Bansal Classes [1]

TEACHING NOTESWORK ENERGY

In XIII 5 Lectures in Bull's Eye 6 Lectures & ACME 6 Lectures including discussion.Syllabus in IIT JEE : Kinetic and potential energy; Work and power; Conservation of mechanicalenergy.

INTRODUCTION :Whatever we have learnt till now is more or less sufficient for solving any problem involving particles. Bydoing force analysis we can find acceleration & then we can find velocity, position, etc. That means weshould be finished with particle dynamics.But it is not so. We will learn a new technique and concept which will make problem solving faster andwill give simple solution to complicated problems. Work Energy Theorem is an extension to Newton'sLaws but much simpler to use (mainly because of absence of vectors).

WORK :When we say 'work' in physics it is different from word 'work' we use in daily life. Work done by force

F

is defined asdW = SdF

,

There are two interpretations of sd .

(1) If the body is moving as a complete unit, sd is the displacement of the body e.g. when we walk on earth,

there is a force of friction on earth. The earth moves as a unit. Each point has same displacement. So sd

is displacement of earth.

(2) If the different points of body have different displacement, sd is the displacement of point of applicationof force. In the example above displacement of different parts of our body are different. So sd

isdisplacement of point of contact i.e. the foot in contact with the ground. So we can say clearly that workby earth on us is also zero.

OPTIONAL

Actually, there are no 2 divisions dW = SdF

, where sd is point of application of force on body. Withthis, you can explain both answers.

i) We can note that work is a scalar quantity.

ii) dW = Cos|Sd||F|

i.e. if component of force is along displacement ( < 90) work is positive otherwisework is negative ( > 90)

QuestionsIn light of these discussions, consider the following true-false questions:

(1)True or False?A boy jumps up into the air by applying a force downward on the ground. The work-kinetic energytheorem W = K can be applied to the boy to find the speed with which he leaves the ground.

This is the teaching module for work energy. This module has to be followed in the class. VK Bansal

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(2)True or False?A balloon is compressed uniformly from all sides. Because there is no displacement of the balloonscenter of mass, no work is done on the balloon.Both of these claims are false. Question (1) refers to a simple, everyday experience that unfortunatelycannot be analyzed by means of traditional physics teaching without the introduction of additional work-like quantities and energy-like equations. The upward force on the boy that projects him into the air is thenormal force on his feet from the ground. The center of mass of the boy indeed moves through anupward displacement. The normal force, however, goes through no displacement in the reference frameof the ground, and therefore no work is done by this force on the boy. The change in the boys kineticenergy does not come from work done on the system of the boy. This is a case of a deformable system.Other cases include a person climbing stairs or a ladder, a girl pushing off a wall while standing on askateboard, and a piece of putty slamming into a wall. In all of these cases, no work is done by thecontact force, because there is no displacement of the point of application of the force

Illustration: A block kept on rough surface is being pulled by force F, as shown

Work by F is positive

Work by friction is negative

Meaning of negative and positive Work

S

(a)

Ball doespositive workon hand

Hand doesnegative workon ball

S

FB on H

S

FH on B

(b) (c)

When you catch a ball as in figure (a), your hand and the ball move together with the same displacement

s (figure b). The ball exerts a force HonBF

on your hand in the same direction as the hand's displacement,

so the work done by the ball on your hand is positive. But by Newton's third law your hand exerts an

equal and opposite force HonBBonH FF

on the ball (figuure c). This force, which slows the ball to astop opposite to the ball's displacement. Thus the work done by your hand on the ball is negative.Because your hand and the ball have the same displacement, the work that your hand does on the ball isjust the negative of the work that the ball does on your hand.Caution: Always specify exactly which force is doing the work, and on what. When you lift a book, youexert an upward force on the book and the book's displacement is upward, so the work done by thelifting force on the book is positive. But the work done by the gravitational force (weight) on a bookbeing lifted is negative because the downward gravitational force is opposite to the upward displacement.

iii) Work depends on reference frame because displacement is relative. (Remember force is not dependenton reference frame)

Illustration: A lift is going up with constant velocity. We will calculate workfrom two reference frames. In both reference frame

N-mg = 0 N = mgWork Done in 2nd ref. frame = 0Work Done in 1st ref. frame = Nut(We can observe that work is dependent on reference frame.)

