1Chapter 6 – Kinetic energy and workI. Kinetic energy.II. Work.III. Work - Kinetic energy theorem.IV. Work done by a constant force- Gravitational forceV. Work done by a variable force.- Spring force.- General.1D-Analysis3D-AnalysisWork-Kinetic Energy Theorem.VI. PowerI. Kinetic energyEnergy associated with the state of motion of an object.)1.7(212mvK =Units: 1 Joule = 1J = 1 kgm2/s2 = N mII. WorkEnergy transferred “to” or “from” an object by means of a force acting on the object. To  +WFrom  -W- Constant force: xxmaF =dvvadavvxx22202202−=→+=Work done by the force = Energy transfer due to the force. Energy: scalar quantity associated with a state (or condition) of one or more objects.)(211)(21202202vvmdmadvvmmaFxxx−=→−==dFWdFKKvvmxxif=→=−=−→ )(212022- To calculate the work done on an object by a force during a displacement,we use only the force component along the object’s displacement. The force component perpendicular to the displacement does zero work.)3.7(cos dFdFdFWx⋅=⋅==ϕ- Assumptions: 1) F=cte, 2) Object particle-like.0909018090→=−→>>+→<ϕϕϕWWUnits: 1 Joule = 1J = 1 kgm2/s2A force does +W when it has a vector component in the same directionas the displacement, and –W when it has a vector component in the opposite direction. W=0 when it has no such vector component.Net work done by several forces = Sum of works done by individual forces.2) Fnet Wnet=FnetdCalculation: 1) Wnet= W1+W2+W3+…Fdcos φII. Work-Kinetic Energy Theorem)4.7(WKKKif=−=∆Change in the kinetic energy of the particle = Net work done on the particleIII. Work done by a constant force- Gravitational force: )5.7(cosϕmgddFW =⋅=Rising object: W= mgd cos180º = -mgd  Fgtransfersmgd energy from the object’s kinetic energy. Falling object: W= mgd cos 0º = +mgd  Fgtransfers mgd energy to the object’s kinetic energy.3IV. Work done by a variable force- External applied force + Gravitational force: )6.7(gaifWWKKK +=−=∆Object stationary before and after the lift: Wa+Wg=0The applied force transfers the same amount ofenergy to the object as the gravitational force transfers from the object.- Spring force: )7.7(dkF−=Hooke’s lawk = spring constant  measures spring’sstiffness. Units: N/mkxFDx−=→1Work done by a spring force:Hooke’s law- Assumptions:• Spring is massless  mspring<< mblock• Ideal spring  obeys Hooke’s law exactly.• Contact between the block and floor is frictionless.• Block is particle-like.2) F(x) ≈ cte within each short ∆x segment.- Calculation:xFxxixf∆xFj1) The block displacement must be divided into many segments of infinitesimal width, ∆x.4Work done by an applied force + spring force:dxkxdxFWxxFWfifixxxxsjs∫∫∑−==⇒→∆⇒∆= )(0222121fisxkxkW −=Ws=0  If Block ends up at xf=xi.0212=−=ifsxifxkWsaifWWKKK +=−=∆Block stationary before and after the displacement: ∆K=0 Wa= -WsThe work done by the applied force displacing the block is the negativeof the work done by the spring force.[ ])(2121222ifxxxxSxxkxkdxxkWfifi−−=−=−=∫Work done by a general variable force:1D-Analysis)10.7()(lim0,,0,,∫=⇒∆=⇒→∆∆⇒∑ ∑∆=∆=∆=∆→∆fixxavgjxavgjjavgjjdxxFWxFWxxmoreionapproximatbetterxFWWxFWGeometrically: Work is the area between the curve F(x) and the x-axis.5Work-Kinetic Energy Theorem - Variable force∫∫==fifixxxxdxmadxxFW )(3D-Analysis)(),(),(;ˆˆˆzFFyFFxFFkFjFiFFzyxzyx===++=kdzjdyidxrdˆˆˆ++=∫∫∫∫++==⇒++=⋅=fifififizzzyyyxxxrrzyxdzFdyFdxFdWWdzFdyFdxFrdFdWdxdtdvmdxma =KKKmvmvdvvmdvmvWififvvvvfifi∆=−=−===∫∫222121== vdxdvdtdxdxdvdtdvmvdvdxvdxdvm =→V. PowerTime rate at which the applied force does work.- Average power: amount of work done in an amount of time ∆t by a force.- Instantaneous power: instantaneous time rate of doing work.)12.7(tWPavg∆=)13.7(dtdWP =Units: 1 Watt= 1 W = 1J/s 1 kilowatt-hour = 1 kW·h = 3.60 x 106J = 3.6 MJ)14.7(coscoscosvFFvdtdxFdtdxFdtdWP⋅=====ϕϕϕxFφ654. In the figure (a) below a 2N force is applied to a 4kg block at a downward angle θ as the block moves rightward through 1m across a frictionless floor. Find an expression for the speed vfat the end of that distance if the block’s initial velocity is: (a) 0 and (b) 1m/s to the right. (c) The situation in (b) is similar in that the block is initially moving at 1m/s to the right, but now the 2N force is directed downward to the left. Find an expression for the speed of the block at the end of the 1m distance.NmgFxFy)(5.0)cos(202vvmKWdFdFWf−=∆==⋅=θNmgFxFysmvJvkgNJmvKsmvcfff/cos12)4(5.0cos)2(25.0/1)(220θθ−=→−=−−=∆→=smvvkgNmvKvafff/cos)4(5.0cos)2(5.00)(220θθ=→==∆→=smvJvkgNsmkgmvKsmvbfff/cos12)4(5.0cos)2()/1()4(5.05.0/1)(2220θθ+=→−=⋅⋅−=∆→=18. In the figure below a horizontal force Faof magnitude 20N is applied to a 3kg psychology book, as the book slides a distance of d=0.5m up a frictionless ramp. (a) During the displacement, what is the net work done on the book by Fa, the gravitational force on the book and the normal force on the book? (b) If the book has zero kinetic energy at the start of the displacement, what is the speed at the end of the displacement?workdoFFOnlyWdNaxgx,0=→⊥xymgNFgyFgxJmNNWmgFgFaFdFWorWWWanetxxnetnetnetxFgxFa31.15.0)7.1432.17(30sin30cos20)(=−=−=−=⋅=−=smvmvJWKKWKbfff/93.05.031.10)(20=→===∆=→=755. A 2kg lunchbox is sent sliding over a frictionless surface, in the positive direction of an x axis along the surface. Beginning at t=0, a steady wind pushes on the lunchbox in the negative direction of x, Fig. below. Estimate the kinetic energy of the lunchbox at (a) t=1s, (b) t=5s. (c) How much work does the force from the wind do on the lunch box from t=1s to t=5s?)10(101ttxparaboladownwardconcaveMotion−=→JsmkgKsmvstaff64.0)/8.0)(2(5.0/8.01)(2===→=JKvstbff005)(==→=JWsKsKKWcff64.064.00)1()5()(−=−=−=∆=2/2.01021021smdtdvatdtdxv−=−==−==)1.0)(4.0(4.0)/2.0)(2(22ttNxFWNsmkgmacteF−−=⋅=−=−===74. (a) Find the work done on the particle by the force represented in the graph below as the particle moves from x=1 to x=3m. (b) The curve is given by F=a/x2, with a=9Nm2. Calculate the work