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Work Introduction The concept of work-energy is universal and is applicable almost in all fields of physics, engineering chemistry, biology etc. However here we analyze the application of work concept in Mechanics. ‘Work’ has much more in it then just a language tool. For example, if a person is holding an object, he gets tired but still does no work. Here we will analyze such myths and also explore the term power. We will also gain an insight into energy approach for solving mechanics problems which were tedious to solve using Newton’s law. WORK: When ever force acting on a body is able to actually move it through some distance in the direction of force, the work is said to be done by the force. The force performing work may be constant or a variable. Work done by CONSTANT FORCE: An object undergoes displacement‘s’ along a straight line while acted on by a force F, the angle between done isq. Then work is W= Fig (1) a) Work done as such has b) Work is scalar c) Unit of d) Dimension of work is [ML2T-2] no relevance quantity work is until (as the commenced force is indicated by dot Joules 1J = maintained. product). 1Kgm 2/s2 Other units of Work: Fig. (2) Discuss this topic Topic Answers Author Views Last message fluid mechanics 9 umang 77 24/02/2007 10:53 rao_prabhat elasticity 9 kash001 124 24/02/2007 10:44 rik_mad stoke's law 3 umang 43 24/02/2007 10:40 devnexus basic summary of kinematics 0 devnexus 6 24/02/2007 10:36 devnexus straight line motion 1 nrki99 25 24/02/2007 01:38 truly FOR EXPERTS 1 deep01 31 24/02/2007 01:24 truly kinematics 7 mandar5227 73 24/02/2007 01:14 truly constrain relation 0 joydas2514795 18 23/02/2007 21:36 joydas2514795 know the answer but not the reason 8 tibu 68 23/02/2007 21:16 aditya_arora04 plzzz help-rotation prob 5 anubhav07 48 23/02/2007 21:00 vish0001 axis of rotation query- please answer quickly !! 1 vish0001 28 23/02/2007 20:39 vish0001 rod 9 neeraj_agarwal_1990 91 23/02/2007 19:08 ankur.kkhurana mechanics 3 sucess_seeker 93 23/02/2007 18:24 bhupesh.mohanty rotational motion 2 swapna.sugunendran 57 23/02/2007 18:19 bhupesh.mohanty QUESTIONS FOR THE EXPERTS 1 deep01 47 23/02/2007 18:07 anuragasp Illustration: 1) A block of mass M is pulled along a horizontal surface by applying a force at an angle with horizontal. Co-efficient of friction between block and surface is . If the block travels with uniform velocity, find the work done by this applied force during a displacement d of the block. Solution: The force acting on the block is shown in figure. Fig (3) As the FCosq Solving block = FSinq+N= moves with uniform velocity the force N Mg (1) and add up ------------------------------- to zero. (1) (2) (2) FCosq F Work = done by W WORK If the = a displacement variable (Where (i.e. remains VARYING then changes) constant for S done interpretation = Area need to d is = BY a (If Graphical Work Area during F.dCosq DONE is force = force this = (Mg-FSinq) work done displacement Ds) 0 of under ) above vs. with F add FORCE: is result: graph. sign. S Fig (4) SPRING: The spring F = force -KX, varies as where Work done (Initial compression/elongation a K lines is function of stiffness in is X0 compression constant or of elongation. spring. on and final spring: compression/elongation is X1) Proof: (Assuming force is parallel to displacement) = Illustration: A block of mass m released from rest onto an ideal non-deformed spring of spring constant ‘K’ from a negligible height. Neglecting the air resistance, find the compression ‘d’ of the spring. Solution: Fig (6) Here there are two forces (Work Net work K= W mg POWER: Power is Power done involved, done must be equal to change 0-0 in as the can and kinetic energy (theorem to = rate at be force. which work is being average weight) be W spring time spring by + defined weight discussed in next = done or energy or is being section). 