This is the teaching module for work energy. This module has to be followed in the class. VK Bansal
TEACHING NOTES WORK – 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 mechanical energy.
INTRODUCTION : Whatever we have learnt till now is more or less sufficient for solving any problem involving particles. By doing force analysis we can find acceleration & then we can find velocity, position, etc. That means we should be finished with particle dynamics. But it is not so. We will learn a new technique and concept which will make problem solving faster and will give simple solution to complicated problems. Work Energy Theorem is an extension to Newton's Laws 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 as dW = F·dS ,
There are two interpretations of ds . (1) If the body is moving as a complete unit, ds 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 ds is displacement of earth. (2)
If the different points of body have different displacement, d s is the displacement of point of application of force. In the example above displacement of different parts of our body are different. So ds is displacement of point of contact i.e. the foot in contact with the ground. So we can say clearly that work by earth on us is also zero.
OPTIONAL
Actually, there are no 2 divisions dW = F·dS , where d s is point of application of force on body. With this, you can explain both answers.
i)
We can note that work is a scalar quantity.
ii)
dW = | F || dS | Cos i.e. if component of force is along displacement ( < 90°) work is positive otherwise work is negative ( > 90°)
Questions In 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 energy theorem W = K can be applied to the boy to find the speed with which he leaves the ground.
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(2)True or False? A balloon is compressed uniformly from all sides. Because there is no displacement of the balloon’s center of mass, no work is done on the balloon. Both of these claims are false. Question (1) refers to a simple, everyday experience that unfortunately cannot be analyzed by means of traditional physics teaching without the introduction of additional worklike quantities and energy-like equations. The upward force on the boy that projects him into the air is the normal force on his feet from the ground. The center of mass of the boy indeed moves through an upward displacement. The normal force, however, goes through no displacement in the reference frame of the ground, and therefore no work is done by this force on the boy. The change in the boy’s kinetic energy 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 a skateboard, and a piece of putty slamming into a wall. In all of these cases, no work is done by the contact 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
FH on B
S (a) Ball does positive work on hand
S
FB on H (b)
S (c)
Hand does negative work on ball
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 FB on H 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 FH on B FB on H on the ball (figuure c). This force, which slows the ball to a stop 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 is just 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, you exert an upward force on the book and the book's displacement is upward, so the work done by the lifting force on the book is positive. But the work done by the gravitational force (weight) on a book being 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 dependent on reference frame) Illustration: A lift is going up with constant velocity. We will calculate work from two reference frames. In both reference frame N-mg = 0 N = mg Work Done in 2nd ref. frame = 0 Work Done in 1st ref. frame = N·ut (We can observe that work is dependent on reference frame.)
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[2]
3.
Calculation of work Case 1: When Force is uniform dW = F·dS
W = dW = F·dS = F · dS W = F·S (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(1–cos) Work Done by force F = FR(Sin Work Done by normal = 0
Ex.
Find work done by gravity and normal when block comes fromA to B Wg = mg(lsin) = (mgsin)l WN = 0 (because displacement is perpendicular to Force)
Q.
[Sol.
l
B
m
A
mg
A particle is moving along a straight line from point A to point B with position vectors 2 ˆi 7 ˆj 3 kˆ and 5 ˆi 3 ˆj 6 kˆ respectively. One of the force acting on the particle is F 20 ˆi 30 ˆj 15 kˆ . Find the work done by this force. [Ans. 315] W F. d d 3ˆi 10ˆj 3kˆ F = 20ˆi 30ˆj 15kˆ
W = 60 + 300 – 45 = 315
Ans.
Case 2: When Force is non–uniform (either magnitude or direction) dW = F·dS W = F·dS ( F cannot come out of integer since it is varying) Ex.
F xˆi y 2 ˆj .Particle moves from (1,2) to (–3,4) dW = F·dS ( dS dxˆi dyˆj ) dW = xdx + y2dy 3
4
x2 xdx y dy W = = 2 1 2
3
2
1
y3 + 3
4
= 2
68 3
* If force is not expressed as function of (x,y,z) then also we can solve problem by expressing force and displacement 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, then work done by spring force. F kxˆi dS (dx )( ˆi ) {dx is +ve as x is increasing} dW = F·dS x1
1
dW = – 0 kx dx = – 2 k x1 Ex. Sol.
