neglect any loss of energy..
Solution: The minimum force needed to lift the lower block is equal to its weight. During upwardmotion the spring will get elongated. If elongation in the spring for just lifting the blockis 0x then
mgkx 0 =
m
k l0
h
m
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k
mgx 0 = ...(i)
From COE
20000 kx
2
1)xl(mg)hl(mg ++=+
200 kx2
1mgxmgh +=
k
gm
2
1
k
)mg(mgh
222
+=
k2
mg3h =
During down ward motion, suppose maximum compression in the spring is x From
COE.
200 kx
21)xl(mg)hl(mg +=+
2kx2
1mgxmgh += 2kx
2
1mgx
k2
mg3mg +=
222 xkkxmg2)mg(3 +=
0)mg(3mgkx2xk 222 =
2
222
k2
)mg(k12)mgk(4mgk2x
+=
2k2
mgk4mgk2 =
k
mg3x = .
Example 8. Two bodies A and B connected by a light rigid bar 10 m long
move in two frictionless guides as shown in the Figure. If B
starts from rest when it is vertically below A, find the
velocity of B when x = 6m. Assume kg200mm BA == and
mc = 100kg.Solution: At the instant, when the bar is as shown in the figure,
x2 + y2 = l2
0dt
dyy2
dt
dxx2 =+ ...(i)
dt
dyy
dt
dxx = ...(ii)
where =dt
dxvelocity of B and =
dt
dyvelocity of A.
c
10 m
A
BO
x
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Applying the law of conservation of energy,
loss of potential energy of A, if it is going down when the rod is vertical to the
position shown in the Figure
8.92002)810(gmA ==C moves down 6 m since B moves 6 m along x-axis
Loss of potential energy of 68.9100)mgh(C ==Total loss of potential energy = 200 68.910028.9 +
J9800108.9100 == .This must be equal to kinetic energy gained
Kinetic energy gained2
C2
BB2
AA )v(m2
1)v(m
2
1)v(m
2
1++=
222
dt
dx100
2
1
dt
dx200
2
1
dt
dy200
2
1
+
+
=
22
dt
dx150dt
dy100
+
=
22
dt
dx150dt
dx
y
x100
+
= from (ii)
22
dt
dx150
dt
dx
8
6100
+
=
2
dt
dx150
16
9100
+= 2
Bv
16
3300=
9800v16
3300 2B =
1B ms9.633
247
33
1698v ==
=
velocity of B at the required moment = 6.9 ms-1.
Example 9. A locomotive of mass m starts moving so that its velocity varies according to the law
s= , where is constant and s is the distance covered. Find the total work done
by all the forces acting on the locomotive during the first t seconds after the beginning
of motion.
Solution: Given v = sDifferentiating w.r.t. t, we get
==
s2dt
ds
s2
1
dt
dv 2/1
2s
s2
2=
=
acceleration,2
a2
=
Now force acting on the Locomotive,
c
A
BO
x
yl
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2mmaF
2== .
Here u = 0
Now, using2at
2
1uts += , we have
4
tt
22
10s
222
2 =
+=
Work done,4
t
2mFsW
222
==
8
tm24
= .
Example 10. The kinetic energy of a particle moving along a circle of radius R depends on the
distance covered S as T 2S= , where is constant. Find the force acting on theparticle as a function of S.
Solution: K.E., T = 2S or22 Smv
2
1=
m
S2v
22 = ...(i)
Differentiating both sides w.r.t. t, we have
vm
S4
dt
dS
m
S4
dt
dvv2
=
=
= v
dt
dSQ
or tam
S2
dt
dv=
=
(Tangential component of acceleration)
Now, centripetal acceleration is
mR
S2
R
va
22
c
== [Using equation (i)]
Net acceleration of the particle is given by2c
2t aaa +=
222
mR
S2
m
S2
+
=
2
R
S1
m
S2
+
=
Now, force acting on the particle is given by,
2
2
R
S1
m
S2mmaF +
==
or 2
2
R
S1S2F += .