0 0 transferred. instantaneous Proof: Remember power is never negative. dW stands for small work done and not difference in work because nothing called ‘difference of work’ exists. Practical unit of power is horse power (hp) 1hp = 746W. Illustration: An elevator has a mass of 1000kg and carries a maximum load of 600kg. A constant frictional force of 4000N retards its upward motion, as shown in figure. What must be the minimum power delivered by the motor to lift the elevator at a constant speed of 3 m/sec. What power must the motor deliver at any instant of time, if it is designed to provide an upward acceleration of 100 m/s2 Fig (7) Solution: Let From a) force applied by T-W-f motor f.b.d 0 = T be T of [because W+f = = P b) V = Ma+W+f = = TV of elevator = at 3 Watts 104 = = Where v is instantaneous speed (Without ‘v’, value of P cannot be evaluated). f lift is constant] (1000+600)´9.8+4000 19680N x T-W-f P be T.VCos 19680 x T (8) force = = 5.9040 = Fig frictional and which power Ma =1600´1+1600´9.8+4000 21280 21280v Watts. needs to be determined ENERGY: A body is said to possess energy if it has the capacity to do work. If some work is done by the body then it looses energy. Energy and work are mutually interconvertible. Various forms of Energy are: 1)Heat 2) Electrical 3) Chemical 4) Mechanical 5) Nuclear. MECHANICAL ENERGY: Kinetic Energy: Kinetic energy of a body is the energy possessed by a body by virtue of its motion. It is the energy associated with moving body. , Where, m-mass of body, v-velocity of Kinetic Energy is measured by the work done by moving against external impressed force, before coming to rest. Relation between Kinetic Energy and body. Momentum: How? A ball falls under gravity from a height 10m, with an initial velocity V0. It hits the ground, looses 50% of its energy after collision and it rises to the same height. What is the value of V0? Solution: Let Let V1 Ratio [ V be of the K be velocity before 50% the velocity after impact and when and after it reaches impact of it hits the the same height ground 10m = K is lost] Potential Energy: Potential Energy of a body is the energy possessed by the body by virtue of its position or configuration. Potential energy is due to the interaction between bodies. Physically potential energy is applicable only to the class of forces where work done against the force gets ‘stored up’ as energy. Such forces are called CONSERVATIVE FORCES. Conservative forces perform work whose value is path independent. Thus work done by conservative force in a closed path is zero. Forces which do not satisfy path independent nature are called NON-CONSERVATIVE FORCES. Mathematically conservative forces F(x) can be written in terms of their Potential energy forces V(x) as For 3-Dimensional Consequently potential energy for case single dimension is given by Potential F = V-0 -Kx here Energy natural length of configuration is considered Stretched as zero of potential spring: energy. = Illustration: A spring of normal length ‘ ’ and spring constant K is fixed on the ground and the other is filled with a smooth ring of mass m which slides on a horizontal rod fixed at a height also equal to l (see fig). Initially the spring makes an angle of 53 0 with horizontal when the system is released from rest then what is the speed of the ring? Solution: In the initial position of the ring as shown in Fig figure the length of the (13) spring Extension Energy stored in spring = This stored energy when released becomes Kinetic energy of the ring, if V is the velocity of the ring, Kinetic energy when it is vertical Equating = two energy WORK ENERGY THEOREM: Work Energy theorem states that, the net change in Kinetic Energy of a particle (system) is equivalent to the sum of work done by all types of external forces be it conservative or non-conservative. Dumb Question: Velocity time graph of particle of mass 2kg moving in a straight line is as shown in figure. Find the work done by all the forces on the particle. fig.14 Solution: Initial velocity of particle V0 = 20 m/s, Final velocity of particle vf, From Work energy theorem = 0-400 Best thing about work energy theorem is that nature and value of force is not at all required for work equation. Proof of Work (Multiplying Energy y and (Segregating Integrating = both dividing the 400. Theorem: by ‘ds’) term) side Illustration: Two bodies A and B connected by a light rigid bar of 10m long and moving in two frictionless guides as shown in the figure. If B starts from rest when it is vertically below A, find the velocity of