Study Material Force & Laws of Motion for AIEEE

7/29/2019 Study Material Force & Laws of Motion for AIEEE

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MODULE 2::NEWTONS LAWS OFMOTION

Syllabus : Newton's laws of motion, Inertial and uniformly accelerated frames of reference; Static and

dynamic friction.

INERTIA :

The property by virtue of which a body retains its state of rest or of uniform motionalong a straight line. For example, we use the word pull the rug; when the rug under someones feet is pulled, his feet move along with the ring where as other part retains itsstationary state and the man falls

MASS :

It is the measure of translational inertia i.e larger the mass, greater the inertia of a

body and to change its state a greater external force has to be applied.

NEWTON'S FIRST LAW :

If the net force acting on a body in a certain direction is zero, then the inertia of thebody in that direction is retained. i.e. if the body is at rest it will not move and if it is movingwith a constant speed in a certain direction it will not change either the direction or themagnitude of speed in that direction. The body will not accelerate in either of the cases.

FORCE :

Broadly, forces can be divided into two categories Field forces Contact forces.Field forces are gravitational, electrostatic, electromagnetic etc., and contact forces arenormal reaction, tension, friction etc. We will discuss a few of them, which are important inthe present context.

Gravitational force at a point is Mg where M is the mass of the body and g is theacceleration due to gravity at the point and it is directed towards the center of the earth nearthe earth.

TENSION : (a) In a string: If a string is tied to a body and is taut then tension acts on thebody. The number of tensions acting on the body is the number of segments joined to thebody. The direction of the tension on the body is along the string away from the point ofattachment to the body.

Illustration 1 :Find the number oftensions acting onthe body m, wedgeM and the wall.

Solution :Number of points of attachment to M is 3

Number of segments of thread attached to 1 is 2,Number of segments of thread attached to 2 is 2 ,

AB

51

H

G

FE

C

43

2

D

I

mM


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Number of segments of thread attached to 3 is 1 ,

So number of tension acting on M is 5

At 1. one horizontally right (A) and one down the inclined plane (B)

At 2. one up the inclined plane (C) and one vertically down wards (D)

At 3. one towards right EOn the smaller block only one segment of thread is connected hence the tension is

only one and is vertically upwards. On the wall there are two points of attachment 4 and5

At 4. one vertically upwards G and one horizontally left F

At 5. one horizontally leftwards I and the other vertically down wards H

B) Tension in spring: In the spring, if the spring is extended then thetension is along the spring and away from the point of attachment tothe body. And if the spring is compressed the tension is towards thepoint of attachment to the body.

C) Tension in the rod: Tension in the rod is same as that in springs.

NORMAL REACTION: It is a surface force. Normalreaction acts at right angles to the tangent at the point of

contact between two bodies. They act on different bodies andare equal and opposite to each other.

FRICTION: The force which opposes relative motion

between two surfaces in contact. It is a self adjusting force.It's magnitude and direction changes in relation to themagnitude and the direction of the applied force (we will bediscussing in further detail about friction later) It is equal tothe force applied on it parallel to the common surface aslong there is no relative motion between the surface only, when the bottom surface is static

a = 0

Illustration 2 :

All the surfaces are rough. Find the direction of friction on M1 , M2 and M3. asshown in the figure.

Solution :

As force is applied on M1, it will try to move towards right. Due to inertia M3 andM2 will be at rest. Surface of M3 w.r.t M1 will

be moving towards left. Therefore the force offriction acts on the mass M3 towards right.Looking from M3 towards surface 31, itsmotion w.r.t M3 is towards right. So, force offriction on surface 31 of M1 is towards left. On

M2

force of friction is towards right. On surface 12 at M1

is towards left.

`

2N

2N1N

1N

3M

2M

1M31f31f

21f12f

3M

2M

1M

kx

kx


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Free Body diagram :-

A free body diagram consists of a diagrammatic representation of all theforces acting on a single body or a subsystem of bodies isolated from its surroundings.

The forces may lie along a line, may be distributed in a plane ( Coplanar) ormay be distributed in space ( non - Coplanar).

Illustration 3 :A thick book is pressed against a wall at the same time a horizontal forceparallel to the wall is applied on the book. Find the direction of friction.

Solution :The forces that produce a tendency to make the bodymove are the weight of the book and the horizontal force.Therefore the resultant direction of motion will be alongthe direction of resultant force. So friction on the book will

be opposite to the resultant force.EQUILIBRIUM: If the net force acting on a body in a certaindirection is zero, then the body continues to move with constantvelocity in that direction or remains at rest in that direction. We say that the body is undertranslated equilibrium when

==

==

0a0f

0a0f

yy

xx

Illustration 4 :A block of mass M = 20 kg hangs by a cord from a knot of negligible mass, which hangs

from a ceiling by means of two other light cords. What are the tensions in the cords?

Solution :

Putting equilibrium condition along x-axis. F.B.D of k

xF = T2 cos 53 - T1 cos 37 = 0

T2 cos 53 = T1 cos 37

T2 =3

4 1T ..(i)

Again putting equilibrium condition along Y- axisT1 sin 37 + T2 sin 530 - T = 0 and T - Mg = 0

T1 sin370 + T2 sin530 = Mg

T1 (0.6) +3

4T1 (0.8) = 200 (ii)

T1 (0.6 +3

23.) = 200

T1 = Nx

1205

3200=

T2 =3

4T1 = 160 N

F

Mg

F

Mg

Friction

tsul tanRe

1T 2T

T

K

20kgM =

0

5 3

0

3 7


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Moment of a force :The moment of a force (or) Torque about any point (or) axis is given as cross product

of position vectorr and force vector

F

=Fxr

= r F Sin , where '' is the angle betweenFandr

= Fr

If the line of action of a force passes through the axis of rotation, itsperpendicular distance from the axis is Zero.

its Torque about that axis is Zero.If = 0 , the body is under rotational equilibrium.

Illustration 5 :ABCD is a square of side 2m and O is its centre. Forces act along the slides asshown in the diagram. Calculate the moment of each force about

a) the point Ab) the point O

Solution: Take clock wise moments as negative(a) Moment of forces about A.

Magnitude of force 2N 3N 4N 5N

ar distance from A 0 0 2m 2mMoment about A 0 0 8 Nm -10Nm

(b) Moment of forces about O

Magnitude of force 2N 3N 4N 5N

ar distance from O 1m 1m 1m 1mMoment about O + 2N - m - 3N- m +4 Nm -5Nm

MOMENTUM :

Momentum is the measure of motion contained in a body. It is given by

= VmP .Where 'm' is the mass and V

is the velocity of the body.

NEWTON'S SECOND LAW :

First law tells us what happens to a body when no external force or zero force acts onit. Second tells us what happens to a body when an external force acts on the body.

When a force or a net force acts on a body along a particular direction, then thereoccurs a change in the momentum of the body in that direction. The rate of change ofmomentum of body in that direction is equal to the applied force in that direction.

dtPdF

= = dtd

vm = m =

dtvd

am


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This implies that acceleration of the body is in the direction along which force isapplied.NEWTON'S THIRD LAW :

For every force applied on a body by an agent, an equal and opposite force acts on

the agent by the body. This is the law of action and reaction.

Applications of Newton's Laws of Motion :

Although each problem based on Newton's laws requires a unique approach thereare a few general rules that are applied in setting up the solutions to all such problems. Acareful study of the illustrations can lead us to a convincing approach of solving problems.

The basic steps are:

Clearly identify the body to be analyzed. There may be more than one body;If they have same acceleration in same direction they may be taken as a singlesystem, otherwise they may be treated separately.

Identify the environment that exert force on the body e.g. surfaces, springs,cords, gravity etc.

Select a suitable inertial (non accelerating) reference frame

Choose co-ordinate system of your convenience from the context of theproblem. Take separate co-ordinate systems for different bodies and systemsof bodies.

Take the body as a point mass and put all the forces vectorially on that point.This makes the free body diagram of the body.

Decompose the forces along the co-ordinate axes of the co-ordinate system

And put Fx = max Fy = may

where Fx = sum of the x-components of all the forces acting on the body and Fy isthe sum total of the Y- components of all the forces acting on the body, ax and ay are theaccelerations of the body along the respective directions.

Application of Newton's First Law :

Illustration 6 :A block of mass M is lying on a rough surface. The force F is applied on the

block at an angle to the horizontal as shown in the figure. Find the normalreaction on the block and the force of friction on the block and hence netforce on the surface of the block, if the block does not move.

Solution :

The tendency of motion is towards right, friction is towards left.Setting the co-ordinate axes as shown in the diagram and putting theequilibrium condition.

Along X-axis Fx = Fcos -fr = 0 fr = Fcos . (1)

along Y-axis Fy = Fsin + N - Mg= 0

N = Mg - Fsin

The net surface force = ( ) ( )22 rfN +

F

F

M

sinF

cosFMg

fr

Y

X

N

blocktheof.D.B.F


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= ( ) ( )22 + cossin FFMg

= ++ 222222 2 cossinsin FFMgFgM

= + sinFMgFgM 2222

NEWTON'S SECOND LAW :

Illustration 7 :

On a smooth fixed wedge two masses M1 and M2 are lying asshown in the figure. Find the acceleration of the masses.