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Example 11. A 0.5 kg block slides from the point A (see figure) on a horizontal track with an initial
speed of 3m/s towards a weightless horizontal spring of length 1 m and force constant 2
Newton/m. The part AB of the track is frictionless and the part BC has the coefficients
of static and kinetic friction as 0.22 and 0.2 respectively. If the distances AB and BD are
2 m and 2.14 m respectively find the total distance through which the block moves
before it comes to rest completely (Take g = 10 m/s2).
A B D C
Solution: Suppose the block comes to rest at the point E i.e. let DE = x. The kinetic energy of the
block is spent in overcoming friction and compressing the spring through a distance
DE = x.
A B D C
m2 2.14 x
E
m m
Kinetic energy of the block2mv
2
1=
J25.235.02
1 2 == ...(i)
As the part AB of the track is frictionless, work done in moving from A to B is zero
Let normal reaction of the block = mg
Coefficient of friction =
Force due to friction along the track BC = mg
105.02.0 == 1 N
Distance through which the block moves against the frictional force
= 2.14 + x m
Work done by block against friction before it comes to rest)x14.2(mg +=
= ( 2.14 + x )J ...(ii)
Let the spring constant = k
Work done by the block in compressing the spring through distance x
2kx2
1=
22 xx22
1 == J ...(iii)
Adding (ii) and (iii) and equating it to (i), we get
25.2xx14.2 2 =++
or 011.0xx 2 =+
or 011x100x100 2 =+or 0)1x10)(11x10( =+
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10
11x = or
10
1x =
Since10
11x ,
m1.0101x ==
Restoring force of the spring = kx
N2.01.02 == ...(iv)Static frictional force of the block
105.022.0mgstatic == 1.1 N ...(v)
From (iv) and (v) it is clear that the static frictional force is grater than the restoring
force of the spring. Therefore the block will not move in the backward direction. Hence
the total distance through which the block moves before it comes to rest completely
2.00 + 2.14 + 0.10 = 4.24 metres.
Example 12. In a spring gun having spring constant
100 N/m, a small ball of mass 0.1 kgis put in its barrel by compressing thespring through 0.05 m as shown inthe figure.(a) Find the velocity of the ball whenspring is released.(b) Where should a box be placed on
ground so that ball falls in it, if the ball
leaves the gun horizontally at a height
of 2 m above the ground (g = 10 m/s2)
Solution: (a) When the spring is released its elastic potential energy is converted into kinetic
energy2
1mv2 , so
22 kx2
1mv
2
1=
v =2
5m/s.
(b) As vertical component of velocity of ball is zero. Time taken by the ball to reach the
ground ,2
gt2
1
h =
t =5
2
g
h2= seconds
So, the horizontal distance travelled by the ball in this time
d = v . t =5
2
2
5 = 1 m
v
h=2 m
d
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Example 13. A smooth, light horizontal rod AB can rotate about
a vertical axis passing through its end A. The rod is
fitted with a small sleeve of mass m attached to the
end A by a weightless spring of length l0and stiffness
k. What works must be performed to slowly get
this system going and reach the angular velocity .
Solution: The mass m rotates in a circle of radius l , which is the extended length of the spring.
Centripital force on m = k(l - l0) = m 2l
or, l =n1
0
l
where n =k
m 2
W =change in KE of m + energy stored in the spring
=2
1m2 2l +
20 )(k
2
1ll
=2
1m
2
00
2
220
n1
k
2
1
)n1(
+
l
ll
W =
+
22
2
2
nk
m
)n1(
k
2
1 l
Example 14. A particle is suspended by a string of length l. It isprojected with such a velocity v along the horizontalthat after the string become slack it flies through itsinitial position. Find v.
Solution: Let the velocity be v at B where the string becomeslack and the string makes angle with horizontal by the law of conservation ofenergy.
)++= sin1(mgvm2
1mv
2
1 22l (i)
or, v2 = v2 2gl (1+ sin ) (ii)By the dynamics of circular motion
mg sin =l
2vm
v2 = gl sin ...(iii)from equation (ii) and (iii) we get
g l sin = v2 2gl (1 + sin ) ...(iv)At B the particle becomes a projectile of velocity v at 90 - with the horizontal.Here, u
x= v sin & v
y= v cos
ax
= 0 & ay
= -g
l cos = v sin t (v)
t =
sinv
cosl& l (1+ sin ) = v cos
22
22
sinv
cosg
2
1
sinv
cos ll
2 sin3 + 3 sin2 - 1 = 0
m
0
A B
v
O
mgmg cos
mg sin
A v
B'v
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sin =2
1is the acceptable solution
v2 = 2gl + 3gl 2
1=
2
g7 l v =
2
g7 l(from equation (iv))
Example 15. A system consists of two identical cubes each of mass m linked together
by a compressed massless spring of stiffness k. the cubes are connected
by a thread which is burnt at a certain moment. At what value of initial
compression of the spring will the lower cube bounce up after thethread is burnt.