B when x=6m. Assume m A = mB =200kg and mC = 100kg. Fig Solution: At any instant Where Including C move Work done by Final Kinetic when the velocity of all down gravity on 6m system = energy of Velocity from gravity only of B at is A three since {Work done the (Speed Now (W ext)net Here bar system of (16) as shown and bodies B in = B Kf = and the required moment the ---------- (2) + 6m Work (KA) C force (16) (1) velocity work external constitute figure --------- moves down A moving the in is CONSERVATION OF Kinetic and Potential energy Total Mechanical energy = K+V changes only if in + of our along done in (KB) must energy which = MECHANICAL mechanical + be is = of (KC) same) doing 6.9m/s energy B system x-axis moving C} theorem work. Ans. ENERGY: system. Energy is converted into other forms like heat, sound etc. Work is being done by non-conservative forces like, frictional force Proof: Let a body undergo displacement Dx under the action of conservative force F. Then from work energy theorem Kf – Ki = F(x) x --------------(1) Since force is conservative hence its potential energy function can also be written as Vf-Vi= -F(x) x ----------------(2) Adding (1) and (2) (Vf+Kf)(Ki+Vi) = F(x) Dx F(x) Dx = 0 Vf+Kf = Ki+Vi Initial Mechanical energy Important Points regarding Conservation Principle: = Final Mechanical energy. If friction and drag are present then mechanical energy will not be conserved Conservation of mechanical energy applies only to isolated systems. Initial and final states of system must be clearly identified before applying conservation principle. Illustration: A block of mass m is pushed against a spring of spring constant K fixed at one end to a wall. The block can slide on a frictionless table as shown in figure. The natural length of the spring is L 0 and it is compressed to half its natural length when the block is released. Find the velocity of the block as a function of its distance x from the wall. Solution: Fig (18) When the Block is released the spring pushes it towards right. The velocity of the block increases till the spring acquires its natural length. There after the block loses contact with the spring and moves with constant velocity. Initially the compression in the spring = When Using the distance the of block from principle the wall becomes of its where x< conservation the compression of is Solving When (L 0-x) energy this the spring acquires We have There after the block continues with this velocity. its natural then length x = L0 CIRCULAR MOTION: Uniform Circular motion: Angular velocity w is constant throughout the motion. The magnitude of velocity also remains constant. Non-Uniform Circular Relation Angular motion: Between Taking a moves an angle velocity v, , small time ‘ ’ changes r, with time. : interval t in which body (very small) Fig (19) Distance Considering moved this time by interval it to on be so v circumference small that is the = distance ( is ) almost = Actual = straight s line. r vector equation: CENTRIPETAL AND TANGENTIAL FORCES: The forces acting on a body could be resolved into two components, one in radial direction and one in tangential direction. The force in radial direction is called CENTRIPETAL FORCE and the other is called TANGENTIAL FORCE. Fig In First part is tangential equation acceleration (20) and second is centripetal acceleration. Illustration: A 0.1 Kg block is undergoing circular motion. What is the range of ‘ ’ for which the particle can perform the circular motion? Solution: f.b.d For of body for wmin, vertical the particle will Equilibrium Similar value of ‘ ’ friction the move and R body will = will have be analysis (22) downwards. body 2R (1) Now For maximum Direction to of For body to rotate in circle, net force towards center of circle must be equal to m From Fig tendency have (2) h tendency in to move downward tan450=h upwards direction. yields. Derivation: Motion in a vertical circle: A particle of mass m is attached to inextensible light string of length l and it is imparted a velocity u in horizontal direction lowest point. Let v be its velocity at point B of circle. Fig Here h By centripetal Now So R (1-Cosq) conservation Necessary The = of force three string substituting is mechanical