PHYS 218sec. 517-520ReviewChap. 6Work and Kinetic EnergyWorksDisplacement vectorFConstant forceStraight-line displacementForce and displacement are in the same direction.,: magnitude of , : magnitude of WFs F s= FsUnit of work1 joule (1 Newton) (1 meter) or 1J 1Nm=× =⋅WorkConstant force in direction of straight-line displacementWFs=Constant force, straight-line displacementcosWFsφ= ⋅ =Fs: force vector, : displacement vector,: angle between the two vectorsφFsVarying x-component of force, straight-line displacement21xxxWFdx=∫Work done on a curved path (general definition of work)222111cosPPPPPPWFdlFdl dφ===⋅∫∫∫FlEvaluation of WorkEx 6.1ˆˆ ˆˆConstant force (160 N) (40 N) , displacement (14m) (11m) .Then (160N) (14m) ( 40N) (11m) 1800 JFij sijWFs= − =+= ⋅ =×+− ×=Ex 6.2()()displacement 20 m, total weight 14,700N,constant force exerted by the tractor 5000N, (angle 36.9 )friction force 3500 NTFfφ=====TFnwfφ()work done by the tractorcos (5000 N)(20 m)cos36.9 80,000J 80 kJwork done by the friction forcecos180 (3500 N)(0 m) 1 70 kJwork done by the normal force work done by the weight 0angleTTfWFsWfsφ== ==== − = −===∵()90total work 10 kJTfnwWW W WW=+++=sWork done by a varying force, straight-line motionm m1x2x2xFs1xF1x2x()xFx1xF2xF1x2x1xF2xFax∆axFbx∆bxF21To calculate the work, divide the displacement into small segments , etc is nearly constant for this small segment, and thenaxax a bx b xxxFWFxFx Fdx∆= ∆ + ∆ +=∫Work is the area under the curve between the initial and final positions in the graph of force as a function of position.Evaluation of Work: Stretched springxxFkx=xFkx= −Hooke's law: , : spring constantFkxk= −Restoring force2 20Work done by the force which stretches the spring ( )when the elongation goes from 0 to a maximum value is11 (work done by the spring is )22Generally, fromXkxXWkxdxkX kX=== −∫122221 to ,11 22xx xxWkxkx=== −Varying x-component of force, straight-line displacement21xxxWFdx=∫Evaluation of Work()221121xxxxx xxxWFdxFdxFxxFs===− =∫∫When Fxis constant21In general, ,so the work is obtained by calculating the above line integral,which is defined on a given path.PPWd= ⋅∫FlKinetic energy and work-energy theoremKinetic energy (K)212Kmv=Work-Energy TheoremWhen v = 0, K = 0The work done by the net force on a particle equals the change in the particle’s kinetic energy21totWKK K= − = ∆This is a very general theorem. This theorem is true regardless of the nature of the force.Kinetic energy at the initial pointKinetic energy at the final pointProof of the work-energy theoremConstant force in direction of straight-line displacementm m1x2xFs1v2v222221212222212121For a constant force (acceleration ),2 211 222avvvv as asvvW Fs mas m s mv mv K Ks−=+ ⇒ =−⇒ == = = − = −When a particle undergoes a displacement, it speeds up if W > 0, slows down if W < 0, and maintains the same speed if W = 0.Proof of the work-energy theoremVarying force, straight-line motion22 2 211 1 12221Start from the definition of acceleration,Then,1122 The work-energy theorem is valid.xx x xxxxxx x vxtot x x x x xxx x vdv dv dv dvdxavvdt dx dt dx dxdvW F dx ma dx mv dx mv dvdxmv mv== = === = == −∴∫∫ ∫ ∫Eliminate time dependence since W is an integral over x.11 2 2At , and at ,xxvv xxvv== ==Ex 6.3 Same as Ex 6.2: suppose that the initial speed v1is 2 m/s. What is the final speed?Examples21We already know that 10kJ. To know the final speed, we use the work-energy theorem.To compute kinetic energy, we need to know the mass. Since the total weight is 14,700 N,14,700 NtottotWWKKwmg== −==()()22211212221500 kg9.8 m/s11Initial kinetic energy 1500 kg 2 m/s 3000 J22Then ,3000 J 10,000 J 13,000 JTherefore,22 13,000 J4.2 m/s1500 kgtotKmvKKWKKvm=== ==+=+ =×== =Use Work-Energy Theorem to calculate speed.Forces on a hammerheadEx 6.4hammerheadI-beamPoint 3Point 2Point 13 m7.4 cmFalling hammerheadyv60Nf =wmg=1212 1212 221 2 2 12Total work done on the hammerhead: ( ) ( ) 5700J.1Work-Energy Theorem gives ( 0)2Therefore,22 (5700 J)7.55 m/s200 kgtottottotWwfsmgfsWKKKmv vWvm→→= − = − == − == =×== =∵Speed of the hammerhead at point 2Forces on a hammerheadEx 6.4 (cont’d)Hammerhead pushing I-beami.e., Point 2 J Point 1y60Nf =wmg=n()()()23 1223 3 21223Total work done on the hammerhead:5700 J1960 N 60 N 79,000 N0.074 mtot tottotWwfnsKKWWnwfs→→→= −− = − = −⇒ = − += − +=Work-energy theoremUsing the previous resultBe careful with the change of unitMotion with a varying forceEx 6.7m1vFriction (µk)k10.1 kg, 20 N/m, 1.5 m/s,Find the maximum distance if 0 and if 0.47,when the object is moving from 0 to the rightkkmk vdxµµ== ====When 0kµ =yxnmgspringFfrF2221 122111Work done by the spring 21Then by Work-Energy theorem; 211Therefore, 10.6 cm22WkdWKK mvmkd mv d vk= −= − = −− = −⇒==When 0kµ ≠221Work done by the friction force Then by Work-Energy theorem; 110.086 m22fr kkWmgdmgd kd mv dµµ= −−−= −⇒=Ex 6.8 Motion on a curved path IRRθsRθ=θdlyxFθTFmgcosT θsinT θ[]0000000The net force is zero ( remains in equilibrium)sin 0, cos 0.This leads totancos cos , and Therefore,tan cos ( ) sincos(1 cosxyFFT FT mgFmgFd Fdl F ds ds RdWmg Rd mgRdmgRmgRθθθθθθθθ θθθθ θθθθ= − ==− ==⋅ == ==== −= −∑∑∫∫∵l)Ex 6.9 Motion on a curved path II (another way to compute the line integral)In Ex 6.8, can be written asˆˆcos sin .All forces acting on the object areˆˆ ˆ ˆ(sin) cos, (), .Therefore,(sin)(cos)(cos)(sin)0sinddds ids jTT iTjwwjFFiTd T ds T dswd w dsFdθθθθθθ θθθ=+= − +=− =⋅ = − +=⋅ = −⋅lllll0cos tan cos sinsin (1 cos )wFwFdsw dswdsWwd wdswRWWθθθθθθ== =⇒= ⋅ = − = −−= −∫∫lWe get the same answer as in Ex 6.8PowerAverage powerThe time rate at which work is done.avWPt∆=∆(Instantaneous) power0limtWdWPtdt∆→∆==∆Unit of powerwatt (W): 1W 1 J/shorsepower (hp): 1 hp 746 W==Writing power in terms of FPFv=
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