Solution : Let us take horizontal line through the C.G. of lower cube as reference level. Let l
be the natural length of the spring.
T.E.initial
= )+ l(mgk2
1 2(i)
Let the spring extend by x after the thread is burnt
T.E.final
= )x(mgkx2
1 2 ++ l
kx = -k or k 2mg (-k not acceptable) kx = k 2mg.
Cube will bounce up when kx mg K 2mg mg
K 3mg = 3mg / k.
k
m
m
k
m
m
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OBJECTIVE
LEVEL - I
1. A force )jxiy(kF += , where k is a positive constant, acts on a particle moving in the xy
plane. Starting from the origin, the particle is taken along the positive x-axis to the point (a,
0), and the parallel to the y-axis to the point (a, a). The total work done by the force on the
particle is
(a) 2ka2 (b) 2ka2
(c) 2ka (d) 2ka .
2. Supposing that the earth of mass m moves around the sun in a circular orbit of radius R,
the work done in half revolution is
(a) RR
mv2 (b) R2
R
mv2
(c) Zero (d) none of these
3. When water falls from the top of a water fall 100 m high
(a) it freezes (b) it warms up slightly
(c) it evaporates (d) there is no change in temperature.
4. A string of mass m end length l rests over a frictionless table with 1/4th of its length
hanging from a side. The work done in bringing the hanging part back on the table is(a) mgl/4 (b) mgl/32
(c) mgl/16 (d) none of these
5. A weight mg is suspended from a spring. If the elongation in the spring is x0, the elastic
energy stored in it is
(a) 0mgx2
1(b) 2mgx
0
(c) mgx0
(d) 0mgx4
1
6. Mechanical Energy is conserved under
(a) Conservative system of forces (b) disipative forces
(c) (a) and (b) (d) none of these
7. The same retarding force is applied to stop a train. If the speed is doubled then the dis-
tance will be
(a) eight times (b) doubled
(c) half (d) four times
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8. Work done in time t on a body of mass m which is accelerated from rest to a speed v in
time t1
as a function of time t is given by
(a)
2
1 tt
v
m2
1
(b)
2
1 tt
v
m
(c)2
2
1
ttt
mv
2
1
(d)
2
2
1
2
tt
vm
2
1
9. A particle moves under the effect of a force F = kx2 from x = 0 to x = 4 the work done by
force is
(a)3
k8(b)
3
k32
(c)3
k64(d)
3
k128
10. Potential energy function describing the interaction between two atoms of a diatomic
molecule is
612 r
b
r
a)r(U =
Force acting between them will be zero when the distance between them would be
(a)
6/1
ba2
(b)
6/1
a2b
(c)
6/1
b
a
(d)
6/1
a
b
11. A ball is thrown up with a certain velocity at angle to the horizontal. The kinetic energyKE of the ball varies with horizontal displacement x as:
(a) (b)
(c) (d)
x
KE
O
x
KE
O
x
KE
O x
KE
O
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12. A body m1
is projected upwards with velocity v1
another body m1
of same mass is projected
at an angle of 450. Both reach the same height. What is the ratio of their kinetic energies at
point of projection
(a) 1 (b) 1/2(c) 1/3 (d) 1/4
13. A 2kg block is dropped from a height of 0.4 m on a spring of force constant 2000N/m. The
max compression of the srpring is:-
(a) 0.1 m (b) 0.2 m
(c) 0.01 m (d) 0.02 m
14. The kinetic energy of body is increased by 300% its momentum will be increased by
(a) 100% (b) 200%
(c) 300% (d) 400%
15. A particle of mass M is moving in a horizontal circle of radius R under the centripetal
force equal to K/R2, where K is constant. The potential energy of the particle is
(a) K/2R (b) -K/2R
(c) K/R (d) -K/R
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LEVEL - II
1. An object of mass m is tied to a string of length l and a variable horizontal
force is applied on it, which initially, is zero and gradually increases
until the string makes an angle with the vertical. Workdone by theforce F is
(a) mgl(1 sin) (b) mgl(c) mgl(1 cos) (d) mgl(1 + cos )
2. A simple pendulum has a string of length l and bob of mass m. When the bob is at its lowest
position, it is given the minimum horizontal speed necessary for it to move in a circular path
about the point of suspension. The tension in the string at the lowest position of the bob is
(a) 3mg (b) 4mg
(c) 5 mg (d) 6 mg
3. A horse pulls a wagon with a force of 360 N at an angle of 60 with the horizontal at a speed
of 10 Km/hr. The power of the horse is
(a) 1000 W (b) 2000 W
(c) 500 W (d) 750 W.