Solution :

From F.B.D of M1 F B D of M1 F.B.D of M2

(Assuming M1 moving downwards)

FY = M1 g - T = M1 aY ----- (1)From F.B.D of M2

== xx aMgMTF 22 sin -------(2) == 02 cosgMNFY N = M2 g cos ------------ (3)

But as the masses are tied to the inextensible string, aY of mass M1 = ax of mass M2 = a

M1 g - T = M1 a

T - aMgM22

=sin

-------------------------------------Adding gM

1

- ( )212

MMagM +=sin

--------------------------------------

a =( )

21

21

MM

sinMMg

+

a is positive if M1 - M2 sin > 0;

a is negative if M1 - M2 sin < 0

a = 0 if M1 - M2 sin = 0

Illustration 8 :Two blocks M1 and M2 are placed on a smooth inclined surface in contact

with each other and a force F parallel to the plane is applied on1

M . Find

the

a) acceleration of the blocks,

b) the contact force between the blocks.

1M

2M

xy

x

y

1M2M

F


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Solution :

We can consider M1 and M2 separately or together because they will have same commonacceleration

Now

( ){ }+ sin.gMMF1

= ( )21 MM + aX

a = aX =21 MM

F

+- gsin

Considering separately, aMgMN 22 sin' =

[ N' = contact force]

aMgMN 22 sin' +=

++= sing

MM

FMsingM

2122

21

2

2122

2

221

MM

gMgMMFMgMgMM

+

++=

sinsinsinsin

21

2

MM

FM

+=

FX = F - N' - M1 g sin = M1 a -----------(1)

FY = N1 - M1 gcos = 0

N1 = M1 g cos ---------- (2)

FX = N' - M2g sin = M2 a ---------- (3)

N2 - M2 g cos = 0

N2 = M2 g cos ---------- (4)

From (1) and (3)

a =( )

21

2

21

21

MM

FMNand

MM

gMMF

+=

+

+'

sin

N( )21 MMofD.B.F +

( )gMM 2+

X

YF

N

F

gM1

xY1N

1

MofDBF ...2N

x

'N

gM2

2MofFBD

Y


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CONSTRAINT RELATIONS :

When two-particle move under certain conditions, aconstraint equation is obtained. When two particles connected byan inextensible taut string move, the components of the velocitiesparallel to the string must be same to avoid the breaking of the

string.

VA cosA = VB cosB

When two bodies movewithout losing contact witheach other, the components ofthe velocities perpendicular tothe surface of contact must besame

VA = VB -VA sin = VBx sin - V By cosAnother constraint relation is developed from the fact that the length of a string is

constant which is used in pulley problems. constraint equation in velocities can beconverted into constraint equation in displacements and accelerations also. In solvingphysics problems, if the number of equations developed is less than the number of variablesassumed, immediately look for any constraint relationship available from the givenconditions.

Illustration 9 :

Two blocks A and B connected by a string and apulley move on a smooth horizontal floor as shown inthe figure. Develop the constraint equation betweenthe velocities of the blocks and the pulley

Solution :

Since the length of the string is constant, the rate at which the length of the stringdecreases between A and P must be equal to the rate at which the length of the stringincreases between P and B.

Approach velocity of A = - Approach velocity of B

- VA cos 450 - VP cos 600 = VB cos300 + Vp cos 600

A

string B

B

BVAV

A

r

contacofsurface

BxV

ByV

90

90

AV

pV

BV

AV

A P

B

060

045

030

A By

V

BxV

B

AV


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FRICTION

As discussed earlier, it is the force which opposes the relative motion between twosurfaces in contact and is a self adjusting force. Broadly the force of friction is of two types1) Static friction 2) Kinetic friction.

STATIC FRICTION :

Static friction is the frictional force which comes into play when there is no relativemotion between surfaces, but there is a tendency for motion. It is a variable force. Itincreases with the force applied on the body, parallel to the surfaces in contact, to a certainmaximum value called limiting friction.

Properties of friction :

When there is no relative motion between two surfaces in contact, then the

static frictional force sf and the component of the forceF that is applied on

the body, parallel to the common surfaces balance each other. The force fs isequal to the component and directed opposite to the component of F.

The magnitude of fs has a maximum value fs (max)= sN, where s is the co-

efficient of static friction and N is the magnitude of the normal force on thebody from the surface If the component of force applied exceed the limitingvalue of friction then the relative motion

between the surfaces begins.

If the body begins to slide along the surface, themagnitude of the frictional force rapidly

decreases to a value fk given by fk = k N, wherek is the coefficient of kinetic friction. If themagnitude of the friction force plotted againstthe force is applied on the body parallel to the surface, we obtain a graph as

shown.

Illustration 10 :A force F is applied on the block as shown in the figure. For whatmaximum value of F a) applied on M b) applied on "m " both the

blocks move together

Solution :(a) When F is applied on 'm'To make both the blocks move together both the blocks should have commonacceleration. External force on the m is F

a =mM

F+

m

M

F


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But M is accelerated at a =( )mM

F

+by the frictional force acting on it.

For a maximum value of F, acceleration is also maximum and force of friction shouldalso be maximum

fs = Ma

mg = M.mM

F

+

F =( )

M

gmmM +

(b) When F is applied m

Then a =mM

F

+

mg = M. mM

F

+ For acceleration of m

F = M+m g

Illustration 11 :A block of mass M is lying on a rough surface and a forceF is applied on it as shown in the figure. The co-efficientof static friction and kinetic friction on the blocks are sand k respectively.a) If the block is at rest find the force of friction on the

block.b) If the block is in motion, find its acceleration

Solution :a) If the block is at rest, net force on it along the horizontal is

zero

fr + F cos = 0 fr = - F cos

b) If the block is in motion, the friction force acting on the

block will be NNow N + Fsin - Mg = 0

N = Mg - F sinForce of friction N = (Mg - F sin)Ma = ( ) sincos FMgF

( ) gM

Fa += sincos

MatFBD

rF

Mg

F

F

M

ks ,


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UNIFORMLY ACCELERATED FRAMES OF REFERENCE AND PSEUDO FORCE.

Suppose an observer fixed to smooth flat car. A box is lying on the car. An observeris standing on the ground. Let the car accelerate towards right. As thecar accelerates horizontally due to smoothness of the surface the box doesnot get any horizontal force but the observer fixed to the car accelerates.For the ground observer the box is at rest, but for the observer on the cartthe box is accelerating towards him. For the observer in the acceleratingframe the box accelerates with out any force and Newton's second Lawfails.

Therefore to make the Newton's second Law applicable, weassume a force, called a pseudo force acting on the bodies in acceleratingframes or non inertial frames. The direction of the force is opposite to thedirection of acceleration of the frame and the magnitude of the force isthe product of mass of the body m and the magnitude of the acceleration

of the frame. Pseudo Force = mass of the body X acceleration of the noninertial frame.

MATHEMATICALLY

Let us discuss the motion of a particle (P) from two frames of references S and S'. S is aninertial frame and S' is a non inertial frame. At an instance position vectors of the particle

with respect to those two frames are

r and

1r respectively. At the same moment position

vector of the origin of S' is

R with respect to S as shown in the figure.From the vector triangle OO'P, we get

= Rrr'

Differentiating this equation twice withrespect to time we get

=

Rdt

d

dt

rd

dt

r'd2

2

2

2

2

2

= Aaa'

Here

a' = acceleration of the particle Prelative to S'

a = Acceleration of the particle relative to S

A = Acceleration of S' relative to S.Multiplying the above equation by m (mass of the particle) we get

= Aaa' mmm ( )real

= FF' -

Am

( )

+=

AFF' mreal

In non-inertial frame of reference an extra force is taken into account to applyNewton's laws of motion. That extra force is called Pseudo force.

O

'O

'O

O

PY

O

S r

R

r'

S'

X

O'X'

Y'


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Illustration 12 :

A block 'm' is lying on the smooth surface of a wedge as shown in thefigure. What should be the acceleration of the wedge such that there isno sliding ofm on M

Solution :From wedge frame, a Pseudo force acts on the block 'm'horizontally towards left.Now drawing the F.B.D of 'm'The block is at rest w.r.t the wedge. Putting equilibriumcondition down the plane.

mg sin - ma cos = 0 a = g tan

Force acting = (M+m) g tan

Considering from ground frame. The block onthe wedge as well as wedge accelerate towards right with same acceleration 'a'.

Illustration 13 :A car is speeding up on a horizontal road with an acceleration 'a'. A ball is suspended fromthe ceiling through a string and it maintains a constant angle with the vertical Find thisangle.

Solution :Considering from the car frame a pseudo force acts on the ball leftwards, and is under equilibrium w.r.t the car.Putting equilibrium condition along and perpendicular the string.

T = mg cos + ma sinand ma cos - mg sin = 0

a = g tan, = tan-1g

a

From ground frame, the bob accelerates with acceleration a along thehorizontal and along the vertical it is at rest or under equilibrium

T cos = mgT sin = ma

a = g tan = tan-1g

a

UNIFORM CIRCULAR MOTION :As discussed in Kinematics, when a body moves in a circular path even though the

magnitude of the speed of the body does not change, due to the change of direction ofvelocity the acceleration of the body is directed towards the centre. For a constant speed V

and radius of curvature of the circular path 'R', the acceleration is given byR

va

2

R = .

Therefore the force acting on the body towards the centre of the circular path as observed

x

ymg

ma

N

mg

ma

M

m

M

m

N

mg

m''onF.B.D.

asinN m=gcosN m=

gtana =

mg

ma

T


13/46

from the ground is the centripetal force =R

mv2or the radial force FR =

R

mv2. If the moving

body is tangentially accelerating in addition to the radial force there will be a tangential

force. If the tangential acceleration is Ta then, the force acting in the body will be

2

T

2

R aamF +=

= m ( )2T

22

ar

mv+

Illustration 14 :The bob of a simple pendulum of length 1 m has a mass 100g and speed of 1.4 m/s at thelowest point in its path. Find the tension in the string at this instant.