provided by the condition does T not = 0 energy resultant arise stuck at and (23) (1) ------------ tension T and depending highest = of point for if completing mgCos on T 0 the circle u, at = , in (B) (At And So highest h from point) 2R (2) = equation Or Or the particle Substituting q = 00 and v = u = position is 6mg. will complete the circle. in equation (3) we get T = 6mg. So in the critical condition tension in the string at lowest Fig (24) Putting T = 0 Substituting in Velocity 0 Now the particle T = h1<h2 before reaching highest equation of (1) particle becomes point for we zero = will 0 leave the but circle if v tension 0 in the string this becomes is zero but velocity is not possible only get when u2-2gh zero or, when Therefore The V h2 particle = if will Further if h1 = h2, oscillate 0 the if velocity but of T the particle 0 particle becomes this < zero leaves but is tension in the possible the circle. string is only not zero. when h1 and tension and velocity both becomes zero simultaneously. Illustration: A ball of mass m slides without friction down a path from height h and then moves in loop of radius exerted on the ball by the track at B and Find the force at C Fig (27) Fig (28) Solution: For On maintaining applying circular energy motion net conservation On centripetal between force must point A be greater and C than we get solving From On Similarly f.b.d of body FB=3mg point solving for mg [h-(R+RCosa)] = (1+Cosb) B (energy equilibrium) FB = 3mg (1-Cosa) Banking of Roads: When vehicle go through turnings, they travel along a nearly circular arc. There must be some force to produce the required centripetal acceleration. Centripetal force is provided to the vehicle by following three ways. By friction only By banking of roads only By friction and banking of roads both. 1) Let a And car of f L= mass m is limiting For moving value a at a By speed of f safe v is = at horizontal N turn So if m and r is fixed the speed of the vehicle should not exceed circular = arc of friction: radius r. mg (N=mg) without sliding, and if v and r is fixed then coefficient of friction should be greater than 2) To By avoid friction outer part of banking road is some of what lifted roads compared to the inner only: part. Fig From figure From At this speeds two (29) NSin we car does (29) = get, not and Tan down slide NCos = mg = even if track is smooth. 3) By friction and banking of road both: If a vehicle is moving on a circular road which is rough and banked also then magnitude of N and direction plus magnitude of friction mainly depends on the speed of the vehicle V- f is outward if v = 0 f is inward if v> f is outward if v < f is zero if v = a) When b) Now the when the car car is is given at a small rest speed v, f Fig ------- is equation upward- (1) (30) becomes Illustration: A turn of radius 20m is banked for the vehicle of mass 200kg going at a speed of 10m/s. Find the direction and magnitude of frictional force acting on a vehicle if it moves with a speed a) 5 m/s b) 15 m/s assume the friction is sufficient to prevent slipping ( g = 10m/s2). Solution: a) As The turn the is banked speed is for speed v decreased = 10 force m/s of therefore friction f acts upwards Fig Substituting b) Substituting , Here v=5 force v=15 m/s, of m/s, (31) friction , m=200kg, f m=200kg will and r=20m act r=20m downwards we get PROBLEMS (EASY) 1) Force acting on a particle varies with x as shown in figure (33) .Calculate the work done by the force as the particle moves from x = 0 to x = 6m. Fig (33) Solution: The work done by the force is equal to the area under the curve from x=0 to x=6.0m. This area is equal to the area of the rectangular section from x = 0m to x = 4.0m plus the area of the triangular section from x = 4.0m to x = 6m. 2) The mass system is kept on sphere. Ball 1 is slightly disturbed. What is the velocity of these balls when it is making angle ‘q’ with horizontal (friction is absent everywhere). Fig (34) Solution: Since they are connected by inextensible string, therefore at any stage the velocity of both particle will be same. Let it be ‘V’. From work energy theorem: (Only 3) What is the minimum value of ‘u’ for completing external circular motion force of particle Fig Solution: as is