4. A particle of mass m is fixed to one end of a light rigid rod of length l and rotated in a
vertical circular path about its other end. The minimum speed of the particle at its highest
point must be
(a) zero (b) lg
(c) lg5.1 (d) lg2
5. A particle of mass m is fixed to one end of a light spring of force constant k and unstretched
length l. The system is rotated about the other end of the spring with an angular velocity ,in gravity free space. The increase in length of the spring will be
(a)k
l2m
(b) 2
2
m
m
-k
l
(c) 2
2
m
m
+
k
l(d) none of these
6. A particle of mass m is moving in a circular path of constant radius r such that its centripetal
acceleration acis varying with time as a
c= k2rt2, where k is a constant. The power delivered
to the particle by the forces acting on it is
(a) 2 mk2r2t (b) mk 2r2t
(c)3
1mk4r2t5 (d) 0
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TT
m
M
d
7. In the figure shown, the net work done by the tension when the bigger
block of mass M touches the ground is
(a) + Mgd
(b) (M + m) gd(c) Mgd
(d) zero
8. In the figure, a ball A is released from rest when the spring is at its
natural (unstretched) length. For the block B, of mass M to leave
contact with the ground at some stage, the minimum mass of A must
be
(a) 2 M
(b) M
(c) M/2
(d) a function of M and the force constant of the spring
9. A man pulls a bucket of water from a well of depth H. If the mass of the rope and that of the
bucket full of water are m and M respectively, then the work done by the man is
(a) (m + M)gh (b)
+ M
2
mgh
(c) gh2
Mm
+(d)
+2
Mm gh
10. A simple pendulum having a bob of mass m is suspended from the ceiling of a car used in a
stunt film shooting. The car moves up along an inclined cliff at a speed v and makes a jump
to leave the cliff and lands at some distance. Let R be the maximum height of the car from
the top of the cliff. The tension in the string when the car is in air is
(a) mg (b) mg R
mv2
(c) mg +
R
mv2
(d) zero.
11. A small block of mass m is kept on a rough inclined surface of inclination fixed in aelevator. The elevator goes up with a uniform velocity v and the block does not slide on the
wedge. The work done by the force of friction on the block in time t will be
(a) zero (b) mgvt cos2(c) mgvt sin2 (d) mgvt sin 2
B M
A
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12. 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)2
1kx2 (b)
2
1kx2
(c)4
1kx2 (d)
4
1kx2
13. A bent bow used for shooting an arrow possesses :
(a) Kinetic Energy (b) Potential Energy
(c) Heat Energy (d) Chemical Energy.
14.A block of mass m moving with speed v compresses a spring through distance x before itsspeed is halved. What is the value of spring constant ?
(a) 2
2
x4
mv3(b) 2
2
x4
mv
(c) 2
2
x2
mv(d) 2
2
x
mv2
15. A ball falls under gravity from a height 10 m with an initial velocity v0. Ithits the ground,
losses 50% of its energy in collision and it rises to the same height. What is the value of v0?
(a) 14 m/s (b) 7 m/s
(c) 28 m/s (d) 9.8 m/s.