Solution :

Drawing the free body diagramThe force acting on the particle along the vertical is

MgTR

Mv2=

T = Mg +L

Mv2= M

+

L

vg

2

= 1.2 N

Illustration 15 :

A 4 kg block is attached to a vertical rod by means of two strings. When thesystem rotates about the axis of the rod, the strings become taut as shown in thefigure.a) How many rev/min must the system make in order that the

tension in the upper cord shall be 60 N?b) What is then the tension in the lower cord ?

Solution :

Let be the angular velocity of rotation of rod.

Along vertical

coscos 21 TmgT += (1)Net force towards center (Centripetal force) substituting the values of cos, sin, r and

m2

m.251

m.251

sinT1

sinT2

cosT1

cosT2

mg

1T

mg

2T


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rmTT 221 sinsin =+ (2) NT 601 = , m = 4kg in (1) and (2), we get;(II) = 3.74 rad/s;

Given : cos =5

4

25.1

1= n = /(2) = 0.6rev/s and

sin =5

3T2 = 10N

r = 1.25 sin = 0.75m

CIRCULAR TURNINGS AND BANKING OF ROADS :

When a vehicle negotiates a curve to do so the vehicle requires a centripetal force.For horizontal road the normal reaction on the body is vertical. The force towards centre is

provided by the friction between the road and the tyre of the vehicle. s

2

fR

Mv =

For a maximum value ofv, fs = N v = Rg

For a greater value of v friction force is insufficient to provide the requiredcentripetal force. Therefore the normal reaction is adjusted to provided therequired extra centripetal force. This is done by raising the outer part of the road.This is called banking of roads.

On a banked road when the speed of a vehicle exceeds a certain value, it tends toskid away from the centre of the circle and the speed is less than a certain value it tends toskid down. At a certain value of the speed the tyres roll with out skidding side wise.

Centrifugal Force :

Centrifugal force is a Pseudo force. By including this force one can use Newton's laws ofmotion in a rotating frame of reference. The magnitude of the Pseudo force is the product ofmass of the body and acceleration of the point at which the body is placed. The direction ofthe force is away from the centre of the rotating force.

* * * *

1.25

1.25

l

r


15/46

WORKED OUT OBJECTIVE PROBLEMS

EXAMPLE : 01

Find the mass of the suspended body, F1 and F2 if the tension in the

inclined string is 20 2 N

a) 2 kg , 20 N, 20N b) 2 2 kg, 20N, 20N

c) 2 kg, 20 2 N, 20Nd) 2 kg, 0N, 0N

Solution :

For the body

T1 = mg

At point AT2 sin45 = T1

T1 = 20 2 x2

1= 20 N

Again T2 cos = F1 = 20 NSimilarly At B

F2 = T2 sin = 20 N

EXAMPLE : 02

A Fireman wants to slide down a rope. The breaking load of the rope is 3/4 th of weightof the man. With what minimum acceleration should the man slide down?

a) g b) g/2 c) g/4 d) g/3

Solution :

On any small part of the rope

T - f = (M)rope a = 0

T = f

Now for the man

Mg - f = Ma

Mg -4

3Mg= ma

a = g/4

EXAMPLE : 03

A balloon is descending with constant acceleration a, less than the acceleration due togravity 'g'. The mass of the balloon with its basket and contents is M.What mass M' fromthe balloon should be released so that the balloon accelerates upward with constantacceleration 'a'?

a)ag

Ma

+

2b)

( )

a

agM +c)

( )

a

agM d)

a

Mg

045

1F

2F

3T

2FB

2T

1F

1T

A

1T

1F

2T1T

mg

Tf =

mg


16/46

Solution :The balloon is accelerating under the up thrust of air and its own weight. ThenMg - U = Ma (1)As the size of the balloon is not changing, up thrust on it remains constant

Then (M-M1

) g - U = - (M - M1

) a . (2)Solving we obtain (1) and (2)

M1 =ag

Ma

+2

EXAMPLE : 04

In the given figure all the surfaces are smooth. F is parallel to the inclined surface. Thecontact force between the masses is

a)mM

MF

+b)

mM

mF

+

c)( )

M

FmM +d)

( )

M

FmM +

Solution :

From (1) and (2)N + Mg sin - F = Ma ..(1)

mgsin - N = ma .(2)-------------------------------------------

(M +m)g sin - F = (M+m)a-------------------------------------------

a = g sin -mM

F

+ N = mg sin - ma

= mg sin - m .

+

mM

Fsing =

mM

mF

+

EXAMPLE : 05A light spring is compressed by0

x and placed horizontally between a

vertical fixed wall and a block which is free to slide over a smooth horizontaltable top as shown in the figure. If the system is released from rest, whichof the following graphs represents the relation between acceleration 'a' of theblock and the distance x traveled by it?

a) b) c) d)

a

O x

a

O x

a

O x

a

O x

M

m

MofF.B.D

Mg

N

F

F

mofF.B.D

mg

NF


17/46

Solution :

At maximum compression, force on the block is maximum andhence the acceleration maximum.

( )xxKF +=0

as F and x are along the same direction. a = - xMK F = K (x0 -

a = +M

K(x0 - x) a =

M

Kx0 +

M

Kx

Hence the graph is a straight line with negative slope and an interceptM

Kx0 on

acceleration axis.

EXAMPLE : 06

wo blocks A and B attached to each other by a mass less spring are kept on a roughhorizontal surface (=0.1) and pulled by a horizontal force F=200 N applied to the blockB. If at some instant, the 10kg mass A has an acceleration 11m/s2, what is the accelerationof the 20 kg mass B

a) 2.5 m/s2 b) 3 m/s2 c) 3.6 m/s2 d) 1.2 m/s2

Solution :

Form M1

T - f1 = M1 a1

T = M1 a1 + M1 g (1)

For M2, F - M2 g - T = M2 a2

( )

2

1112

M

gMaMgMF += a2

=( )20

1010101110102010200 xx.xxx. +

2

a = 3 m/s2

EXAMPLE : 07

A cart of mass M has a block of mass " m " in contact with it is as

shown in the figure. The coefficient of friction between the block and

0xx

1f

1N

T

gM1

kgM 101 = kgM 202 = N200A B

2MorF

2N

2FT

g2m

F

M m

a


18/46

the cart is . What is the minimum acceleration of the cart so that the block m does notfall

a) g b) g/ c) /g d)M

gm

Solution :

The mass " m " experiences a Pseudo force back wards and it is under equilibrium along thehorizontal from the cart frame (non - inertial). It is under equilibrium due to normal reactionfrom the cart and the Pseudo force. Along vertical it is under equilibrium due to mg and thefriction between m and M. For minimum acceleration friction between the surfaces should

be maximum i.e. limiting one

Now N - ma = O (1)

mg = N .(2)

a =g

( Think from ground frame )

EXAMPLE : 08

The coefficient of friction of all the surfaces is The string andthe pulley are light . The blocks are moving with constant speed.Choose the correct statement

a) F = 9 mg b) T1 = 2 mg c) T1 = 3 mg d) T2 = 4 mg

Solution :

For block 2m

T1 - (2mg) = 0

T 1 = 2 mg ..(1)

For block 3 m

f1 + T1 + f2 - F = 0

F = 2mg + 2 mg + 5 mg

F = 9 mg

T2 = 2T1 = 4 mg

EXAMPLE : 09

A uniform circular ring of mass per unit length and radius R is rotating with angularvelocity '' about its own axis in a gravity free space Tension in the ring is

1f

F1T2f

2N

1N

3mg

3mofF.B.D

1T

2T

F

A

m2

m3B


19/46

a) Zero b) 1/2 R22 c) R22 d) R 2

Solution :

Consider a small elemental part of the ring which makes at the center angle .

Then the small part is moving in a circular path with angular speed '

'. Thenecessary centripetal force is provided by the component of the tension towardsthe centre.

i.e. dm R2 = 2 T sin/2

As is very small sin /2 = /2

dm = R

R R2 = T

T = R22

EXAMPLE : 10

A single flexible wire P R Q fixed at P and Q passes through a smoothring R. which revolves at a constant speed in the horizontal circle ofradius 'a' as shown in the figure. What is the speed of the revolution.

a) ag b) ag2 c) ag5 d) ag3

Solution :As the smooth ring is in circular motion, the centripetal force is provided by the c

T cos0

53 + T cos 370 = a

Mv 2

(i)

Along the vertical the ring is under equilibrium.Hence T sin530 + Tsin 370 = Mg (ii)

From (i) and (ii) v = ag

SINGLE ANSWER TYPELEVEL - 11. A wedge is moving horizontally with uniform acceleration a towards right and a

block of mass m stays freely at rest on its smooth inclined surfaceas shown in fig. What is the acceleration of the wedge ?

A) g cot B) g cos C) g sin D) g tan

2. Three forces start acting simultaneously on a particle moving with velocity v .These forces are represented in magnitude and direction by the three sides of atriangle ABC(as shown). The particle will now move with velocity

A) less than v B) greater than v

a

2mv

mg

T

T

mgtheofF.B.D

a

P

Q

R

037

053


20/46

C) |v| in the direction of the largest force BC D) v , remaining unchanged.

3. A man slides down a light rope whose breaking strength is times his weight ( mg C) < mg D) either (b) or(C) depending on the acceleration of the chain.