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3. Calculation of workCase 1: When Force is uniform

dW = SdF

W = dW = SdF

= F

Sd

W = SF

(This is true only for uniform forces)

W =

Ex. A block of mass m is taken from A to B along spherical bowl.Work Done by gravity = mgR(1cos) Work Done by force F = FR(SinWork Done by normal = 0

Ex. Find work done by gravity and normal when block comes from A to BWg = mg(lsin) = (mgsin)l

mg

lm A

BWN = 0 (because displacement is perpendicular to Force)

Q. A particle is moving along a straight line from point A to point B with position vectors k3j7i2 and

k6j3i5 respectively. One of the force acting on the particle is k15j30i20F

. Find thework done by this force. [Ans. 315]

[Sol. d.FW

k3j10i3d

F = k15j30i20

W

= 60 + 300 45 = 315 Ans.

Case 2: When Force is nonuniform (either magnitude or direction)

dW = SdF

W = SdF

( F

cannot come out of integer since it is varying)

Ex. jyixF 2

.Particle moves from (1,2) to (3,4)

dW = SdF

( jdyidxSd

)dW = xdx + y2dy

W =

3

1

4

2

2dyyxdx =

3

1

2

2x

+

4

2

3

3y

= 368

* If force is not expressed as function of (x,y,z) then also we can solve problem by expressing force anddisplacement in same format.

Solved Example 4, 6 of HCV

[Home work : HCV (part-1) Chapter-8, Ex-1 to 9 ]

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SPRING FORCE :Natural length of spring is l0.Similarly, when we compress spring by x1 from natural length, thenwork done by spring force.

ikxF

)i)(dx(Sd

{dx is +ve as x is increasing}

dW = SdF

dW = 1x

0

dxkx = 21

k 21x

Ex. Find work done by spring if we compress it further by x2.

Sol. kxiF

Sd

= dx( i )

dW = SdF

= kx dx

W = k )xx(

x

12

1

dxx = 21

k[(x2 + x1)2 21x ]

WorkEnergy TheoremThe total work done on a body by external forces is related to the body's displacement-that is, tochanges in its position. But the total work is also related to changes in the speed of the body. To see this,consider figure, which shows several example of a block sliding on a frictionless table. The forces actingon the block are its weight w , the normal force n and the force F

exerted on it by the hand.In figure (a) the net force on the block is in the direction of its motion. From Newton's second law, thismeans that the block speeds up ; from equation, this also means that the total work Wtot done on the

block is positive. v

(a)

F

n

wW > 0tot

The total work is also positive in figure (b), but only the component F cos contributes to Wtot. Theblock again speed up, and this same component F cos is what causes the acceleration.

v

(b) F

n

wW > 0tot

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The total work is negative in figure (c) because the net force opposes the displacement ; in this case the

block slows down. v

(c)

F

n

wW < 0tot

The net force is zero in figure (d), so the speed of the block stays the same and the total work done onthe block is zero . We can conclude that when a particle undergoes a displacement, it speed up ifWtot > 0, slow down if Wtot < 0, and maintains the same speed if Wtot = 0.

(d)

vF

wW = 0tot

A block sliding on a frictionless table.(a) The net force causes the speed to increase and does positive work.(b) Again the net force causes the speed to increase and does positive work.(c) The net force opposes the displacement, causes the speed to decrease, and does negativework.(d) The net force is zero and does no work, and the speed is constant.

Derivation:For a particle F

=ma

SdF

= SddtVdm

Work Done by resultant Force = SdF = m vdvf

i

v

v

Summation of work by all the forces = )SdF(

= 2i2f mv2

1mv21

W = kf kii.e. Sum of work done by all the forces on a particle is equal to change in kinetic energy of the particleSo the kinetic energy of a particle is equal to the total work that was done to accelerate it from rest to itspresent speed.The kinetic energy of a particle is equal to the total work that particle can do in the process of beingbrought to rest. This is why you pull your hand and arm backward when you catch a ball. As the ballcomes to rest, it does an amount of work (force times distance) on your hand equal to the ball's initialkinetic energy. By pulling your hand back, you maximize the distance over which the force acts and sominimize the force on your hand.