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LEVEL - 1(CBSE)(REVIEWYOURCONCEPTS)
1. A body of mass 2 kg initially at rest moves under the action of an applied horizontal force of
7 N on a table with coefficient of kinetic fiction = 0.1. Compute the
(a) work done by the applied force in 10 s,
(b) work done by friction in 10s
(c) work done by the net force on the body in 10 s
(d) change in kinetic energy of the body in 10 s.
2. The potential energy function for particle executing linear
simple harmonic motion is given by 2/kx)x(V 2= . Where
k is the force constant of the oscillator. For k = 0.5 N m-1,the graph of )x(V versus x is shown in Figure. Show that
a particle of total energy 1 J moving under this potential
must turn back when it reaches x = 2 m.
3. A body is initially at rest. It undergoes one-dimensional motion with constant
acceleration. Show that power delivered to it at time t is proportional to t.
4. A body constrained to move along the z-axis of a coordinate system is subject to a
constant force F given by
Nk3j2iF ++=
where k,j,i are unit vectors along the x y and z-axis of the system respectively. What is
the work done by this force in moving the body a distance of 4 m along the z-axis?
5. A rain drop of radius 2 mm falls from a height of 500 m above the ground. It falls with
decreasing acceleration (due to viscous resistance of the air) until at half its original height,
it attains its maximum (terminal) speed, and moves with unifrom speed thereafter. What is
the work done by the gravitational force on the drop in the first and second half of its
journey? What is the work done by the resistive force in the entire journey if its speed on
reaching the ground is 10 ms-1
?
6. A particle of mass 0.5 kg travels in a straight line with velocity 2/3axv = where a = 5 12/1 sm .What is the work done by the net force during its displacement from x = 0 to x = 2 m?
7. A large family uses 8 kW of power. (a) Direct solar energy is incident on the horizontal
surface at an average rate of 200 W per square meter. If 20% of this energy can be converted
to useful electrical energy, how large an area is needed to supply 8 kW? (b) Compare this
area to that of the roof of a typical house.
V(x)
x
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8. A 1 kg block situated on a rough incline is connected to a
spring of spring constant 100 Nm-1 as shown in Figure. The
block is released from rest with the spring in the unstretched
position. The block moves 10 cm down the incline beforecoming to rest. Find the coefficient of friction between the
block and the incline. Assume that the spring has negligible
mass and the pulley is frictionless.
9. A pump on the ground floor of a building can pump up water to fill a tank of volume 30m3
in 15 min. If the tank is 40 m above the ground, and the efficiency of the pump is 30% , how
much electric power is consumed by the pump?
10. A bolt of mass 0.3 kg falls from the ceiling of an elevator moving down with an uniform
speed of 7 ms-1. It hits the floor of the elevator (length of the elevator = 3m) and does not
rebound. What is the heat produced by the impact. Would your answer be different if the
elevator were stationary?
11. A particle of mass 0.01 kg travels along a space curve with velocity given by 1msk16i4 + .
After some time, its velocity becomes 1msj20i8 + due to the action of a conservative force.
Calculate the work done on particle during this interval of time
12. A particle moves in a straight line with retardation proportional to its displacement. Show
that its loss of kinetic energy for any displacement x is proportional to x2 .
13. 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 d. If the same body attached to
the same spring is allowed to fall suddenly, what would the maximum stretching in this case
14. A spring obeys Hookes law with a force constant k. It required 4 J of work to stretch it
through 10 cm beyond its unstretched length. Calculate (a) value of force constant k and (b)
the extra work required to stretch it through additional 10 cm.
15. Show that the kinetic energy acquired by a body of mass m in traveling a certain distance
starting from rest, under a constant force is independent of m.
1 kg
37
k = 100 N/m
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LEVEL - II(BRUSHUPYOURCONCEPTS)
1. A ring of mass m slides on a smooth vertical rod ; attached to the ring is alight string passing over a smooth peg distant a from the rod, and at the other
end of the string is a mass M ( > m). The ring is held on a level with the peg
and released :
Show that it first comes to rest after falling a distance : 22 mM
mMa2
.
2. A small body A starts sliding from the height h down an
inclined groove passing into a half - circle of radius h/2 (see
figure). Assuming the friction to be negligible, find the velocity
of the body at the highest point of its trajectory (after breaking
off the groove).