5. Two blocks of masses m1 and m2 are placed in contact with each otheron a horizontal platform. The coefficient of friction between theplatform and the two blocks is the same. The platform moves with anacceleration. The force of interaction between the blocks is

A) zero in all cases B) zero only if m1 = m2C) nonzero only if m1 > m2 D) nonzero only if m1 < m2

6. A car starts from rest to cover a distance s. The coefficient of friction between the

road and the tyres is . The minimum time in which the car can cover the distance isproportional to

A) B) C) 1/ D) 1/

7. A car is moving in a circular horizontal track of radius 10m with a constant speed of10m/s. A plumb bob is suspended from the roof of the car by a light rigid rod oflength 1m. The angle made by the rod with the vertical isA) zero B) 300 C) 450 D) 600

8. A car moves along a horizontal circular road of radius r with velocity v. Thecoefficient of friction between the wheels and the road is . Which of the followingstatements is not true ?

A) The car will slip if > rg B) The car will slip if rg

v2

D) the car will slip at a lower speed, if it moves with some tangential acceleration,than if it moves at constant speed.

9. A block rests on a rough floor. A horizontal force which increases linearly with time

(t), begins to act on the block at t = 0. Its velocity (v) is plotted against t. which of thegiven graphs is correct ?

10. A man running along a straight road leans a little in the forward direction. The anglebetween the vertical line and the line joining the man's center of gravity with the

point of support is 0. For the man not to slip, the coefficient of friction satisfies thecondition given by


21/46

A) tan 0 B) tan 0 C) tan20 D) tan2011. A body rests on a rough horizontal plane. A force is applied to the body directed

towards the plane at an angle with the vertical. If is the angle of friction then forthe body to move along the plane

A) > B) < C) = D) can take upany value12. In the shown system, m1 > m2. Thread QR is holding the system. If this thread is

cut, then just after cutting.A) Acceleration of mass m1 is zero and that of m2 is directed upward.B) Acceleration of mass m2 is zero and that of m1 is directed downwardC) Acceleration of both the blocks will be same.

D) Acceleration of system is given by

+

21

21

mm

mmkg, where k is a spring factor.

13. A mass is resting on a rough plank. At initial instant a horizontal impulse isapplied on the mass. If the velocity of mass at instant t is v and displacement uptothis instant is S, then correct graph is

A) B) C) D)

14. A body is moving down along an inclined plane of inclination with horizontal. Thecoefficient of friction between the body and the plane varies as = x/2, where x is thedistance moved down the plane :

A) the instantaneous acceleration of the body, down the plane is g (2sin - x cos)B) the body will accelerate if tan > 2x.C) the body will accelerate in all the conditions.D) the body will first accelerate an then decelerate.

15. A block is placed on the top of a smooth inclined plane of inclination kept on thefloor of a lift. When the lift is descending with a retardation a, the block is released.The acceleration of the block relative to the incline is :

A) g sin B) a sin C) (g-a) sin D) (g+a)sin

16. The velocity of a bullet is reduced from 200m/s to 100m/s while traveling through awooden block of thickness 10cm. Assuming it to be uniform, the retardation will beA) 15 x 105m/s2 B) 10 x 104m/s2 C) 12 x 104m/s2 D) 14.5 m/s2

17. An open knife edge of mass m is dropped from a height 'h' on a wooden floor. If the bladepenetrates upto the depth d into the wood, the average resistance offered by the wood tothe knife edge is

A) mg

+d

h1 B) mg

2

d

h1

+ C) mg

d

h1 D) mg

18. A block of mass 4kg is placed on a rough horizontal plane. A time dependenthorizontal force F = kt acts on the block, k = 2N/s2. The frictional force between the

block and plane at time t = 2 sec is: ( = 0.2)A) 4N B) 8N C) 12N D) zero


22/46

19. If the banking angle of a curved road is given by tan-1

5

3and the radius of

curvature of the road is 6m then the safe driving speed should not exceed : (g =10m/s2)

A) 86.4 km/h B) 43.2 km/h C) 21.6 km/h D) 30.4 km/h20. Which of the following statements about the centripetal and centrifugal forces is

correct ?A) Centripetal force balances the centrifugal forceB) both centripetal force and centrifugal force act on the same body.C) Centripetal force is directed opposite to the centrifugal forceD) Centripetal force is experienced by the observer at the centre of the circular pathdescribed by the body.

21. A particle is moving in a circle of radius R in such a way that at any instant thenormal and tangential components of its acceleration are equal. If its speed at t = 0 isv0. The time taken to complete the first revolution is

A)ov

R B)ov

R (1-e-2) C)ov

R e-2 D)ov

R2

22. A particle of mass m is suspended from a fixed point O by a string of length l. Att = 0, it is displaced from equilibrium position and released. The graph, whichshows the variation of the tension T in the string with time 't', may be :

23. Given, 0FFF 321 =++ Angle between 1F and 2F is 800, between 2F and

3F is 1300. Then angle between 1F and 3F should be :

A) 300 B) 1500

C) 750D) will depend on magnitudes of 1F , 2F and 3F

24. Two identical balls 1 and 2 are tied to two strings as shown infigure. They are rotated about point O, Ball 1 is observed from

ball 2, centrifugal force on ball 1 is 1F . Similarly ball 2 is

observed from ball 1 and centrifugal force on ball 2 is2

F then :

A) 21 FF = = 0 B) 21 FF > C) 21 FF = 0 D) 21 FandF are

antiparallel25. In the figure shown, block of mass 2 kg is very long. Force of friction

on 1 kg block isA) always towards rightB) always towards leftC) is first towards right and then towards leftD) is first towards left and then towards right

26. A time varying force is applied on a block placed over a rough

surface as shown in figure. Let be the angle between contact forceon the block and the normal reaction, then with time, will


23/46

A) remain constantB) first increase to a maximum value (say max) and then becomes constant in a value lessthan max.C) first decrease to a minimum value (say min) and then becomes constant in a value morethan max.D) none of these

27. A block of mass m is attached with a massless instretchable string. Breakingstrength of string is 4 mg. Block is moving up. The maximum acceleration andmaximum retardation of the block can be :A) 4g, 3g B) 4g, gC) 3g, g D) 3g, 4g string

28. A frame is rotating in a circle with varying speed v = (2t-4) m/s, where t is inseconds. An object is viewed from this frame. The pseudo forceA) is maximum at 2 second B) is minimum at 2 secondC) is zero at 2 second D) data is insufficient

29. If the horizontal acceleration a0 of the surface is less than that requiredto keep the block stationary, as shown in figure, thenA) The block m accelerates downwardsB) The block M accelerates towards rightC) The tension in the string is less than mg D) All the above

30. Two men A and B of masses M and M + m respectively start simultaneously from theground and climb with uniform acceleration up from the free ends of a mass lessinextensible rope which passes over a smooth pulley at a height h from the ground.Which man will reach the pulley first?A) A B) B C) Both reach simultaneously D) Datainsufficient

31. Two blocks of masses m1 and m2 are connected with a mass less spring and placedover a plank moving with an acceleration 'a' as shown in figure. The

coefficient of friction between the blocks and platform is A) Spring will be stretched, if a > gB) Spring will be compressed, if a gC) Spring will neither be compressed nor be stretched for a gD) Spring will be natural length under all conditions.

KEY1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

D D B C A D C C D B A A B D D A A A C C

21 22 23 24 25 26 27 28 29 30 31

B D B B D B C B D A D

LEVEL II1. A heavy uniform rope of length is pulled by a constant

force F as shown in fig. The tension in the string at a distancex from the end where the force is applied is

A) F B) F

xC) F

+

x1 D) F

l

x1

2. A balloon of gross weight W is falling vertically downward with constantacceleration a. The amount of ballast Q that must be thrown out in order to give the

balloon an equal upward acceleration a is such that


24/46

A) Q = W/2 B) Q = 2 W( )ga

a

+

C) Q = W( )

a

ga D) Q = W

( )

g

ga

3. Two masses of 10kg and 20kg respectively are connected by a massless spring kepton a horizontal smooth table. A force of 200N acts on the 20kg mass. At a giveninstant if the 10kg mass has an acceleration of 12m/s2. What is the acceleration of20kg mass ?A) zero B) 10m/s2 C) 4m/s2 D) 12m/s2

4. A 2kg mass pulls horizontally on a 3kg mass by means of a lightly stretched spring.If at one instant the 3kg mass has an acceleration towards 2kg mass of 1.8m/s2 theacceleration of 2kg mass isA) 1.2m/s2 B) 3.6 m/s2 C) 2.7 m/s2 D) zero

5. An elivator is moving up with uniform acceleration a. At some instant (t = 0), a loosebolt drops from its ceiling. If the distance between ceiling and the floor of theelevator is 't', what is the time t taken by the bolt to hit the floor ?

A)g

2lB)

( )ag

2

+l

C)( )ag

2

l

D)( )ag2 +

6. Two monkeys P and Q of masses M and m (>M) hold a light rope passing over asmooth fixed pulley. P and Q climb up the rope so that the acceleration of Qupward is double that of P downward. The tension in the rope is

A) gmM

Mm2

+B) g

mM2

Mm3

+C) g

m2M

Mm3

+D) g

m2M2

Mm3

+

7. A block of mass m is placed over a block B of mass 2m. The block Bcan move on a smooth horizontal surface. If the coefficient of

friction between the blocks A and B is , then the minimum force Rrequired to initiate sliding motion in block A is

A) mg B) 3mgC) 3/2mg D) 3mg /

8. The acceleration of the blocks A and B shown in figure are

A)13

g5,

13

g5B)

13

g10,

13

g10

C)13

g5,

13

g10D)

7

g10,

7

g5

9. If the acceleration of block m1is a1 and that of block m2 is a2 in fig., thenA) a1= 2a2B) a2= 2a1C) a1= g = a2D) a1 = a2 = g/2

10. A monkey of mass 20kg is holding a vertical rope. The rope will not break when amass of 25kg is suspended from it but will break if the mass exceeds 25kg. Whatis the maximum acceleration with which the monkey can climb up along the rope? (g = 10m/s2)A) 2.5m/s2 B) 5 m/s2 C) 10 m/s2 D) 25 m/s2

11. A man of mass m stands on a frame of mass M. He pulls on a light rope, which

passes over a pulley. The other end of the rope is attached to the frame. For thesystem to be in equilibrium, what force must the man exert on the rope ?