2
Find work done by spring if we compress it further by x2. dS = dx(– ˆi ) F kxi dW = F·dS = –kx dx ( x 2 x1 )
W = –k
x dx = – 12 k[(x2 + x1)2 – x12 ]
x1
Work–Energy Theorem The total work done on a body by external forces is related to the body's displacement-that is, to changes 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 acting on 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, this means that the block speeds up ; from equation, this also means that the total work Wtot done on the n
F
block is positive.
(a) v
w Wtot > 0
The total work is also positive in figure (b), but only the component F cos contributes to Wtot. The block again speed up, and this same component F cos is what causes the acceleration. n
(b) v
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F w Wtot > 0
[4]
The total work is negative in figure (c) because the net force opposes the displacement ; in this case the n F
block slows down.
(c) v
w Wtot < 0
The net force is zero in figure (d), so the speed of the block stays the same and the total work done on the block is zero . We can conclude that when a particle undergoes a displacement, it speed up if Wtot > 0, slow down if Wtot < 0, and maintains the same speed if Wtot = 0.
(d) F w Wtot = 0
v
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 negative work. (d) The net force is zero and does no work, and the speed is constant. Derivation: For a particle
F =m a dV F·dS = m ·dS dt v
Work Done by resultant Force =
f F dS = m v dv
vi
1 2 1 2 Summation of work by all the forces = ( F·dS) = mv f mv i 2 2 W = kf – ki
i.e. Sum of work done by all the forces on a particle is equal to change in kinetic energy of the particle So the kinetic energy of a particle is equal to the total work that was done to accelerate it from rest to its present speed. The kinetic energy of a particle is equal to the total work that particle can do in the process of being brought to rest. This is why you pull your hand and arm backward when you catch a ball. As the ball comes to rest, it does an amount of work (force times distance) on your hand equal to the ball's initial kinetic energy. By pulling your hand back, you maximize the distance over which the force acts and so minimize the force on your hand. Q. [Sol.
A 60 gm tennis ball thrown vertically up at 24 m/s rises to a maximum height of 26 m. What was the work done by resistive forces? 1 wg + wres = (0 – mu2) 2 1 –mgh + wres = – mu2 2
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wres = 0.06 × 10 × 26 – = – 1.68 J Q.
1 × 0.06 × 24 × 24 2 ]
A force of (3 ˆi 1.5ˆj) N acts on a 5 kg body. The body is at a position of ( 2 ˆi 3ˆj) m and is travelling at 4 ms–1. The force acts on the body until it is at the position (ˆi 5ˆj) m. Assuming no other force does
[Sol.
work on the body, the final speed of the body. Given Mass of the body = 5 kg Force F = 3ˆi 1.5ˆj
[Ans. 10 ms–1]
Now displacement s = { (ˆi 5ˆj) ( 2ˆi 3ˆj) } m = ( iˆ 8ˆj) m From Work Energy principle 1 W = F · s = m(v2 – u2) 2
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 gravity Work done for x displacement 1 W = mgx – kx2 2 for max. displacement velocity should become zero ki = 0 kf = 0 1 mgx – kx2 = 0 2 2mg x= k
Applying Work Energy theorem on system We 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 systems that 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 pushes against the wall, setting himself in motion to the right. The forces acting on his are his weight w , the upward normal forces n1 and n 2 exerted by the ground on his skates, and the horizontal force F exerted on him by the wall. There is no vertical displacement, so w , n1 and n 2 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 is applied (the man's hands) do not move. Thus the force F also does no work. So where does the man's kinetic energy come from ? The difficulty is that it's simply not correct to represent the man as a single point mass. For the motion as we've described , different parts of the man's body must have different motions ; his hands are stationary against the wall while his torso(upper body) is moving away from the wall. The various parts of his body interact with each other, and one part can exert forces and do work on another part. Therefore the total kinetic energy of this composite system of body parts can change, even though no work is done by forces applied by bodies(such as the wall) that are outside the system. This would not be possible with a 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 particles together 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 done by static friction, tension and normal contact force i.e. by action and reaction on a system will always be zero.) According to newton’s laws, the body moves under the influence of the external force only. Internal forces do not accelerate the body. But work can still be done by them, even if they do not accelerate the body. In the case of work energy theorem, we should be cautious. Do not forget to take into account the work done by the internal forces. To illustrate these points about work, let us take a simple daily life example. Suppose a boy of mass 40 kg is walking on a rough ground with a uniform acceleration of 1 m/s2. We wish to find the work done on the 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 force is friction force between him and the ground. We can imagine trying to walk on a smooth surface. We will slip! 