3. An ideal massless spring can be compressed by 1 m by a
force of 100 N. This same spring is placed at the bottom of
a frictionless inclined plane which makes an angle = 300
with the horizontal. A 10 kg mass is released from rest at
the top of the incline and is brought to rest momentarily
after compressing the spring 2 meters.
(a) Through what distance does the mass slide before
coming to rest ?
(b) What is the speed of the mass just before it reaches the spring ?
4. A spring of mass m and stiffness k is fitted to a block of mass
M. The system is moving with a constant velocity v on a
smooth horizontal surface. If the system collides with a wall,
find the maximum compression of the spring before it re-
coils, assuming that the total energy is conserved.
5. A heavy particle hangs from a point O, by string of length a. It is projected horizontally with
a velocity v such that v2 = (2 + 3 ) ag, show that the string becomes slack when it has
described an angle cos-1(-1/ 3 ).
6. A stone with weight w is thrown vertically upward into the air with initial speed v0. If a
constant force f due to air drag acts on the stone throughout its flight.
(a) Show that the maximum height reached by the stone is]w/f1[g2
vh
2
0
+=
(b) Show that the speed of the stone upon impact with the ground is
2/1
0fw
fwvv
+
=
300
mM
v
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7. A heavy particle hangs by a inextensible string of length a from a fixed point and is then
projected horizontally with a velocity )gh2( . If (5/2)a > h >a, Prove that circular motion
ceases when the greatest height ever reaches by the particle above the point of projection is
(4a - h)(a + 2h)2/(27a2).
8. A particle is projected, along the inside of a smooth fixed sphere, from the lowest point, with
a velocity equal to the due to falling freely down the vertical diameter of the sphere. Show
that the particle will leave the sphere and afterwards pass vertically over the point of projec-
tion at a distance equal to (25/32) of the diameter.
9. Show that a particle projected with velocity )ag2( from the lowest point of a vertical
circle of radius a and moving inside it will just reach the end of the horizontal diameter;
while if projected with velocity )ag5( , it will just reach the highest point. Prove that the
reaction at any point in the first case is proportional to the depth below the horizontal diam-
eter and in the second case to the depth below the highest point.
10. An automobile of mass m accelerates, starting from rest, while the engine supplies
constant power P, show that
(a) The velocity is given as a function of time by 2/1)m/pt2(v =
(b) The position is given as a function of time by2/3
2/1
tm9
P8s
= .
11. A particle of mass m moves along a circle of radius R with a normal acceleration varying
with time as 2n bta = , where b is a constant. Find the time dependence of the power developed
by all the forces acting on the particle, and the mean value of this power developed by all the
forces acting on the particle, and the mean value of this power averaged over the first t
seconds after the beginning of motion.
12. Two bars of masses m1and m
2connected by a weightless spring of stiffness k (figure) rest on
a smooth horizontal plane.
1 2x
Bar 2 is shifted a small distance x to the left and then released. Find the velocity of the centre
of inertia of the system after bar 1 breaks off the wall.
13. In a certain two-dimensional field of force the potential energy of a particle has the form
22 yxU += , where and are positive constants whose magnitudes are diferent. Find out :(a) whether this field is central:
(b) what is the shape of the equipotential surfaces and also of the surfaces for which the magni-
tude of the vector of force F = const.
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14. One end of a spring of natural length h and spring constant k
is fixed at the ground and the other is fitted with a smooth
ring of mass m which is allowed to slide on a horizontal rod
fixed at a height h (see figure). Initially, the spring makes anangle of 37 with the vertical when the system is released
from rest. Find the speed of the ring when the spring becomes
vertial.
15. A car is travelling on a level road with speed v0
at the instant when the brakes lock, so the
tires slide rather than roll.
(a) Use the work energy theorem to calculate the minimum stopping distance of the car in
terms of v0, the acceleration of gravity g, and the coefficient of kinetic friction
kbetween
the tires and the road.
(b) The car stops in a distance of 98.3 m if v0
= 90 km/h. What is the stopping distance if
v0
= 60 km/h ? Assume that k
remains the same, so that the friction force remains the
same.