25/46

A) ( )gmM2

1+

B) (M+m)gC) (M-m)g

D) (M+2m)g12. Block A is placed on block B, whose mass is greater than that of A. There isfriction between the blocks, while the ground is smooth. A horizontal forceP, increasing linearly with time, begins to act on A. The accelerations a1 anda2 of A and B respectively are plotted against time (t). Choose the correctgraph.

A) B) C) D)

13. A car C of mass m1 rests on a plank P of mass m2. The plankrests on a smooth floor. The string and pulley are ideal. Thecar starts and moves towards the pulley with acceleration.A) If m1 > m2, the string will remain under tensionB) If m1 < m2, the string will become slackC) If m1 = m2, the string will have no tension, and C and Pwill have accelerations of equal magnitude.

D) C and P will have acceleration of equal magnitude if m1m2.

14. A frame of reference F2 moves with velocity v with respect to another frame F1.

When an object is observed from both frames, its velocity is found to be 1v in F1 and

2v in F2. Then, 2v is equal to

A) vv1 + B) vv1 C) 1vv D)1

11

v

vvv

15. A block of mass m slides down an inclined plane which makes an angle with thehorizontal. The coefficient of friction between the block and the plane is . The force exerted

by the block on the plane is

A) mgcos B) 12 + mgcos C)1

cosmg

2 +

D) mgcos

16. In the figure, the block A of mass m is placed on the block B ofmass 2m. B rests on the floor. The coefficient of friction between A

and B as well as that between the floor and B is . Both blocks are

given the same initial velocity to the right. The acceleration of Awith respect to B is


26/46

A) zero B) g to the left C) g to the right D)2

1g to the

right17. A particle of mass M is moving with acceleration a0 as measured by an observer 1

standing in a frame of reference moving with a uniform velocity. Another observer 2is standing in a frame of reference moving with acceleration a. Select wrongstatement from the following.A) Observer 1 measures the force acting on the body as Ma0.B) Observer 2 measures the force acting on the body as Ma.C) Observer 2 measures the force acting on the body as Ma0 - Ma.D) Observer 1 is standing in an inertial frame of reference and observer 2 is standingin a non-inertial frame of reference.

18. In the pulley arrangement show, the pulley P2 is movable. Assuming coefficient

of friction between m and surface to be , the minimum value of M for which mis at rest is

A) M = m/2 B) m = M/2 C) M = m/2 D) m = M/219. Block B moves to the right with a constant velocity v0. The velocity of

body A relative to B is

A)2

vo , towards left B)2

vo , towards right

c2

v3 o , towards left D)2

v3 o , towards right

20. Consider the shown arrangement. Assume all surfaces to be smooth. If 'N' representsmagnitudes of normal reaction between block and wedge thenacceleration of 'M' along horizontal equals.

A) M

sinN along + ve x-axis B) M

cosN along - ve x-axis

C)M

sinN along - ve x-axis D)

Mm

sinN

+

along - ve x-axis

21. Block A of mass m rests on the plank B of mass 3m which is free toslide on a frictionless horizontal surface. The coefficient of friction

between the block and plank is 0.2. If a horizontal force of magnitude2mg is applied to the plank B, the acceleration of A relative to theplank and relative to the ground respectively, are :

A) 0,2

gB) 0,

3

g2C)

5

g,

5

g3D)

5

g,

5

g2

22. A block is gently placed on a conveyor belt moving horizontally with constant speed.After t = 4s, the velocity of the block becomes equal to the velocity of the belt. If the

coefficient of friction between the block and the belt is = 0.2, then the velocity of theconveyor belt isA) 2ms-1 B) 4ms-1 C) 64ms-1 D) 8ms-1

23. A block of mass M rests on a rough horizontal surface as shown.

Coefficient of friction between the block and the surface is . A force F =Mg acting at angle with the vertical side of the block pulls it. In whichof the following cases the block can be pulled along the surface :

A) tan B) tan

2 C) cot D) cot

2


27/46

24. A body is moving down a long inclined plane of angle of inclination . Thecoefficient of friction between the body and the plane varies as = 0.5x, where x isthe distance moved down the plane. The body will have the maximum velocity whenit has travelled a distance x given by :

A) x = 2tan B) x =tan

2 C) x = 2 cot D) x =cot

2

25. A block of mass M is sliding down the plane. Coefficient of static friction is sand kinetic friction is k. Then friction force acting on the block isA) s Mg cos B) (F+Mg)sinC) k (F+Mg) cos D) (Mg+F)tan

26. A pendulum of mass m hangs from a support fixed to a trolley. The direction

of the string when the trolley rolls up a plane of inclination with accelerationa is :

A) 0 B) tan-1

C) tan-1+

cosg

singaD) tan-1

g

a

27. A block X of mass 4kg is lying on another block Y of mass 8kg as shown in the figure.The force f acting on X is 12N. The block X is on the verge of slippingon Y. The force F in newton along with f necessary to make both X andY move simultaneously will be :A) 36 B) 56C) 60 D) 24

28. A monkey of mass 20kg is holding a vertical rope. The rope will notbreak when a mass of 25kg is suspended from it but will break if the mass exceeds

25kg. What is the maximum acceleration with which the monkey can climb up alongthe rope ? (g = 10m/s2)A) 25m/s2 B) 2.5 m/s2 C) 5 m/s2 D) 10 m/s2

29. A smooth incline plane of length L having inclination with the horizontal is inside alift which is moving down with a retardation a. The time taken by a body to slidedown the inclined plane from rest will be :

A)( ) + sinag

L2B)

( ) sinag

L2C)sina

L2D)sing

L2

30. A particle of mass m begins to slide down a fixed smooth sphere from the top. Whatis its tangential acceleration when it breaks off the sphere ?

A) 2g/3 B)3

g5 C) g D) g/3

31. A person wants to drive on the vertical surface of a large cylindrical wooden 'well'commonly known as 'deathwell' in a circus. The radius of the well is R and thecoefficient of friction between the tyres of the motorcycle and the wall of the 'well' is

S. The minimum speed, the motorcylist must have in order to prevent slipping,should be :

A)s

Rg

B)

Rg

s C)R

gs D)g

R

s

32. A wet open umbrella is held vertical and it is whirled about the handle at a uniform

rate of 21 revolutions in 44 seconds. If the rim of the umbrella is a circle of 1 metre in


28/46

diametre and the height of the rim above the floor is 4.9m. The locus of the drop is acircle of radius :

A) 5.2 m B) 1m C) 3m D) 1.5m

33. A block of mass m is at rest on wedge as shown in fig. Let N be the normal reaction

between two and 'f' the force of friction, then choose the wrong option.A) N = mg cos B) N cos + f sin = mgC) N sin = f cos D) none of these

34. A block of mass m = 1 kg has a speed v = 4 m/s at = 600 on a circular track of radiusR = 2m as shown in fig. Size of the block is negligible. Coefficient offriction between block and the track is = 0.5. The force of friction

between the two isA) 10N B) 8.5NC) 6.5N D) 5N

35. In the figure shown, there is no relative motion between the two blocks.Force of friction on 1 kg block is

A) zeroB) 3NC) 6ND) 7N

36. A block is projected upwards on a rough plane at 20

m/s. Let t be the time in going up, then t is equal toA) 4 secB) 2 secC) 8 secD) 6 sec

37. A ladder of mass 10kg is held at rest against a smooth wall on a roughground as shown in figure. The normal reaction between the wall and theladder is

A) 50N B) 50 3 N

C)3

50N D) 100N

38. In the above problem, the downward acceleration of the block m is

A)Mm

)ag(m 0

+

B)

Mm

)ag(M 0

+

C)

Mm

aMgm 0

+

D)

MmaMgm 0++

39. A smooth sphere of weight 'W' is supported in contact with a smooth vertical wall bya string to a point on its surface, the end being attached to a point on the wall. If thelength of the string is equal to the radius of the sphere, then the tension in the stringand reaction of the wall are

A) T =3

2W, R =

3

WB) T =

3

1W, R =

3

2W

C) T =3

2W, R =

3

2W D) T =

3

1W, R =

3

1W


29/46

40. A body projected along an inclined plane of angle of inclination 300 stops aftercovering a distance x1. The same body projected with the same speed stops aftercovering a distance x2, if the angle of inclination is increased to 600. The ratio x1/x2 is

A) 1 B) 3 C) 2 D) 2

41. A 2 kg block is connected with two springs of force constants k1 = 100 N/m and k2 =300 N/m as shown in figure. The block is released from rest with the springsunscratched. The acceleration of the block in its lowest position is (g = 10m/s2)A) Zero B) 10 m/s2 upwardsC) 10 m/s2 downwards D) 5 m/s2 upwards

42. Three identical blocks are suspended on two identical springs one below the otheras shown in figure. If thread is cut that supports block 1, then initiallyA) The second ball falls with zero acceleration

B) The first ball falls with maximum accelerationC) Both (A) and (B) are wrong D) Both (A) and (B) correct.