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 work is W = F .S = FS cos. What is wrong here? Let us examine the motion of the boy more closely. When the boy walks, he does not slip his feet on the ground. Rather, he places one foot on the ground, lifts another foot and moves that foot further. The force of friction is not acting on the moving foot. It is acting on the foot which is in contact with the ground. So this is the case of a body where different parts of the body having different displacements. `
Just now we said that in such cases, the work done by the force is W = F .S = FS cos. where S is the displacement of the point of application of the force. So although the boy as a whole is moving, the point at which friction force is being applied is not moving. On close examination, we can say that it is the muscles of the body who are rotating the legs, imparting the energy to the boy. Their work can be estimated with the help of the work energy theorem. V2 = U2 + 2as = 0 + 2 × 1 × 1
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Wint = K =
1 × 2
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 by the boy reflects in his food consumption. This should be the same from every frame of reference. Frictional work Consider a block sliding across a horizontal table and eventually coming to rest due to the frictional force exerted by the table. As the kinetic energy of the block decreases, there is a corresponding increase in the internal energy of the system of block + table. This increase in internal energy might be observed as a slight increase in the temperature of the surfaces of the block and the table. It is a common observation that kinetic friction between two surfaces causes an increases in the temperature, as for example in the case of holding a piece of metal against a grinding wheel or applying the brackes to an automobile or a bicycle (in which case both the brakes and the sliding tires can become warmer). You can even observe that effect by rubbing your hands together. We might be tempted to write the magnitude of the work done by the frictional force as the product of the 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 textbook and lecture discussions that the work done by friction on a block sliding on a surface is W = –fkd, where fk is the force of kinetic friction on the block and d is the distance through which the block moves relative to the surface. The negative sign indicates that the friction force is in the opposite direction to the displacement. This expression for work is then incorporated into the work-kinetic energy theorem for the block. This approach ignores the fact that the displacement of the block is not the same as the many displacements of the friction force at a large number of contact points. This latter displacement is complicated and involves deformations of the lower surface of the block. IT HAS TO BE NOTED THAT THE WORK DONE BY FRICTION CANNOT BE CALCULATED BECAUSE WE CANNOT FIND THE DISPLACEMENT AT LARGE NUMBER OF CONTACT POINT. EVEN IN THE CASE OF STATIC FRICTION, THERE IS VERY SMALL DISPLACEMENT AT THE CONTACT POINT SO THE WORK DONE BY STATIC FRICTION IS ALSO NOT ZERO. BUT FOR TEACHING IN JEE, WE WILL ASSUME THE FOLLOWING FOR FINDING THE WORK DONE BY FRCTION. A student may have little difficulty with W = –fkd, based on his or her understanding of evaluating the work done by any force by performing a path integral over the path followed by the object. In the case of a block sliding over a stationary surface, the friction force is always oppositely directed to each infinitesimal displacement of the block. For a constant friction force, this integral reduces to the product of 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 bodies
W = 30 = kEsys =
wA
=
5×2
wB
=
15 × 2 +(–5) × 2
1 1 × 10v2 + × 5v2 2 2
v=2
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(II) We know work done by static friction will be zero because action–reaction will be in opposite direction but displacement of contact point will be same. Thus f A ·dSA f B ·dSB 0 because dSA = d S B but f A = fB Thus 15 × 2 = (1/2)× 10v2 + (1/2)× 5v2 v=2 METHOD OF VIRTUAL WORK : The method of virtual work for finding constraint relation is very useful in complicated situations where visual inspection is difficult and number of strings is more. Step–I : Constraint forces are those forces whose work on the entire system is zero. To apply this method we should write the tension acting on each block. Step–II: Displace each of the movable bodies in +ve direction by SA.SB etc. Here we need not bother whether these displacements are physically possible or not. Automatically the analysis will tell the relationship between them. Step–III : Find the work done by tension on each of the bodies. The sum total of all these works should be zero.