shown gravity) in figure (35)? (35) Choosing If V1 zero is of the velocity potential at energy ‘A’ then at from level energy of A. conservation 4) A block is projected horizontally on rough horizontal floor. The coefficient of friction between the block and the floor is m. The block strikes a light spring of stiffness K with velocity v0. Find the maximum compression of the spring. Fig (36) ������� Solution: Since the block slides and the spring is compressed through a distance ‘x’ the net retarding force acting on it. = Þ F Work = done by (kx net +mN) force for Fig = the (37) (mmg+kx) displacement x, 5) In figure 38-A and 38-B AC, DE and EF are fixed inclined planes BC = EF and AB = DE = y. A small block of mass m is released from rest from the point A. It slides down AC and reaches C with a speed VC. The same block is released from rest from the point D, it slides down DEF and reaches the point F with speed VF. The coefficient of kinetic friction between the block and the surface AC and DEF is a, calculate VC and VF. Fig Solution: a) ME This Loss at A in = mgY+0 ME is and equal (As b) In if to (38) VC work is done Cosa this the velocity against = at friction C i.e. x/s) situation, 6) A the Solution: locomotive of mass m starts moving so that its velocity v is according to the law where a is constant and s is distance covered. Find the total work done by all the forces acting the locomotive during first t seconds after the beginning of motion. Given, w.r.t differentiating Accelerating now force Here ‘t’ we acting on locomotive u Now get =0 using we have 7) The Kinetic energy of a particle moving along a circle of radius R depends on the distance covered S and T = as2, where a is constant. Find the force acting on the particle as a function of S. Solution: Differentiating both sides w.r.t ‘t’ we have a Now force = acting on the particle is given by 8) A ball suspended by a string of length 20 cm is fixed to the free end of the pivoted rod of length 40 cm as shown in the figure (39). The rod is made to rotate in a horizontal plane with constant angular speed. The string makes an angle q = 300 with the vertical axis. Find the angular speed of the rotation? Fig (39) = angular speed T = Tension in the string. Ball moves in horizontal circle TCosq = mg Along radial direction, TSinq Using (1) with radius = mrw2 r = L1+L2Sinq and along vertical direction, ----------(1) ----------(2) (2) w = 3.398 rad/s 9) A smooth, light rod AB can rotate about a vertical axis passing through its end A. The rod is fitted with small sleeve of mass m attached to the end A by a weightless spring of length l0, stiffness k. What work must be performed to slowly get this system going and the angular velocity w? Fig Solution: The mass Centripetal \W = m rotates force change in in a on circle of radius = m KE of (40) l, m+ which k energy is extended (l- stored length l0) of in = the spring. mw2l the spring 10) A spring gun having spring constant 100 N/m, a small ball of mass 0.1Kg is placed in its barrel by compressing the spring through 0.05m as shown in figure (41) a) Find the velocity of the ball when spring is released b) Where should a box is placed on ground so that ball falls in it, if the ball leaves the gun horizontally at a height of 2m above the ground. Fig (41) Solution: When the spring is released its elastic potential energy is converted into kinetic energy. As vertical component of velocity of ball is zero. Time taken by the ball to reach the ground, So, the horizontal distance traveled by the ball in this time PROBLEMS (MODERATE) 1) A hemispherical bowl of radius R = 0.1m is rotating about its own axis (which is vertical), with an angular velocity w. A particle of mass 10-2Kg on the frictionless inner surface of the bowl is also rotating with the same w. The particle is at a height h from the bottom of the bowl. a) Obtain the relation between h and w. What is the minimum value of w needed, in order to have a non-zero value of h? b) It is designed to measure g (acceleration due to gravity) using this step-up, by measuring h accurately. Assuming that R and w are known precisely and that the least count in the measurement of h is 10-4m, what is its minimum possible error Dg in the measured