Q. A 60 gm tennis ball thrown vertically up at 24 m/s rises to a maximum height of 26 m. What was the workdone by resistive forces?

[Sol. wg + wres = (0 21

mu2)

mgh + wres = 21

mu2

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wres = 0.06 10 26 21

0.06 24 24

= 1.68 J ]

Q. A force of )j5.1i3( N acts on a 5 kg body. The body is at a position of )j3i2( m and is travelling at

4 ms1. The force acts on the body until it is at the position )j5i( m. Assuming no other force does

work on the body, the final speed of the body. [Ans. 10 ms1][Sol. Given

Mass of the body = 5 kgForce F

= j5.1i3

Now displacement s = { )j3i2()j5i( } m = )j8i( mFrom Work Energy principle

W = sF

= 21

m(v2 u2)

v = 10 m/s ]

Ex. A block is connected to spring while spring is in relaxed state. Find maximum extension of spring.

Sol. Forces acting on block are spring and gravityWork done for x displacement

W = mgx 21

kx2

for max. displacement velocity should become zeroki = 0 kf = 0

mgx 21

kx2 = 0

x = kmg2

Applying Work Energy theorem on systemWe have been careful to apply the work energy theorem only to bodies that we can represent as particles- that is, as moving point masses. The reason is that new complexeties appear for more complex systemsthat have to be represented in terms of many particles with different motions.Here's an exmaple.

F

n1 n2

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Consider a man standing on frictionless roller skates on a level surface, facing a rigid wall. He pushesagainst the wall, setting himself in motion to the right. The forces acting on his are his weight w , theupward normal forces 1n

and 2n exerted by the ground on his skates, and the horizontal force F

exerted on him by the wall. There is no vertical displacement, so 1n,w and 2n

do not work. The force

F is the horizontal force that accelerates him to the right, but the parts of his body where that force isapplied (the man's hands) do not move. Thus the force F

also does no work. So where does the man'skinetic energy come from ?The difficulty is that it's simply not correct to represent the man as a single point mass. For the motion aswe've described , different parts of the man's body must have different motions ; his hands are stationaryagainst the wall while his torso(upper body) is moving away from the wall. The various parts of his bodyinteract with each other, and one part can exert forces and do work on another part. Therefore the totalkinetic energy of this composite system of body parts can change, even though no work is done byforces applied by bodies(such as the wall) that are outside the system. This would not be possible witha system that can be repesented as a single point particle.While using Work Energy Theorem for a system or applying work energy equation on many particlestogether we must remember that work due to all the forces (external & internal) must be written.

Otherwise use this equation seperately on individual particles. (Although total work doneby static friction, tension and normal contact force i.e. by action and reaction on a system willalways be zero.)

According to newtons laws, the body moves under the influence of the external force only. Internalforces do not accelerate the body. But work can still be done by them, even if they do not accelerate thebody. In the case of work energy theorem, we should be cautious. Do not forget to take into account thework done by the internal forces.To illustrate these points about work, let us take a simple daily life example. Suppose a boy of mass 40kg is walking on a rough ground with a uniform acceleration of 1 m/s2. We wish to find the work done onthe boy when he moves a distance of 1 m starting from rest.Sol: It is clear that the horizontal acceleration of the boy is possible only by an external force. This forceis friction force between him and the ground. We can imagine trying to walk on a smooth surface. We willslip!

This friction force f = ma = 40 N.But who does work on the boy? Obviously not the friction force. Then none of us need to eat

anything, cars need not be supplied any petrol! Friction cannot do any work. But we studied that workis

W = F.S = FS cos.

What is wrong here? Let us examine the motion of the boy more closely. When the boy walks, he doesnot slip his feet on the ground. Rather, he places one foot on the ground, lifts another foot and moves thatfoot further. The force of friction is not acting on the moving foot. It is acting on the foot which is incontact with the ground. So this is the case of a body where different parts of the body having differentdisplacements.

` Just now we said that in such cases, the work done by the force is W = F.S

= FS cos. where S is thedisplacement of the point of application of the force. So although the boy as a whole is moving, the pointat which friction force is being applied is not moving. On close examination, we can say that it is themuscles of the body who are rotating the legs, imparting the energy to the boy. Their work can beestimated with the help of the work energy theorem.V2 = U2 + 2as = 0 + 2 1 1

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Wint = K = 12

40 2 = 40 J

Observe the same thing from the frame of the boy and explain it.The work done by internal forces should be same as before. This should be! After all the work done bythe boy reflects in his food consumption. This should be the same from every frame of reference.