KEY1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22

D B C C B C C C C A A C ABCD B B A B A B C D D23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42

D A C C A B A B B A D C C B C C A B B D


30/46

LEVEL III1. The pulleys shown in fig. are smooth and light. The acceleration of A is 'a' upwards

and the acceleration of C is b downwards, then the acceleration of B is

A) ( )ba

2

1 up B) ( )ba

2

1+ up

C) ( )ba2

1+ down D) ( )ab

2

1 up

2. Three light strings are connected at the point P. A weight W issuspended from one of the strings. End A of string AP and endB of string PB are fixed as shown. In equilibrium PB is horizontal and PA makesan angle of 600 with the horizontal. If the tension in PB is 30N then the tension inPA and weight W are respectively.

A) 60N, 30N B) 60/ 3 N, 30/ 3 N

C) 60N, 30 3 N D) 60 3 N, 30 3 N

3. A force F is applied to hold a block of mass m on an inclined plane making an

angle with the horizontal. The force F is perpendicular to the plane. Thecoefficient of friction between the plane and the block is . The minimum force Fnecessary to keep the block at rest on the inclined plane is

A) mg sin B) mg cos C)

mgsin D)

mg(sin - cos)

4. In the fig, the blocks A, B and C of mass m eachhave accelerations a1, a2 and a3 respectively. F1and F2 are external forces of magnitudes '2mg'and 'mg' respectively.

A) a1 = a2 = a3B) a1 > a3 > a2C) a1 = a2 , a2 > a3D) a1 > a2 , a2 = a3

5. A block of mass m is at rest on an inclined surface having a coefficient of

friction > tan as shown in the figure. The horizontal acceleration whichshould be given to the inclined plane, so that the force of friction between the

block and the plane becomes zero, is :

A) g sin, leftward B) g tan, leftward C) g cot, rightward D) g tan,

rightward6. Two blocks are in contact on a frictionless table. One has mass m and the other 2m. A

force F is applied on 2m as shown in the figure. Now the same force F isapplied from the right on m. In the two cases respectively, the force ofcontact between the two blocks will be :

A) same B) 1 : 2C) 2 : 1 D) 1 : 3

7. All surfaces shown in fig are smooth. For what ratio m1 : m2 : m3,system is in equilibrium. All pulleys and strings are massless,A) 1 : 2 : 1

B) 2 : 2 : 1C) 2 : 1 : 2


31/46

D) 1 : 2 : 28. In the diagram shown in figure, both pulleys and strings are

massless. The acceleration of 2kg block isA) 2.5m/s2()

B) 5m/s2

()C) 7.5m/s2()D) 10m/s2()

9. In the diagram shown in figure, wedge of mass M isstationary. Block of mass m = 2kg is slipping down.

Force of friction on the wedge is (g=10m/s2)A) 5 3 N B) 10N

C) 5N D) 10 3 N

10. Two different blocks of equal mass 'm' are released from two positions as shown infigure. Net force on the block at bottommost in case (i) is say F 1 and in case (ii) is sayF2. then :

A) F1 = F2 B) F1 > F2 C) F1 < F2 D) datainsufficient

11. The acceleration a of the frame for which the uniform slender rodmaintains the orientation shown in the figure is (Neglect the friction atthe contact between trolley and rod)

A) 2 g B) 3 g C) g/ 3 D) 2g

12. A cyclist rides along the circumference of a circular horizontal track of

radius R. The coefficient of friction

=0

R

r1

, where0

is a constant and r is thedistance form the center of the circle. The maximum velocity of the cyclist is

A)2

gR0B) gR0 C)

2

g0D)

2

gR0

13. In the diagram shows the blocks A, B and C weight 3 kg, 4 kg and 8 kgrespectively. The coefficient of sliding friction is 0.25 between any twosurfaces. A is held at rest by mass less rigid rod fixed to the wall while B andC are connected by a flexible cord passing round a fixed frictionless pulley.Assuming that the arrangement shown in the diagram B on C and A on B ismaintained all throughout, then force necessary to drag C along the

horizontal surface to the left at constant speed will beA) 4 kg B) 8 kg C) 16 kg D) 12 kg


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14. Two bodies of mass m and 4 m are attached with string as shown in thefigure. The body of mass m hanging from a string of length l is

executing oscillations of angular amplitude 0 while the other body is atrest. The minimum coefficient of friction between the mass 4 m and the

horizontal surface should be

A)

3

cos2 0 B) 2 cos2

2

0 C)

2

cos1 0 D)

4

cos3 0

KEY

1 2 3 4 5 6 7 8 9 10 11 12 13 14

D C D D B B B A A C B A B D

MULTIPLE ANSWER TYPE QUESTIONS

1. A light vertical chain is used to haul up an object of mass M attached to its lowerand. The vertical pull applied has a magnitude F at t = 0 and it decreases at a

uniform rate of f Nm-1

over a distance s through which the object is raisedA) The acceleration of the object is

M

MgfyFwhen the object is raised through a

distance y ( < s)B) The acceleration of the object is constant

C) The object has a velocity

2

fsMgF

M

s2when it has been raised through a

distance s

D) The object has velocityg

swhen it has been raised through a distance s.

2. A lift is moving downwards. A body of mass m kept on the floor of the lift is pulledhorizontally. If is the coefficient of friction between the surfaces in contact thenA) frictional resistance offered by the floor is mg when lift moves up with a uniformvelocity of 5 ms-1

B) frictional resistance offered by the floor is mg when lift moves up with a uniformvelocity of 3 ms-1

C) frictional resistance offered by the floor is 5 m when life accelerates down withan acceleration of 4.8 ms-2

D) frictional resistance (f) offered by the floor must lie in the range 0 f < 3. A weight W can be just supported on a rough inclined plane by a force P either

acting along the plane or horizontally. The angle of friction is and is the anglewhich incline makes with the horizontal

A) The incline makes an angle with the horizontal twice the angle of friction i.e. =2B) The incline make an angle with the horizontal equal to the angle of friction i.e. =

C) The ratio of the force to the weight isW

P= cot

D) The ratio of the force to the weight isW

P= tan

4. A block of mass m is placed at rest on a rough horizontal surface with the coefficient

of friction .If the block is pulled by a force F at an angle with the horizontal, then


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A) minimum force is required when is equal to the angle of friction ()

B) minimum force is required when =2

-

C) the magnitude of minimum force is mg sin

D) the magnitude of minimum force is mg cos5. Imagine that the horizontal floor as shown in figure lies in the x - yplane. A block of mass m is placed at rest on the origin. The

coefficient of friction between the block and the surface is . Twoforce Fx and Fy act on the block along the positive x-axis and thepositive y-axis, respectively. If the block remains stationary, then

A) normal reaction, kmgN =

B) friction force, )jFiF(f yx +=

C) mg|f|

D) 2y2x FF|f| +=

6. Imagine the situation in which a horizontal force F is applied on thewedge. If F0 is the force required to keep the body stationary,Choose the correct statement(s)

A) If F F0, the block remains stationary with respect to wedgeB) If F < F0, the block slides down the wedgeC) If F > F0, the block slides up the wedgeD) If F = F0, the block is accelerating with respect to ground

7. In the figure two blocks M and m are tied together with an inextensible string. Themass M is placed on a rough horizontal surface with coefficient of

friction and the mass m is hanging vertically against a smoothvertical wall.Imagine a situation in which the given arrangement is placed inside

an elevator that can move only in the vertical direction and comparethe situation with the case when it is placed on the ground. When theelevator accelerates downward with a0, thenA) the friction force between the block M and the surface decreases

B) the system can accelerate with respect to the elevator even when m < MC) the system does not accelerate with respect to the elevator unless m > MD) The tension in the string decreases

8. Imagine the situation in which the given arrangement is placed inside atrolley that can move only in the horizontal direction, as shown infigure. If the trolley is accelerated horizontally along the positive x-axiswith a0, then choose the correct statement(s).

A) There exists a value of a0 = at which friction force on block M becomes zeroB) There exists two values of a0 = at which the magnitudes offriction acting on block M are equalC) The maximum value of friction force acts on the block M at two accelerations a1

and a2 such that a1 + a2 = 2D) The maximum value of friction is independent of the acceleration a0

9. If a horizontal force F is applied on the bigger block M0 as shown infigure then mark out the correct statement(s).A) For any value of F all the blocks will be accelerating with respect to ground

B) If m > M, there exists a unique value of F for which M and m arestationary with respect to M0

C) For a non-zero value of F, the normal reaction between m and M0 isalso non-zero


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D) The friction force between M and M0 is independent of F10. Imagine the situation as shown in figure in which the block of

mass M is not pulled by a block of mass m, but instead it is beingpulled by a constant force F equal to the weight of the block m,

i.e., F = mg. Compare the results of this case B with the previouscase A in which a block of mass m is suspended.If the surface is given a horizontal acceleration a0 along thepositive x-axis, then mark out the correct alternative(s).A) The value of a0 to keep the block stationary with respect to surface in case B ismore than that in case AB) In both the cases the values of a0 is same to keep the block stationaryC) The tension in the string is equal to mgD) The tension in the string depends on a0

11. If the acceleration of block m1 is a1 and that of block m2 is a2 in figure,then

A)12

21mmgma+=

B)12

22m4m

gma+=

C)21

22

mm4

gma+

= D)12

21

m4m

gm2a+

=

12. A painter is raising himself and the crate on which he stands with an accelerationof 5 m/s2 by a massless rope-and-pulley arrangement. Mass of the painter is 100kg and that of the crate is 50 kg. If g = 10 m/s2, then theA) tension in the rope is 2250 NB) tension in the rope is 1125 NC) contact force between painter and floor is 750 ND) contact force between painter and floor is 375 N