Assume that m1 moves a distance S1 down and m2 moves a distance S2 down.{ This is not physically possible, but we are dealing with vectors here. If the displacements are in opposite directions, the answer will be negative for them. W1 = F.S Fs cos = TS1.cos180º = –TS1 Since the pulley is massless, the tension in teh string connecting m2 to the pulley can be found out Newton’s law for pulley
f = 2T W2 = 2TS2.cos180º = –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 that is as good, becuase sum of two zeroes will also be zero. –TS1 –2 TS2 = 0 S1 + 2S2 = 0 V1 + 2V2 = 0 a1 + 2a2 = 0 Principle of virtual work seems to be more complicated, but once we get an understanding of it, it becomes 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 velocity of the ring and the velocity of the block at any position? What will be the distance that the ring moves before coming to rest?
m
a y
h
y–H H
Sol.
To find the relationship between the velocities of the block and the ring, we will use the concept of virtual work. We have already studied that the total work done by tension on a system is always zero. Assuming small displacements of the bodies when the angle made by the string is ( displacement is assumed to be small so that the angle made by the string does not change appreciably) –TSRcos –TSB = 0 SRcos + SB = 0 vRcos + vB = 0 Here we should be careful, the relationship between the small displacements is same as that of the velocities because we have divided the entire relationship by small time interval dt. But to obtain the relationship between accelerations, we have to differentiate this expression which will involve derivative of 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 1 2
1 2
mgh – MgH = mv R 2 mvB2 The interesting thing to note here is that even if mass of the ring is less than that of the block, the ring will go 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 = . a2 h2 y H
a 2 h2 a H h 2 H 2 2aH 2
MH 2 H 2aH m 2aM 2 H 2 M m2
h
2aMm M 2 m2
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 C has descended 2m Sol. Here work is done by kinetic friction between A & B so it will not cancel out. But by tension on A & C will cancel out.
wA = T × 2 – 1 × 2 wC = 100 × 2 – T × 2 Total work = 100 × 2 – 1 × 2 99 × 2 =
1 1 × 10v2 + × 1 × v2 2 2
99 2 2 | v = 6 m/s | A and C 11 Finding displacement of B aB = 0.5 ms–2, u = 0, t from A and C t = 2/3
v2 =
s=
F·S
1 1 4 1 × × = m 2 2 9 9
1 1 1 = × 2 × v2 or vB = ms–1 9 2 3 you can see that work done by kinetic friction on A & B is not cancelling out completely.
1×
Note : Thus except tension, normal & static friction even if we write work because of action & reaction on a system 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.
Ex.
In this case work done by force of gravity is same for taking body from A to B by any path (You can show it mathematically) * In conservative forces total work done for around closed path = 0 [2] Non–conservative forces: Forces for which calculation of work depends on path not just on initial and final position eg. friction. WfI = –mgl WfII = –mg(3l) F = xy ˆi + xy ˆj Calculate work required to take particle from (0,0) to (2,2) (give it w/o path) Then remind them it cannot be calculated w/o path then show
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w OP w I
II OAP
[11]
POTENTIAL ENERGY : It can be defined only for conservative forces. Definition : It is defined as negative of work done by conservative forces Formula: F represents force for which we are writing potential energy dU = – F·dS 2 dU = – F·dS 2
1
U2 – U1 = – F·dS 2
thus
1
1
Purpose: By defining PE we can avoid repeated calculation of work for conservative forces and since PE depends only on position (initial and final), we can directly write effect of conservative forces in terms of their respective PE's We will define PE for gravity and spring Gravity
h
ˆ ˆ U2 – U1 = mg ( j)dh ( j) 0
U2 – U1 = mgh U3 – 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 calculus emphasise that if we have tried to find work due to gravity directly then it would have been very difficult as compared to the solution we are giving.
Ex.
Sol.
Chain is on the verge of slipping, find the velocity of the chain, when it has slipped.