Frictional workConsider a block sliding across a horizontal table and eventually coming to rest due to the frictional forceexerted by the table.As the kinetic energy of the block decreases, there is a corresponding increase in the internal energy ofthe system of block + table. This increase in internal energy might be observed as a slight increase in thetemperature of the surfaces of the block and the table. It is a common observation that kinetic frictionbetween two surfaces causes an increases in the temperature, as for example in the case of holding apiece of metal against a grinding wheel or applying the brackes to an automobile or a bicycle (in whichcase both the brakes and the sliding tires can become warmer). You can even observe that effect byrubbing your hands together.We might be tempted to write the magnitude of the work done by the frictional force as the product ofthe frictional force times the displacement through which the object moves : |Wf| = fs.

If r in the definition of work is identified as the displacement of the object, it often follows in textbookand lecture discussions that the work done by friction on a block sliding on a surface is W = fkd, wherefk is the force of kinetic friction on the block and d is the distance through which the block moves relativeto the surface. The negative sign indicates that the friction force is in the opposite direction to thedisplacement. This expression for work is then incorporated into the work-kinetic energy theorem forthe block.

This approach ignores the fact that the displacement of the block is not the same as the many displacementsof the friction force at a large number of contact points. This latter displacement is complicated andinvolves deformations of the lower surface of the block. IT HAS TO BE NOTED THAT THEWORK DONE BY FRICTION CANNOT BE CALCULATED BECAUSE WE CANNOTFIND THE DISPLACEMENT AT LARGE NUMBER OF CONTACT POINT. EVEN INTHE CASE OF STATIC FRICTION, THERE IS VERY SMALL DISPLACEMENT AT THECONTACT POINT SO THE WORK DONE BY STATIC FRICTION IS ALSO NOT ZERO.BUT FOR TEACHING IN JEE, WE WILL ASSUME THE FOLLOWING FOR FINDINGTHE WORK DONE BY FRCTION.

A student may have little difficulty with W = fkd, based on his or her understanding of evaluating thework done by any force by performing a path integral over the path followed by the object. In the caseof a block sliding over a stationary surface, the friction force is always oppositely directed to eachinfinitesimal displacement of the block. For a constant friction force, this integral reduces to the productof the force and the length of the path (not the displacement).

Eg. The force 15 N pulls the lower block for 2m, Find final speed.

Sol.(I) For individual bodieswA = 5 2

wB = 15 2 +(5) 2

W = 30 = kEsys = 21

10v2 + 21

5v2

v = 2

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(II) We know work done by static friction will be zero because actionreaction will be in opposite direction

but displacement of contact point will be same. Thus 0SdfSdf BBAA

because ASd

= BSd

but

Af

= Bf

Thus 15 2 = (1/2) 10v2 + (1/2) 5v2v = 2

METHOD OF VIRTUAL WORK :The method of virtual work for finding constraint relation is very useful in complicated situations wherevisual inspection is difficult and number of strings is more.

StepI : Constraint forces are those forces whose work on the entire system is zero. To apply this methodwe should write the tension acting on each block.

StepII: Displace each of the movable bodies in +ve direction by SA.SB etc. Here we need not botherwhether these displacements are physically possible or not. Automatically the analysis will tellthe relationship between them.

StepIII : Find the work done by tension on each of the bodies. The sum total of all these works should bezero.

Assume that m1 moves a distance S1 down and m2 moves a distance S2 down.{ This is not physicallypossible, but we are dealing with vectors here. If the displacements are in opposite directions, theanswer will be negative for them.

W1 = F.S Fscos

= TS1.cos180= TS1

Since the pulley is massless, the tension in teh string connecting m2 to the pulley can be found outNewtons law for pulley

f = 2TW2 = 2TS2.cos180 = 2TS2

W1 + W2 = 0 This principle that the work done by the string is 0 is called the principle of virtual work.

