lecture12

KMEM 1168 Engineering MechanicsLecture 12: Kinetic of Particles: Work and Energy12.1 Introduction 12.2 Work of a Force12.3 Principle of Work & Energy12.4 Applications of the Principle of Work & Energy 12.5 Power and Efficiency 12.6 Sample Problem 13.1 12.7 Sample Problem 13.1 12.8 Potential Energy12.9 Conservation of Energy 12.10 Sample Problem 13.6 12.11 Sample Problem 13.312. 1 IntroductionPreviously, problems dealing with the motion of particles were solved through the fundamentalequation of motion a m F rr =∑. This chapter introduces 2 additional methods of analysis.•Method of work and energy : directly relates force, mass, velocity and displacement. 222111V T U V T +=++−•Method of impulse and momentum : directly relates force, mass, velocity, and time.2121L Fdt L t t =+∫∑12.2 Work of a Force• Differential vector dr is the particle displacement •Work of the force isdzF dy F dx F ds F rd F dU z y x ++==•=αcos r r• Work is a scalar quantity, i.e., it has magnitude and sign but not direction.• Dimensions of work are Force x Length. Units are,1 J (Joule) = 1 Nm• Work of a force during a finite displacement,• Work is represented by the area under the curve of F t plotted against s .• Work of a constant force in rectilinear motion,•Forces which do not do work (ds = 0 or cos a = 0):a) reaction at frictionless pin supporting rotating bodyb) reaction at frictionless surface when body in contact moves along surface c) weight of a body when its center of gravity moves horizontally12.3 Principle of Work & Energy•Consider a particle of mass m acted upon by a force•Integrating from A 1to A 2,•The work of the force is equal to the change in kinetic energy of the particle.• Units of work and kinetic energy are the same:12.4 Applications of the Principle of Work & Energy•The bob is held at point 1 and if we wish to find the velocity of pendulum bob at point 2, we have to consider the method of work and energy. Force P acts normal to path and does no work.•In this method, unlike the method of Newton 2nd Law, we can find velocity without having to determine expression for acceleration and integrating.•All quantities are scalars and can be added directly. Forces which do no work (eg. In this case is the tension in the cord P)are eliminated from the problem. Principle of work and energy cannot be applied to directly determine the acceleration of the pendulum bob.•The tension in the cord is required to supplement the method of work and energy with an application of Newton’s second law.•As the bob passes through A 2,12.5 Power and Efficiency•Power = rate at which work is done.vF dt r d F dt dU r r r r •=•==•Dimensions of power are work/time or force x velocity. Units for power are• Efficiency is the ratio of the power output to the power input,12.6 Sample Problem 13.1An automobile weighing 4000 N is driven down a 5oincline at a speed of 88 m /s when the brakes are applied causing a constant total breaking force of 1500 N.Determine the distance traveled by the automobile as it comes to a stop.• Evaluate the change in kinetic energy.()()m N 15488008810/4000sm 88221212111⋅====mv T v0022==T v•Determine the distance required for the work to equal the kinetic energy change.()()()()x x x U 11515sin 4000150021−=°+−=→()0115115488002211=−=+→x T U Tm 6.1345=x12.7 Sample Problem 13.2Two blocks are joined by an inextensible cable as shown. If the system is released from rest, determine the velocity of block A after it has moved 2 m.Assume that the coefficient of friction between block A and the plane is μk= 0.25 and that the pulley isweightless and frictionless.• Apply the principle of work and energy separately to blocks A and B .Block A:()()()()()()()()()221221221120024902220:N490196225.0N 196281.9200v F vm F F T U T W N F W C A A C A k A k A A =−=−+=+======→µµBlock B:()()()()()()()()2212212211300229402220:N 294081.9300v F vm W F T U T W c B B c B =+−=+−=+==→•When the two relations are combined, the work of the cable forces cancel. Solve for the velocity.()()()()22120024902v F C =−()()()()221300229402v F c =+−()()()()()()2212215004900300200249022940vv=+=− s m 43.4=v12.8 Potential EnergyThere are 2 kinds of potential energyi) Gravitational Potential Energy, V g ii) Elastic Potential Energy, V eGravitational Potential Energy (V g )• Work of the force of gravity W ,2121y W y W U −=→• Work is independent of path followed, it depends only on the initial and final values of W(dy).()()2121g g V V U −=→•Choice of datum from which the elevation y is measured is arbitrary. But always choose the lower position as the datum to avoid negative potential energy. Units of work and potential energy are the same:J m N )(=⋅==dy W V gElastic Potential Energy, V e• Work of the force exerted by a spring dependsonly on the initial and final deflections of the spring,2221212121kx kx U −=→• The potential energy of the body with respect tothe elastic force,()()2121221e e e V V U kxV −==→•Note that the preceding expression for V e is valid only if the deflection of the spring is measured from its undeformed position.=→21U 221kx12.9 Conservation of Energy•Conservation of energy equation,constant2211=+=+=+V T E V T V T• When a particle moves under the action of conservative forces, the total mechanical energy (E) is constant.•Friction forces are not conservative. Total mechanical energy of a system involving friction decreases. Mechanical energy is dissipated by friction into thermal energy or heat.ll W V T W V T =+==11110()ll l W V T V W g g Wmv T =+====22222212022112.10 Sample Problem 13.6A 20-N collar slides without friction along a vertical rod as shown. The spring attached to the collar has an undeflected length of 4 cm and a constant of 3 N/cm. If the collar is released from rest at position 1, determine its velocity after it has moved 6 cm to position 2.• Apply the principle of conservation of energy betweenpositions 1 and 2.Position 1:()()0024cmN 24483112212121=+=+=⋅=−==T V V V kx V g e ePosition 2:()()()()222221222212221102021cm N 6612054cmN 120620cmN 544013v mv T V V V Wy V kx V g e g e ==⋅−=−=+=⋅−=−==⋅=−==•Conservation of Energy:cmN 66cm N 240222211⋅−=⋅++=+v V T V T↓=s m 5.92v12.11 Sample Problem 13.3A spring is used to stop a 60 kg package which is sliding on a horizontal surface. The spring has a constant k = 20 kN/m and is held by cables so that it is initially compressed 120 mm. The package has a velocity of 2.5 m/s in the position shown and the maximum deflection of the spring is 40 mm. Determine(a) the coefficient of kinetic friction between the package and surface(b) the velocity of the package as it passes again through the position shown.(a) Use principle of work and energy equation,NF g N µ==60222111V T U V T +=++→021)(212122212+∆=−∆+x k s N x k mu µ()()222)04.012.0)(20000(2)04.06.0)(60()12.0)(20000(215.26021+=+−+g µ 20.0=µ(b) Apply the principle of work and energy for the rebound of the package.333222V T U V T +=++→ 2323222121)(210x k mv s N x k ∆+=−∆+µ2232)12.0)(20000(21)60(21)64.0)(60)(2.0()16.0)(20000(21+=−v g s m v /11.13=NB: Part B demonstrates that the final velocity at 3 is less than the initial velocity at 1. This is due to the loss of energy due to friction. The total mechanical energy is not conserved.。