3Mg 4 3Mg 3Mg = =3 4 4 work done by friction force when chain completely slip off the table. df = dmg
f=
L/4
dw = df x
0
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M dx gx L
[12]
L/4
x2 3Mgl = 2 32 0 Now decrease in PE = inc. in KE + wf Mg wf = 3 L
l/4 f
u=0
x dx
1 PEi – PEf = mv2 + wf 2
3l 4
9 Mgl Mgl 1 3Mgl – = Mv2 + 32 2 2 32
7Mgl 1 3Mgl = Mv2 + 32 2 32 1 4 Mgl Mv2 = 2 32 1 gl 2 As we have learnt from previous problem if some forces are acting on a body W1 + W2 + ... + Wn = KEf – KEi if some of them are conservative and others are non–conservative then for conservative forces we can write PE Wc + Wnc = KEf – KEi –(Uf – Ui)} + Wnc = KEf – KEi Wnc = KEf – KEi + (Uf – Ui) Term on RHS is often called mechanical energy. Emphasise that effect of a force can either be written as work on LHS or it can come as PE on RHS
v=
Optional Eg. Find vel. of A & B when A is about to touch the ground. Also verify that work done by tension on the whole system and N between A and B is zero. mA = 5 kg mB = 10 kg Sol. |V| = |u| Net speed of block v (–37°)
u
Vb =
u 2 u 2 2u 2 cos 37 =
2u 2 2u 2
4 =v 5
2 5
By energy conservation
m v
Dec. in P.E. of block = Increase in KE of wedge + block mgh. =
1 1 mv2 + mvb2 2 2
mgh =
1 1 2 mv2 + mv2 2 2 5
5 × 10 × 2 =
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u m
2
370
1 1 2 10v2 + 5 × v2 2 2 5
[13]
5 × 10 × 2 =
T
12 2 v 2
T cos
T T sin
12 2 5 × 10 × 2 = v 2
x
v=
50 3
velocity of wedge = 5
2 m/s 3
velocity of block = v
2 = 5
50 2 3 5 T
20 5 = =2 m/s 3 3 work done by tension (1) on wedge w = (T – T cos ) x (2) on the block T cos (x – x cos ) – T sin x sin T × cos – Tx Net w = TX – T x cos + Tx cos – Tx = 0 By normal reaction between A & B (1) on the wedge N sin . x (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 Net work done by normal raction
T sin
T cos
x cos x x x sin
N N sin N x
N cos
N cos N N sin x cos x
x
x sin
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 zero potential energy (This is a universal assumption). In gravity we can take any point as reference and assign it any value of potential energy. Uf – Ui = – F·dS f
i
xi
ˆ ˆ Uf – 0 = kx ( i )(dx ) i o
U=
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1 kx 2 2 1
[14]
xi ˆ ˆ Uf – Ui = – F·dS = kx i (dx )( i ) f
for compression
i
o
U=
1 2 kx 2
Thus if spring is either stretched or compressed from natural length by x the potential energy is
1 2 kx . 2
Emphasise that for solving problems of spring always measure distances from nature length.
Ex. [Sol.
Find how much m will rise if 4m falls away. Blocks are at rest and in equilibrium Applying WET on block of mass m wg + wsp = kf – ki Let finally displacement of block from equilibrium is x 5mg 1 x + k – mg 2 k
25 m 2 g 2 1 k 2 – kx2 = 0 2
1 2 15 m 2 g 2 kx + mgx – =0 2 2k
x=
3mg k
displacement from initial is Ex.
5mg 3mg 8mg + = k k k
Find velocity of ring when spring becomes horizontal
m = 10 kg k = 400 Nm–1 A
Sol.
C
37°
4m
3m=h B
m = 10 kg k = 400 N/m natural length of spring = 4m decreasing in PE = inc. in KE 1 1 k × 1 + mgh = mv2 2 2 1 1 × 400 × 12 + 10 × 10 × 3 = × 10V2 2 2 2 200 + 300 = 5V 5V2 = 500
V = 100 m/s
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= 10 m/s
[15]
EX. (a) A 2 kg block situated on a smooth fixed incline is connected to a spring of negligible mass, with spring constant k = 100 Nm–1, via a frictionless pulley. The block is released from rest when the spring is unstretched. How far does the block move down the incline before coming (momentarily) to rest? What is its acceleration 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 the incline 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° =
1 2 ks 2
3 1 = ×100×s2 on solving s = 0.24 m 5 2 accelerating at its lowest point
2×10×s×
(b)
3 100 0.24 2 10 ks mg sin 37 5 a= = = 6 m/s2 a = 6 m/s2 m 2 Work done by gravity + work done by friction = Energy stored in spring mg s sin37° – mg cos 37° × s = mg sin 37° –
1 2 ks 2
1 ks = mg cos 37° 2
3 1 4 – ×100×s = ×2 ×10× 5 2 5 given s = 0.20 m 12 50 s = 16 1 = ] 8
2 ×10 ×
Ex.