Here we are actually using the fact that the work done by the two strings on the total system is 0. But thatis as good, becuase sum of two zeroes will also be zero. TS1 2 TS2 = 0 S1 + 2S2 = 0 V1 + 2V2 = 0 a1 + 2a2 = 0Principle of virtual work seems to be more complicated, but once we get an understanding of it, itbecomes a very easy tool.

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Ex. In the figure shown, the ring starts moving down from rest. What will be the relation between the velocityof the ring and the velocity of the block at any position? What will be the distance that the ring movesbefore coming to rest?

a

ym

yHh

H

Sol. To find the relationship between the velocities of the block and the ring, we will use the concept of virtualwork. We have already studied that the total work done by tension on a system is always zero. Assumingsmall displacements of the bodies when the angle made by the string is ( displacement is assumed to besmall so that the angle made by the string does not change appreciably)

TSRcos TSB = 0SRcos + SB = 0

vRcos + vB = 0Here we should be careful, the relationship between the small displacements is same as that of thevelocities because we have divided the entire relationship by small time interval dt. But to obtain therelationship between accelerations, we have to differentiate this expression which will involve derivativeof cos also because is also a variable.Since tension does not do any work, only work here is done by the force of gravity.

Wg = K1 + K2

mgh MgH = 2 2R B1 1mv mv2 2

The interesting thing to note here is that even if mass of the ring is less than that of the block, the ring willgo down. Explain.

At the position of rest, vB = 0. So from the equation of constrained motion, vR is also 0.mgh MgH = 0.also from the geometry, the original length of the string is a + y = .

2 2a h y H

2 2

2 2

22

2

2 2

a h a H

h H 2aH

MH H 2aHm

2aMHM m

2 2

2aMmhM m

It is clear from the equation that the physically admissible solution is available only when M>m. If M

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(OPTIONAL)Ex. mA = 1 kg, mB = 2 kg, mC = 10kg

Find velocity of A, B & C when Chas descended 2m

Sol. Here work is done by kineticfriction between A & B so it will notcancel out. But by tensionon A & C will cancel out.

wA = T 2 1 2

wC = 100 2 T 2Total work = 100 2 1 2

99 2 = 21

10v2 + 21

1 v2

v2 = 112299

| v = 6 m/s | A and C

Finding displacement of BaB = 0.5 ms2, u = 0, t from A and Ct = 2/3

s = 21

21

94

= 91

m

SF

1 91

= 21

2 v2 or vB = 31

ms1

you can see that work done by kinetic friction on A & B is not cancelling out completely.

Note : Thus except tension, normal & static friction even if we write work because of action & reaction on asystem it is not necessary that total work will be zero.

NATURE OF FORCES :[1] Conservative forces: Forces for which calculation of work is

independent of path taken by body. e.g. gravity,spring.

In this case work done by force of gravity is same for taking body from A to B by any path (Youcan show it mathematically)

* In conservative forces total work done for around closed path = 0[2] Nonconservative forces: Forces for which calculation of work

depends on path not just on initial and final position eg. friction.WfI = mglWfII = mg(3l)

Ex. F

= xy i + xy jCalculate work required to take particle from (0,0) to (2,2) (give it w/o path) Then remind them it cannotbe calculated w/o path

then show OAPOP III ww

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POTENTIAL ENERGY :It can be defined only for conservative forces.Definition : It is defined as negative of work done by conservative forcesFormula: F

represents force for which we are writing potential energy

dU = SdF

2

1

dU = 2

1

SdF

thus U2 U1 = 2

1

SdF

Purpose: By defining PE we can avoid repeated calculation of work for conservative forces and sincePE depends only on position (initial and final), we can directly write effect of conservative forces in termsof their respective PE'sWe will define PE for gravity and springGravity

U2 U1 = h

0

)j(dh)j(mg

U2 U1 = mghU3 U1 = mgh(similarly)

Emphasise that by definition we can only find difference of PE not absolute value.If we assume U1 = 0 then U2 = mgh U3 = mgh

Ex.1 If chain starts slipping find its KE when chain becomes completely straight

Hint: Wg = (KEf KEi)But Wg = (Uf Ui)Uf + Ui = KEf KEi

KEf + Uf = KEi + Ui find U by using calculusemphasise that if we have tried to find work due to gravity directly then it would have been very difficultas compared to the solution we are giving.