LEVEL - III(CHECKYOURSKILLS)
1. A body with zero initial velocity slips from the top of an inclined plane forming an angle with the horizontal. The coefficient of friction between the body and the plane increases
with the distance l from the top according to the law bl= . The body stops before it reachesthe end of the plane. Determine the time t from the beginning of motion of the body to the
moment when its comes to rest.
2. In the reference frame K two particles travel along the x-axis one of mass m1
with velocity
v1, and the other of mass m
2with velocity v
2. Find:
(a) The velocity v of the reference frame k in which the cumulative K.E. of these particles
is minimum.
(b) The cumulative K.E. of these particles in the k frame
3. A thin rim of mass m and radius r rolls down an inclinedplane of slope , winding thereby a thin ribbon of lineardensity (shown in the figure). At the initial moment, the rim
is at a height h above the horizontal surface.
Determine the distance s from the foot of the inclined plane
at which the rim stops, assuming that the incline plane
smoothly changes into the horizontal plane.
4. A small particle of mass m initially at A (see Figure) slides down a frictionless surface
AEB. When the particle is at the point C, show that the angular velocity and the force
h
r
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exerted by the surface are
r
sing2 = and = sinmg3F
5. A chain AB of length l is loaded in a smooth horizontal tube so that
its fraction of length h hangs freely and touches the surface of the
table with its end B. At a certain moment, the end A of the chain is
set free, with what velocity will this end of the chain slip out of the
tube?
6. Figure shows a smooth track, a part of which is a circle of a radius
R. A block of mass m is pushed against a spring constant k fixed
at the left end and is then released. Find the initial compression of
the spring so that the block presses the track with a force mg when
it reaches the point P, where the radius of the track is horizontal.
7. Three identical spring A, B and C each of natural length l and
spring constant K are connected to a point mass m as shown in the
figure. A and B are horizontal and C is vertically fixed with rigid
supports. What is the work done by the external agent in slowly
lowering the mass m till it attains equilibrium when the springs A
and B make an angle 2 sin1 3/5 between them. Neglect the masses
of the springs.
8. A small sphere tied to the string of length 0.8m is describing a vertical circle so that the
maximum and minimum tensions in the strings are in the ratio 3:1. The fixed end of the
string is at a height of 5.8m above ground.
(a) Find the velocity of the sphere at the lowest position.
(b) If the string suddenly breaks at the lowest position, when and where will the sphere hit
the ground? (take = 10 m/s2)
9. A body of mass m was slowly hauled up the hill. (fig) by a
force F which at each point was directed along a tangent to thetrajectory. Find the work performed by this force, if the height
of the hill is h, the length of its base l, and coefficient of
kinetic friction k.
10. A small bar resting on a smooth horizontal plane is attached by
threads to a point P (fig.) and by means of a weightless pulley, to
a weight B possessing the same mass as the bar itself. Besides,
the bar is also attached to a point O by means of a light non-
deformed spring of length l0
= 50 cm and the stiffness
K = 5 mg/l0
, where m is the mass of the bar. The thread PA having
E
m
l
h
B
l0
P
O
A
m
BA r
E
C
A
h
B
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m
O
l0
m m
l
N
F
mg
R
B /4
m1
been burned, the bar starts moving. Find its velocity at the moment when it is breaking off
the plane.
11. A horizontal plane supports a plank with a bar of mass m = 1.0kg placed on it and attached by a light elastic non-deformed
cord of length l0
= 40 cm to a point O (fig.). The coefficient of
friction between the bar and the plank equals k = 0.20. The
plank is slowly shifted to the right until the bar starts sliding
over it. It occurs at the moment when the cord deviates from
the vertical by an angle = 300. Find the work that has beenperformed by that moment by the friction force acting on the
bar in the reference frame fixed to the plane.
12. A block of mass m connected with a light spring of natural
length l0
and stiffness k as shown in the figure. The spring
is relaxed the block is pulled by an external force slowly.
Find the work done by the external agent till the block
break off the surface.
= 1
0kl
mgGiven .
13. The figure shows a ball A of mass m connected to a light spring
of stiffness k. Another identical ball B is connected with the ball
A by a light inextensible string as shown in the figure. Other end
of the spring is fixed. Initially the spring is in relaxed position.A vertical force F acts on B such that the balls move slowly.