value of g? Solution: Fig a) Þ N Along Using mRw2Cosq b) \Dg |Dg|min Along x-direction, NSinq = y-direction, NSinq = = mRw2 NCosq (1) m m = (AC) (RSinq) ---------------------------- mg and = From = (3), = |Dh| g w2min = -Dh = (R-h) (42) w2 w2 (1) (2) (2) mg 2 w2 2) A small body is placed on the top of a smooth hemisphere of radius R. When the sphere is given a uniform horizontal acceleration a0 the body starts sliding down. a) Find the velocity of body relative to sphere at the instant of loosing contact b) Find the angle Æ between radius vector drawn to the body from center of sphere at the time of loosing contact if a0 = g Fig (43) a) Drawing f.b.d in frame of reference of particle. From For just Eliminating b) 2g work loosing a0 Equating (1-Sinq) energy contact N from + = 0 then (1) (1) 2a0Cosq theorem using (1) (2) and and = (2) gSinq - and (2) gives gives a0Cosq 3) A small bar A resting on a smooth horizontal plane is attached by threads to a point P as shown in figure. And by P means of a weightless pulley to weight B possessing the same mass as the bar itself. Beside, the bar is also attached to a point O by means of a light non deformed spring of length l0 = 50cm and with spring constant K, then find the velocity? Fig (44) Solution: Fig (45) Drawing N Þ From the f.b.d = of KxCosq KxCosq (1) A for = just breaking = off =0 mg and (2) mg Mass Since Now they B are would have connected with inextensible from moved string hence same velocity of energy A distance and B would downwards be same. conservation, 4) A horizontal plane support a stationary vertical cylinder of radius R and disc A attached to the cylinder by a horizontal thread AB of length l0. An initial velocity v0 is imported to the disc as shown. How long will it move along plane until it strikes against the cylinder? The friction is assumed to be absent. Fig (46) Solution: Fig (47) ds If the = is the angle path turned by (l0-R string traversed till the disc comes by in ) contact disc d with cylinder the length of Dumb Question: How is l0 = R ? Ans: Suppose the final picture is fig.(48). The whole thread rolls over the cylinder, then as obvious the length l 0 = R . Fig (48) How the motion does become a projectile motion? Ans: Note that at point B, the string becomes slack, so there is no tension acting over the particle. Now the particle has some velocity and gravity is acting over it, so it starts projectile motion at B and ends at A. PROBLEMS (HARD) 1) A particle is suspended by a string of length ‘l’. This is projected with such a velocity v along the horizontal such that after the string becomes slack it flies through its initial position. Find V? Solution: Fig (49) Let the velocity be v1 at B where the string becomes slack and the string makes angle q with horizontal, by the law of conservation of energy. Here From glSinq At B Here ax \lCosq Þ \ centripetal = the force is provided by equation (2) and v2-2gl (1+Sinq) particle becomes a projectile of Ux = v1Sinq and = 0 and = 2Sin3q+3Sin2q-1 = ½ Sinq component of (3) velocity v we ---------90-q = = at Uy ay v1Sinq = is weight get (4) horizontal v1Cosq -g t 0 solution acceptable 2) System consists of two identical slabs each of mass m linked by compressed weightless spring of stiffness K as shown in figure. The slabs are also connected by a thread which is burnt at a certain moment. a) Find at what value of Dl the initial compression of spring, the lower slab will bounce up after the thread is burned through b) If the upper block is compressed by than what is the max displacement of center of mass in upward direction? Fig Solution: When Kx the lower block is just (50) about to jump, then mg = From energy conservation b) When the lower block is just about to move up, center of mass will be displaced to maximum for this Work energy theorem gives, On yields Only solving force acting on it this is 2mg (both masses taken single system) KEYWORDS: Work Spring force Power Energy Mechanical Energy Kinetic Energy Potential Energy Conservative forces Work Energy Theorem Conservation of Mechanical Energy Uniform and Non-Uniform Circular motion Angular Velocity Angular Acceleration Centripetal and Tangential forces Radius of Curvature Vertical Circular motion Banking