Ex. Chain is on the verge of slipping, find the velocity of the chain, when it has slipped.

Sol. f = 4Mg3

4Mg3

= 4Mg3

= 3

work done by friction force when chain completely slip off the table.df = dmg

dw = df x 4/L

0

gxdxLM

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wf = 3 LMg

4/L

0

2

2x

= 32

Mg3 l

3l 4

l/4u = 0

dx

xfNow decrease in PE = inc. in KE + wf

PEi PEf = 21

mv2 + wf

32Mg9 l

2

Mgl = 2

1Mv2 + 32

Mg3 l

32Mg7 l

= 21

Mv2 + 32Mg3 l

21

Mv2 = 32Mg4 l

v = lg21

As we have learnt from previous problem if some forces are acting on a bodyW1 + W2 + ... + Wn = KEf KEi

if some of them are conservative and others are nonconservative then for conservative forces we canwrite PE

Wc + Wnc = KEf KEi(Uf Ui)} + Wnc = KEf KEiWnc = KEf KEi + (Uf Ui)

Term on RHS is often called mechanical energy. Emphasise that effect of a force can either bewritten as work on LHS or it can come as PE on RHS

Optional Eg. Find vel. of A & B when A is about to touch theground. Also verify that work done by tension on thewhole system and N between A and B is zero.

mA = 5 kg mB = 10 kg Sol. |V|

= |u|

Net speed of block

( 37)u

v

Vb = 37cosu2uu222 =

54u2u2 22 = v

52

By energy conservationm

370

mu

2vDec. in P.E. of block = Increase in KE of wedge + block

mgh. = 21

mv2 + 21

mvb2

mgh = 21

mv2 + 21

mv2 52

5 10 2 = 21

10v2 + 21

5 52

v2

Bansal Classes [14]

5 10 2 = 212

v2

5 10 2 = 212

v2

T cos

TT sin

T

x

v = 3

50

velocity of wedge = 325 m/s

velocity of block = 52v =

52

350

= 320

= 352 m/s

work done by tension

T T sin

x cos

x

x sin x

T cos

(1) on wedgew = (T T cos ) x(2) on the blockT cos (x x cos ) T sin x sin T cos TxNet w = TX T x cos + Tx cos Tx = 0By normal reaction between A & B(1) on the wedge

N sin . x

N

N

N sin

x N cos

(2) on the block N sin (x cos ) + ( N cos x sin )= N sin x + N x sin cos N x sin cos = N sin x

N cos

xx

N

x cos N sin

x sin

Net work done by normal raction

N sin x N sin x= 0

Solved Example 7 & 9 of HCV[Home work : HCV (part-1) chapter-8 Ex.36 to 40]

SPRING :In case of spring natural length of spring is assumed to be reference point and always assigned zeropotential energy (This is a universal assumption). In gravity we can take any point as reference andassign it any value of potential energy.

Uf Ui = f

i

SdF

Uf 0 = ix

o

i)dx)(i(kx

U = 21

kx12

Bansal Classes [15]

for compression Uf Ui = f

i

SdF

= ix

o

)i)(dx(ikx

U = 21

kx2

Thus if spring is either stretched or compressed from natural length by x the potential energy is 21

kx2.

Emphasise that for solving problems of spring always measure distances from nature length.

Ex. Find how much m will rise if 4m falls away. Blocks are at rest and in equilibrium

[Sol. Applying WET on block of mass mwg + wsp = kf kiLet finally displacement of block from equilibrium is x

mg

xkmg5

+ 21

k

2

22

kgm25

21

kx2 = 0

21

kx2 + mgx k2

gm15 22 = 0

x = kmg3

displacement from initial is kmg5

+ kmg3

= kmg8

Ex. Find velocity of ring when spring becomes horizontal

m = 10 kgk = 400 Nm1

Sol. 4 m3m=h

37BC

A

m = 10 kgk = 400 N/mnatural length of spring = 4mdecreasing in PE = inc. in KE

21

k 1 + mgh = 21

mv2

21

400 12 + 10 10 3 = 21

10V2

200 + 300 = 5V25V2 = 500V = 100 m/s = 10 m/s

Bansal Classes [16]

EX. (a) A 2 kg block situated on a smooth fixed incline is connected to aspring of negligible mass, with spring constant k = 100 Nm1, via africtionless pulley. The block is released from rest when the spring isunstretched. How far does the block move down the incline beforecoming (momentarily) to rest? What is itsacceleration at its lowest point?