What is the work done by the force in pulling the ball B till that
ball A reaches at the top of the cylindrical surface the ball A
remains in contact with the surface and coefficient of friction
between the surface and the ball A is .
14. A weightless horizontal rigid rod along which two ball of the
same mass m can move without friction rotates at a constant
angular velocity about a vertical axle. The ball are connectedby a weightless spring of rigidity k, whose length in the
undeformed state is l0. The ball which is closer to the vertical
axle is connected to it by a similar spring. Determine the lengths of the springs. Under what
conditions will the balls move in the circle ?
15. Two blocks P and Q of mass 2 m and m respectively are
connected by a massless string and are at rest as shown in
figure all pulleys are ideal and the surface is frictionless. Find
the velocity of the block P at point A and B when the system
is released from rest. [at A, thread from P to pulley is vertical]
A
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LEVEL - IV (IIT-JEE PROBLEMS)
(JUDGEYOURSELFAT JEE-LEVEL)
1. A particle is suspended vertically from a point O by an inextensible massless string
of length L. A vertical line AB is is at a distance of LS from O as shown. The object
is given a horizontal velocity u. At some point, its motion ceases to be circular and
eventually the object passed through the line AB. At the instant of crossing AB, its
velocity is horizontal. . Find u. [I.I .T. 1999]
2. A cart is moving along x direction with a velocity of 4 m/s. A person on the cart throws a stone with
a velocity of 6 m/s relative to himself. In the frame of reference of the cart, the stone is thrown in y-
z plane making an angle of 300 with vertical z-axis. At the highest point of its trajectory, the stone
hits an object of equal mass hung vertically from branch of a tree by means of a string of length L.
A completely inelastic collision occurs in which the stone gets embedded in the object. Determine
(a) The speed of the combined mass immediately after the collision with respect to an observer
on the ground.
(b) The length L of the string such that the tension in the string becomes zero when the string
becomes horizontal during the subsequent motion of the combined mass.
[ I.I .T. 1997 New]
3. A spherical ball of mass m is kept at the highest point in the space
between two fixed, concentric spheres A and B (see Figure). Thesmaller sphere A has a radius R and the space between the two
spheres has a width d. The ball has a diameter very slightly less
than d. All surfaces are frictionless. The ball is given a gentle
push (towards the right in the figure). The angle made by the
radius vector of the ball with the upward vertical is denoted by (Shown in the figure).
(a) express the total normal reaction force exerted by the spheres on the ball as a function of
angle .(b) let N
Aand N
Bdenote the magnitudes of the normal reaction forces on the ball exerted by the
spheres A and B, respectively. Sketch the variations of NA and NB as functions of cos in therange 0 < by drawing two separate graphs in your answer book, taking cos on thehorizontal axes.
[I.I .T. 2002]
L
L8
O A
u
B
RO
d
Sphere B
Sphere A
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SOLUTION (OBJECTIVE)
LEVEL - I
1. (c)2. (c)
3. (b)
4. (b)
5. (a)
6. (a)
7. (d)
8. (d)
9. (c)
10. (a)
11. (c)12. (b)
13. (a)
14. (a)
15. (c)
LEVEL - II
1. (c)
2. (d)
3. (c)
4. (a)
5. (b)
6. (b)
7. (d)
8. (c)
9. (b)
10. (d)
11. (c)
12. (d)
13. (b)14. (a)
15. (a)
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SOLUTION (SUBJECTIVE)
LEVEL - I (CBSE)
1. (a) 882 J (b) -247 J
(c) 635 J (d) 635 J Work done by the net force on a body equals change in its K.E.
4. 12 J
5. 0.082 J in each half, 0.163 J
6. 50 J
7. (a) 200 m2 (b) comparable to the roof of a large house of dimension 14 m 14 m
8. 0.125 9. 43.6 KW
10. 8.82 J for both cases.
11. 0.96 J 13. 2 d
14. (a) 800 N (b) 12 J
LEVEL - II
2. v =3
gh
3
2
3. (a) 4 m (b) s/m52
4. vk3
mM3 +
11. P = mRat
=2
mRat
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12. )mm/(kmxv 212c +=
13. (a) No;
(b) ellipses whose ratio of semiaxes is = /b/a ; also ellipses, but with a/b = / .
14.m
k
4
h