13. The spring balance A reads 2 kg with a block m suspended from it. A balance Breads 5 kg when a beaker filled with liquid is put on the pan of the balance. Thetwo balances are now so arranged that the hanging mass is inside the liquid asshown in figure. In this situationA) the balance A will read more than 2 kgB) the balance B will read more than 5 kgC) the balance A will read less than 2 kg and B will read more than 5 kgD) balance A and B will read 2 kg and 5 kg respectively

14. A uniform bar AB of weight 100 N is hinged at A to a vertical wall andheld in horizontal position by a cord BC as shown in figure. If T is the

tension in the cord, R is the reaction at the hing and is the angle whichthe reaction R makes with the rod AB, thenA) T = 100 N B) R = 50 N

C) R = 100 N D) = 300

15. A block of mass 10 kg is placed at a distance of 4.5 m from the rear end of a longtrolley as shown in figure. The coefficient of friction between the block and thesurface below is 0.3. If the trolley starts with anacceleration of 4 m/s2 and g = 10 m/s2 then

A) the block will not slideB) the block will slide backward with acceleration


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C) the block will fall from trolley after 3sD) the block falls after the trolley has moved 18 m

16. In the figure, the blocks are of equal mass. The pulley is fixed. In

the position shown, A moves down with a speed u, and vB = thespeed of BA) B will never lose contact with the groundB) The downward acceleration of A is equal in magnitude to thehorizontal acceleration of B

C) B = u cos D) B = u /cos 17. A cart moves with a constant speed along a horizontal circular path. From the cart, a

particle is thrown up vertically with respect to the cartA) The particle will land somewhere on the circular pathB) The particle will land outside the circular pathC) The particle will follow an elliptical path

D) The particle will follow a parabolic path18. A block of weight 9.8 N is placed on a table. The table surface exerts an upward forceof 10 N on the block. Assume g = 9.8 m/s2A) The block exerts a force of 10 N on the tableB) The block exerts a force of 19.8 N on the tableC) The block exerts a force of 9.8 N on the tableD) The block has an upward acceleration

19. A particle of mass 70 g, moving at 50 cm/s, is acted upon bya variable force opposite to its direction of motion. The force Fis shown as a function of time t.A) Its speed will be 50 cm/s after the force stops acting

B) Its direction of motion will reverseC) Its average acceleration will be 1 m/s2 during the interval in which the force actsD) Its average acceleration will be 10 m/s2 during the interval in which the force acts.

20. A monkey of mass m kg slides down a light rope attached to a fixed spring balance,with an acceleration a. The reading of the spring balance is W kg. [g = accelerationdue to gravity]A) The force of friction exerted by the rope on the monkey is m (g - a) N

B) m =ag

gW

C) m = W

+

g

a1 D) The tension in the rope is Wg N

21. Two masses of 10 kg and 20 kg are connected by a light spring asshown. A force of 200 N acts on a 20 kg mass as shown. At acertain instant the acceleration of 10 kg is 12 ms-2A) At that instant the 20 kg mass has an acceleration of 12 ms-2B) At that instant the 20 kg mass has an acceleration of 4 ms-2C) The stretching force in the spring is 120 ND) The collective system moves with a common acceleration of 30 ms -2 when theextension in the connecting spring is the maximum


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22. In the pulley system shown the movable pulleys A, B and C have mass m each,D and E are fixed pulleys. The strings are vertical, light and inextensible. Then,

A) the tension through out the string is the same and equals T =3

mg2

B) pulleys A and B have acceleration3g each in downward direction and pulley

C has acceleration3

gin upward direction

C) pulleys A, B and C all have acceleration3

gin downward direction

D) pulley A has acceleration3

gin downward direction and pulleys B and C

have acceleration3

geach in upward direction

23. A small mirror of area A and mass m is suspended in a vertical plane by a light

string. A beam of light of intensity I falls normally on the mirror and the string isdeflected from the vertical by an angle . Assuming the mirror to be perfectlyreflecting we have

A) radiation pressure equal toc

I2B) radiation pressure equal to

c2

I

C) tan =mgc

AI2D) tan =

mgc2

AI

24. Two masses m1 and m2 are connected by a light string which passes over the top ofan inclined plane making an angle of 300 with the horizontal such that one mass restson the plane and other hangs vertically. It is found that m1 hanging vertically candraw m2 up the full length of the incline in half the time in which m2 hangingvertically draws m1 up the full length of the incline. Assuming the surfaces in contactto be frictionless and pulley to be light and smooth, we haveA) the ratio of acceleration in the two cases respectively as 4

B) the ratio of acceleration in the two cases respectively as4

1

C) the ratio of masses m1 and m2 as2

3

D) the ratio of masses m1 and m2 as3

2

25. A particle slides down a smooth inclined plane of elevation . The incline is fixed

end to end in an elevator of base length l accelerating up with acceleration a0.Assume at t = 0 the particle is at the top of the incline then,

A) the particle has to travel a length l cos with acceleration (g + a0) sin down the

incline in a time+ 2sin)ag( 0

l

B) the particle has to travel a lengthcos

lwith acceleration g sin down the incline

in a time2sina

2

0

l


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C) the particle has to travel a lengthcos

lwith acceleration ( g + a0) sin down the

incline in a time

)2sin()ag(

4

0 +

l

D) the incline offers a normal reaction m (a0 + g) cos to the block so that it remainsin contact with the incline

26. A block A of mass 500 g is placed on a smooth horizontal tablewith a light string attached to it. The string passes over a smoothpulley P at the end of the table (as shown) and is connected toother block B of mass 200 g. Initially the block A is at a distance of200 cm from the pulley and is moving with a speed of 50 cms -1 tothe left. At t = 1 sA) the block is at a distance of 90 cm from the pulley PB) the block is at a distance of 110 cm from the pulley P

C) the block has a velocity of 230 cms-1 towards leftD) the block has a velocity of 230 cms-1 towards right

27. Two blocks of masses 2.9 kg and 1.9 kg are suspended from a rigid support Sby two inextensible wires each of length 1 m as shown in figure. The upperwire has negligible mass and lower wire has a uniform mass of 0.2 kgm-1.The whole system of blocks, wires and support have an upwardacceleration of a0 = 0.2 ms-2. Taking g = 9.8 ms-2 we haveA) tension at the midpoint of upper rope is 50 NB) tension at the midpoint of upper rope is 40 NC) tension at the midpoint of lower rope is 19.6 ND) tension at the midpoint of lower rope is 20 N

28. Two masses m and M (m < M) are joined by a light string passing over a smoothand light pulley (as shown)

A) The acceleration of each mass is gmM

mM

+

B) The tension in the string connecting masses is gmM

mM2

+

C) The thrust acting on the pulley is gmM

mM4

+

D) The centre of mass of the system (i.e M and m) moves down with an acceleration

of gmMmM

2

+

29. The spheres A and B as shown have mass M each. The strings SAand AB are light and inextensible with tensions T1 and T2respectively. A constant horizontal force F = Mg is acting on B.For the system to be in equilibrium we have

A) tan = 1 B) tan = 0.5

C) T2 = 2 Mg D) T1 = 5 Mg

30. A block of weight W is suspended from a spring balance. The lower surface of theblock rests on a weighing machine. The spring balance reads W1 and the weighing

machine reads W2. (W, W1, W2 are in the same unit)A) W = W1 + W2 if the system is at rest


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B) W > W1 + W2 if the system moves down with some accelerationC) W1 > W2 if the system moves up with some accelerationD) No relation between W1 and W2 can be obtained with the given description of thesystem

31. A simple pendulum with a bob of mass m is suspended from the roof of a car movingwith a horizontal acceleration aA) The string makes an angle of tan-1 (a/g) with the vertical

B) The string makes an angle of tan-1

g

a1 with the vertical

C) The tension in the string is m 22 ga + D) The tension in the string is m

22 ag

32. In the previous question, the blocks are allowed to move for sometime, after which Mis stopped momentarily (brought to rest and released at once). After this,A) both blocks will move with the same accelerationB) the string will become taut (under tension) again when the blocks acquire the samespeedC) the string will become taut again when the blocks cover equal distancesD) at the instant when the string becomes taut again, there may be some exchange ofimpulse between the string and blocks

33. A block of mass 1 kg moves under the influence of external forces on a roughhorizontal surface. At some instant, it has a speed of 1 m/s due east and anacceleration of 1 m/s2 due north. The force of friction acting on it is FA) F acts due west B) F acts due southC) F acts in the south-west directionD) The magnitude of F cannot be found from the given data

34. A block of mass m is placed on a rough horizontal surface. The coefficient of friction

between them is . An external horizontal force is applied to the block and itsmagnitude is gradually increased. The force exerted by the block on the surface is R

A) The magnitude of R will gradually increase B) R mg 12 +

C) The angle made by R with the vertical will gradually increase

D) The angle made by R with the vertical tan-135. A man pulls a block heavier than himself with a light rope. The

coefficient of friction is the same between the man and the ground, andbetween the block and the groundA) The block will not move unless the man also moves

B) The man can move even when the block is stationaryC) If both move, the acceleration of the man is greater than the acceleration of the

blockD) None of the above assertions is correct

36. A car C of mass m1 rests on a plank P of mass m2. The plank restson a smooth floor. The string and pulley are ideal. The car startsand moves towards the pulley with accelerationA) If m1 > m2, the string will remain under tensionB) If m1 < m2, the string will become slackC) If m1 = m2, the string will have no tension, and C and P will have accelerationsof equal magnitude