(b) The experiment is repeated on a rough incline. If the block is observed to move 0.20 m down along theincline before it comes to instantaneous rest, calculate the coefficient of kinetic friciton.

[Ans. (a) s = 0.24 m, a = 6 m/s2, (b) x = 1/8][Sol.(a) Applying work-energy theorem

mgs sin 37 = 21

ks2

210s 53

= 21

100s2 on solving s = 0.24 m

accelerating at its lowest point

a = m37sinmgks

= 2

5310224.0100

= 6 m/s2 a = 6 m/s2

(b) Work done by gravity + work done by friction = Energy stored in spring

mg s sin37 mg cos 37 s = 21

ks2

mg sin 37 21

ks = mg cos 37

2 10 53

21

100s = 2 10 54

given s = 0.20 m

16s5012

=

= 81

]

Ex. Draw Ux graph

Finding F from U

)U(F

represents kz

jy

ix

Ex. If U = 4x2y + 2yz2 find force

* if U depends on only one variable lets say r, then rdrdUF

Ex. U = 4r3 find forceSol. F

= 12 r2( r )

Bansal Classes [17]

EQUILIBRIUM :

If F and U are dependent only on one variable )x(dxdUF

Thus, if we say equilibrium 0F

dxdU

= 0

i.e. maxima or minima of PE representsequilibrium. Maxima is unstableequilibrium and minima is stableequilibrium

minima

maxima

neutral

x

u(x)

i.e. 22

dxUd

< 0 maxima unstable equilibrium

2

2

dxUd

> 0 minima stable equilibrium

2

2

dxUd

= 0 neutral equilibrium x1 x2 x

F

x1 is unsatble & x2 is stable.

(Optional)

To check whether a force is conservative. Explain this by the concept of potential energy.

0

FFF

zyx

kji

zyx

iz

F

yF yz

+ jz

FxF xz

+ ky

Fx

Fxy

= 0

POWER :Rate of doing work is called power delivered by the force.

Poweravg = tW

= timetotalworktotal

Powerinst = dtdW

= )SdF(dtd

= dtSdF

= VF

P = |V|)cos|F(|

= (comp. of F

along V

)Speed

Ex. If power delivered by net force if P0 find velocity as function of time (t = 0, vel. = u).

Sol. P = vdtdvm

Pdt = dvmv

Solved Example 3 of HCVHome work Q.No. 19 to 27, 41 to 51

Bansal Classes [18]

VERTICAL CIRCULAR MOTION:

EX. What is the minimum speed to reach B and C.

(B)21 mu2 = mg (R)

u = gR2Solve for (C) like this

(C)21 mu2 = mg (R)

u = gR4 . This is wrong.Why?

At any with vertical.

N = mg cos = R

mV2

N = mg cos + R

mV2

This equation is valid through out for > 2 as cos will go negative and comp. of mg will act in the

direction of normal. For O < gR5The body will freely move in a circle and 'N' will never be zero.

Case V: gR5 > u > gR2 .The normal will become zero some where between B and C. At this point v O. It will leave circularmotion and will become projectile because symmetry will no more be there as in the next instant velocitywill decrease further for which N should be negative which is not possible and so it will leave circularmotion and will have projectile motion.

Bansal Classes [20]

For a mass tied by a string about O.Here instead of normal 'Tension' is the worrying factor.T = O String is slack and 'm' wil leave circular motion.

All previous cases are valid similarly.

Case I: u < gR2 Pendulum

Case II: u = gR2 will reach B and come back.

Case III: gR4 > u > gR2 . The body will continue moving in circular motion as tension of a rod can gonegative which is allowed as then the rod instead of pulling the body will push it.

A P Q P A P' Q'Case IV: u = gR4

v = 0, T veThe body will stop at the top.

Case V: u > gR4 Forever will do circular motion.

Q. Block kept on a fixed smooth sphere give this as a classroom exercise to students.

(a)Find at which block will break off.(b) Initial velocity, so that block breaks off in initial position itself.

Solved Example 14 of HCV[Home work : Remain questions of chapter-8 ]