Draw U–x graph
Finding F from U F ( U)
ˆ ˆ ˆ i j k represents x y z Ex.
If U = 4x2y + 2yz2 find force dU * if U depends on only one variable lets say r, then F rˆ dr
Ex. Sol.
U = 4r3 find force F = 12 r2(– rˆ )
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EQUILIBRIUM : dU If F and U are dependent only on one variable F ( xˆ ) dx dU Thus, if we say equilibrium F 0 =0 dx i.e. maxima or minima of PE represents equilibrium. Maxima is unstable equilibrium and minima is stable equilibrium
u(x)
maxima
neutral
d 2U < 0 maxima unstable equilibrium dx 2
i.e.
d 2U > 0 minima stable equilibrium dx 2 d 2U = 0 neutral equilibrium dx 2 x1 is unsatble & x2 is stable.
minima x
F
•x 1
x2
x
(Optional) To check whether a force is conservative. Explain this by the concept of potential energy.
ˆi
ˆj
kˆ
x
y
0 z
Fx
Fy
Fz
Fz Fy F F ˆi + z x ˆj + x z y z
Fy Fx kˆ x y = 0
POWER : Rate of doing work is called power delivered by the force. Poweravg =
W total work = total time t
dS dW d Powerinst = = ( F·dS) = F· = F·V dt dt dt P = (| F | cos ) | V | = (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.
dv P = m v Pdt = mv dv dt
Solved Example 3 of HCV Home work Q.No. 19 to 27, 41 to 51
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VERTICAL CIRCULAR MOTION: EX.
What is the minimum speed to reach B and C. (B)
1 mu2 = mg (R) 2
u = 2gR Solve for (C) like this (C)
1 mu2 = mg (R) 2
u = 4gR . This is wrong.Why? At any with vertical. N = mg cos =
mV 2 R
N = mg cos +
mV 2 R
This equation is valid through out for > direction of normal. For O < <
2
2
as cos will go negative and comp. of mg will act in the
N will never be zero as both mg cos and
mV 2 R
are positive.
Hence it will be contact and will have circular motion. Using work energy – mg (R(1–cos )) =
mV 2 2
2 mu 2
–
mu 2 2
V2 = m 2 mgR (1 cos ) V2 = u2 – 2gR (1 – cos ) V2 = u2 – 2gR + 2g R cos N = mg cos + = mg cos + N=
m(u 2 2gR 2gR cos R mu 2 R
– 2 mg + 2 mg cos
m 2 [u – 2gR + 3g R cos ] R
O<<
2
Normal will not become zero. If we want to find minimum value to reach B there is no need to see the equation of normal all that matters is speed. at = 2 O = u2 – 2gR + 2gR (O) u = 2gR
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Case I:
u = 2gR it will just reach B.
Motion:
A B A D A B At B, N = O but it wil not loose constant.
Case II:
u<
2gR The body will not reach B but its velocity will become zero before B. e.g. Let u = gR O = gR– 2gR + 2gR cos 1 2
cos =
at 60° the body will stop. The body will not remain stationary as its tangential acceleration will not be zero. >
what if
2
Here the normal will become zero before velocity. This is why 4gR was wrong as we were considering speed and not normal where as to reach C it is necessary that 'N' does not become zero. Find minimum speed to reach C. O = u2 – 5gR [ = ] u = 5gR Case III:
u = 5gR v2 = 5gR – 2gR – 2gR = gR v = gR
Minimum possible value of 'N' and 'v' is at 'C'.
v = gR As mg =
mv2 R
is valid
so the body will continue moving in circular motion. u = 5gR implies the body has just completed circular motion. Note: We check for '' as cos has maximum negative value. If N is not O at this point then for all < the normal will never be zero. Case IV: u > 5gR The body will freely move in a circle and 'N' will never be zero. Case V: 5gR > u > 2gR . The normal will become zero some where between B and C. At this point v O. It will leave circular motion and will become projectile because symmetry will no more be there as in the next instant velocity will decrease further for which N should be negative which is not possible and so it will leave circular motion and will have projectile motion.
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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 < 2gR – Pendulum
Case II:
u = 2gR will reach B and come back.
Case III: 4gR > u > 2gR . The body will continue moving in circular motion as tension of a rod can go negative 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 = 4gR v = 0, T –ve The body will stop at the top.
Case V:
u > 4gR 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 ]
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