D) C and P will have accelerations of equal magnitude if m1m2


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37. A man tries to remain in equilibrium by pushing with his hands and feet against twoparallel walls. For equilibrium,A) he must exert equal forces on the two wallsB) the forces of friction at the two walls must be equal

C) friction must be present on both wallsD) the coefficients of friction must be the same between both walls and the man38. Two me of unequal masses hold on to the two sections of a light rope passing

over a smooth light pulley. Which of the following are possible?A) The lighter man is stationary while the heavier man slides with someaccelerationB) The heavier man is stationary while the lighter man climbs with someaccelerationC) The two men slide with the same acceleration in the same directionD) The two men slide with accelerations of the same magnitude in oppositedirections

39. A body is kept on a smooth inclined plane having an inclination of 1 in x. Then,A) slope of inclined plane is

x

1

B) slope of inclined plane is1x

1

2

C) for the body of mass m to remain stationary relative to the incline, the incline

must offer a normal reaction of mg1x

x

2

D) for the body to remain stationary relative to the incline, the incline must be given

a horizontal acceleration of 1x

g

2

KEY

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15ABCD AD AC ABCD BCD ACD ABD AC BC AD BD BC ACD CD AD

16 17 18 19 20 21 22 23 24 25 26 27 28 29 30BD AD AB ABD ABD AB AC AC CD BD AD ABCD ABCD AC AC

31 32 33 34 35 36 37 38 39ACD AD ABCD ABC ABCD AC ABD BCD BCD

* * *

COMPREHENSION TYPE QUESTIONS

Passage I (Q.No: 1 to 5):An engineer is designing a conveyor system for loadinglay bales into a wagon. Each bale is 0.25 m. wide, 0.50 mhigh, and 0.80 m long (the dimension perpendicular tothe plane of the figure), with mass 30.0 Kg. The center ofgravity of each bale is at its geometrical center. The

coefficient of static friction between a bale and theconveyor belt is 0.60, and the belt moves with constant


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speed. The angle of the conveyor is slowly increased. At some critical angle a balewill tip (if it doesn't slip first), and at some different critical angle it will slip (if itdoesn't tip first).

1. Find the first critical angle (In the same conditions) at which it tips

A) = tan-1

0.50 B) = tan-1

0.60 C) = tan-1

0.40 D) = tan-1

0.202. Find the second critical angle (in the same conditions) at which it slips.

A) = tan-1 0.50 B) = tan-1 0.60 C) = tan-1 0.12 D) = tan-10.70

3. Find the first critical angle at which it tips if the coefficient of friction were 0.40 ?


4. Find the second critical angle at which it slips if the coefficient of friction were 0.40 ?


5. Which statement/s is/are correctA) At smaller angle it will tip first ifs = 0.60B) At smaller angle it will tip first if s = 0.40C) At larger angle it will tip first if s = 0.60D) At larger angle it will tip first if s = 0.40

Passage II: (Q.No: 6 to 10):A moving company uses the pulleysystem in figure 1 to lift heavy crates upa ramp. The ramp is coated with rollersthat make the crate's motion essentially

frictionless. A worker piles cinderblocks onto the plate until the platemoves down, pulling the crate up theramp. Each cinder block has mass 10 kg. The plate has mass 5 kg. The rope is nearlymassless, and the pulley is essentially frictionless. The ramp makes a 300 angle withthe ground. The crate has mass 100 kg.Let W1 denote the combined weight of the plate and the cinder blocks piled on theplate. Let T denote the tension in the rope. And let W2 denote the crate's weight.

6. What is the smallest number of cinder blocks that need to be placed on the plate inorder to lift the crate up the ramp ?A) 3 B) 5 C) 7 D) 10

7. Ten cinder blocks are placed on the plate. As a result, the crate accelerates up theramp. Which of the following is true ?A) W1 = T = W2 sin 300 B) W1 = T > W2 sin 300C) W1 > T = W2 sin 300 D) W1 > T > W2 sin 300

8. The ramp exerts a "normal" foce on the crate, directed perpendicular to the ramp'ssurface. This normal force has magnitude.A) W2 B) W2 sin 300 C) W2 cos 300 D) W2 (sin 300+ cos 300)

9. The net force on the crate has magnitude.A) W1 - W2 sin 300 B) W1 - W2 C) T - W2 sin 300 D) T - W2

10. After the crate is already moving, the cinder blocks suddenly fall off the plate.Which of the following graphs best shows the subsequent velocity of the crate, afterthe cinder blocks have fallen off the plate ? (up-the-ramp is the positive direction)


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PASSAGE - III (Q.No: 11 to 16):Two physics student were going to appear at the

International Physics Olympiad, scheduled to start at 11.00a.m. They left for the examination centre on a car in themorning at 8.00 a.m. It is being known to them thatduring the complete journey the car accelerates or

decelerates at the same constant rate. Both of them decidedto verify some of the concepts of physics on the way. Theyare having a light string and two identical small balls, say B1 and B2, each of mass 'm'. In thecar Joe sits on the left end and Becky sits on the right end with a separation of 1 metre

between them, as shown in the figure. Joe ties the ball B1 with the string whose one end isfixed to the ceiling of the car as shown. At 9:30 a.m. Joe notices the string attached to the ballB1 to be inclined with the vertical towards Becky. At the same instant Becky throws the ballB2 vertically upward with respect to herself such that it does not touch the ceiling of the car.Again at 10:15 a.m. when the car was heading towards the examination centre, Joe finds thatthe string, which was vertical, has just started deflecting from the vertical. He quickly cutsthe string and finds that the ball lands in Becky's hards!.

11. At 9:30 a.m. the tension in the string is:A) equal to (mg) B) greater than (mg) C) less than (mg) D) needs moreinformation

12. At 9:30 a.m., the car is :A) moving with uniform velocity B) acceleratingC) decelerating D) temporarily at rest

13. The ball B2 will :A) fall in hands of Becky B) fall infront of BeckyC) fall behind Becky D) land depending upon the speedof car

14. The magnitude of acceleration or retardation of car is :A) 8 m/s2 B) 4 m/s2 C) 2 m/s2 D) 1 m/s2

15. The path of ball B1, after being separated from the cut string, as observed by Becky is:A) Circle B) Straight line C) Parabola D) Ellipse

16. As observed by Becky the acceleration of ball 'B1' after the string is cut, will be:A) equal to 'g' B) greater than 'g' C) less than 'g' D) zero

A) B) C) D)


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Passage - IV (Q.No: 17 to 21):

A sufficiently long plank of mass 4 kg is placed on a smooth horizontal surface. Asmall block of mass 2 kg is placed over the plank and is being acted upon by a timevarying horizontal force F = (0.5t), where 'F' is in Newton and 't' is in second asshown in fig. (a). The coefficient of friction between the plank and the block is given

as == ks . At time t = 12 sec, the relative slipping between the plank and theblock is just likely to occur.If the force F acting on 2 kg block is removed and the system (plank + block) is givenhorizontal velocity 'V0, as shown in fig. (b), this system strikes a mass less spring ofspring constant k = 120 N/m fixed at the end of the relative slipping occurs between

the plank and the block.17. The coefficient of friction is equal to :

A) 0.10 B) 0.15 C) 0.20 D) 0.3018. The acceleration (a) versus time (t) graph for the plank and the block shown in figure

(a) is correctly represented in :

19. The average acceleration of the plank in the time interval 0 to 15 sec. in fig. (a) willbe:A) 0.20 m/s2 B) 0.30 m/s2 C) 0.40 m/s2 D) 0.60 m/s2

20. The magnitude of frictional force '' rf developed on 2 kg block versus compression'x' of the spring from its natural length, as in fig. (b), is best represented in:

A) B) C) D)

A) B) C) D)


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21. The maximum possible value of '' 0V , as in fig. (B), upto which no relative slipping

occurs between the plank and the block will be:

A) sm/10.0 B) sm/20.0 C) sm/30.0 D) sm/40.0

Passage - V (Q.No: 22 to 26)

A physicist decided to find the frictioncoefficient between a plank and a block. Hisexperimental setup is as shown in the above figurethrough which he could determine the coefficientof static friction and coefficient of kinetic friction,

i.e.,s and k , between the plank AB and the

block. The length of the plank is measured to be 4 metres and the mass of the blockis 'M'.

The block is first placed on the plank and then plank is slowly inclined. Whenthe height 'h', as shown in figure, becomes 2 metre, the relative slipping between the

block and the plank is just likely to occur. Now this height 'h' is further increased to2.4 metres and then the block is released from rest at the position 'B' and the timetaken by the block to reach the position 'A' is measured to be 2 seconds by a stop-watch.

22. The friction coefficients,s and k , are:

A) 3/1=s and k = 0.5 B) 3=s and 5.0=k

C) 3/1=s and k = 0.6 D) s = 0.75 and 5.0=k

23. The variation of frictional force ( ),rf between the block and the plank, versus angle

( ) made by the plank with horizontal is correctly represented in :

24. When h = 2.4 m, the block is projected from position A with a velocity of 10m/s up

the plane. The velocity of the block when it reaches to position B will be:A) zero B) 52 m/s C) 132 m/s D) won't

reach B25. If h = 1 m, then the frictional force applied by the surface of plank on the block will

be:A) mg B) 0.5 mg C) 0.6 mg D) 0.25 mg

26. If h = 2.4 m, and a force 'F' is applied on the block parallel to the plane of the plank,then the range of the force 'F' for which the block will remain in equilibrium is

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