Chapter 7 – Kinetic Energy And Work - Physics

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Chapter 7 – 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.PowerEnergy: scalar quantity associated with a state (or condition) of one ormore objects.I. Kinetic energyEnergy associated with the state of motion of an object. K 1 2mv2(7.1)Units: 1 Joule 1J 1 kgm2/s2 N mII. WorkEnergy transferred “to” or “from” an object by means of a force acting onthe object.To WFrom -W- Constant force:v 2 v02 2a x d a x Fx ma xv 2 v022d111m(v 2 v02 ) ma x d m(v 2 v02 )2d21 m(v 2 v02 ) K f K i Fx d W Fx d2Fx ma x Work done by the force Energytransfer due to the force.1

- 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. Theforce component perpendicular to the displacement does zero work. W Fx d F cos ϕ d F dF(7.3)cos φdϕ 90 W- Assumptions: 1) F cte, 2) Object particle-like.180 ϕ 90 WUnits: 1 Joule 1J 1 kgm2/s2ϕ 90 0A force does W when it has a vector component in the same directionas the displacement, and –W when it has a vector component in theopposite direction. W 0 when it has no such vector component.Net work done by several forces Sum of works done by individual forces.Calculation:1) W net W 1 W 2 W 3 2) Fnet W net Fnet dII. Work-Kinetic Energy Theorem K K f K i W(7.4)Change in the kinetic energy of the particle Net work done on the particleIII. Work done by a constant force- Gravitational force: W F d mgd cos ϕ(7.5)Rising object: W mgd cos180º -mgd Fg transfersmgd energy from the object’s kinetic energy.Falling object: W mgd cos 0º mgd Fg transfersmgd energy to the object’s kinetic energy.2

- External applied force Gravitational force: K K f K i Wa Wg(7.6)Object stationary before and after the lift: W a W g 0The applied force transfers the same amount ofenergy to the object as the gravitational forcetransfers from the object.IV. Work done by a variable force - Spring force:F kd(7.7)Hooke’s lawk spring constant measures spring’sstiffness.Units: N/mHooke’s law1D F x kxWork done by a spring force:- Assumptions: Spring is massless mspring mblockIdeal spring obeys Hooke’s law exactly.Contact between the block and floor is frictionless.Block is particle-like.Fx- Calculation:1) The block displacement must be divided intomany segments of infinitesimal width, x.xi xxfxFj2) F(x) cte within each short x segment.3

xfxfiiWs F j x x 0 Ws x F dx x ( kx) dxxf 1 WS k x dx k x 2xi 2 [ ]Ws 11k xi2 k x 2f221Ws k x 2f2xfxi 1 k ( x 2f xi2 ) 2 W s 0 If Block ends up at xf xi.if xi 0Work done by an applied force spring force: K K f K i Wa WsBlock stationary before and after the displacement: K 0 W a -W s The work done by the applied force displacing the block is the negativeof the work done by the spring force.Work done by a general variable force:1D-Analysis W j F j ,avg xW W j F j ,avg xbetter approximation more x, x 0xf W lim F j ,avg x W F ( x) dx x 0(7.10)xiGeometrically: Work is the area between the curve F(x) and the x-axis.4

3D-Analysis F Fx iˆ Fy ˆj Fz kˆ ;Fx F ( x), Fy F ( y ), Fz F ( z ) dr dx iˆ dy ˆj dz kˆrxyz dW F dr Fx dx Fy dy Fz dz W dW Fx dx Fy dy Fz dzffffrixiyiziWork-Kinetic Energy Theorem - Variable forcexfxfW F ( x) dx ma dxxixidvma dx mdxdt mdvv dx mvdvdx dv dv dx dv v dt dx dt dx vfvfW mv dv m v dv vivi1 2 1 2mv f mvi K f K i K22V. PowerTime rate at which the applied force does work.- Average power: amount of work done in an amount of time t by a force.Pavg W t(7.12)- Instantaneous power: instantaneous time rate of doing work.P P dWdtF(7.13) dW F cos ϕ dx dx F cos ϕ Fv cos ϕ F vdtdt dt φx(7.14)Units: 1 Watt 1 W 1J/s1 kilowatt-hour 1 kW·h 3.60 x 106 J 3.6 MJ5

54. In the figure (a) below a 2N force is applied to a 4kg block at a downwardangle θ as the block moves rightward through 1m across a frictionless floor. Findan expression for the speed vf at the end of that distance if the block’s initialvelocity is: (a) 0 and (b) 1m/s to the right. (c) The situation in (b) is similar in thatthe block is initially moving at 1m/s to the right, but now the 2N force is directeddownward to the left. Find an expression for the speed of the block at the end ofthe 1m distance. W F d ( F cos θ )dNFxFxFymgW K 0.5m(v 2f v02 )NFymg( a) v0 0 K 0.5mv 2f( 2 N ) cos θ 0.5(4kg )v 2f v f cos θ m / s(b) v0 1m / s K 0.5mv 2f 0.5 (4kg ) (1m / s ) 2 (c ) v0 1m / s K 0.5mv 2f 2 J(2 N ) cos θ 0.5(4kg )v 2f 2 J (2 N ) cos θ 0.5(4kg )v 2f 2 J v f 1 cos θ m / s v f 1 cos θ m / s18. In the figure below a horizontal force Fa of magnitude 20N is applied to a 3kgpsychology 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, thegravitational force on the book and the normal force on the book? (b) If the bookhas zero kinetic energy at the start of the displacement, what is the speed at theend of the displacement?yx N d W 0Only Fgx , Fax do workN(a ) W WFa x WFg xFgxmgFgyor Wnet Fnet dFnet Fa x Fg x 20 cos 30 mg sin 30 Wnet (17.32 N 14.7 N )0.5m 1.31J(b) K 0 0 W K K fW 1.31J 0.5mv 2f v f 0.93m / s6

55. A 2kg lunchbox is sent sliding over a frictionless surface, in the positivedirection of an x axis along the surface. Beginning at t 0, a steady wind pusheson the lunchbox in the negative direction of x, Fig. below. Estimate the kineticenergy of the lunchbox at (a) t 1s, (b) t 5s. (c) How much work does the forcefrom the wind do on the lunch box from t 1s to t 5s?Motion concave downward parabolax 1t (10 t )10dx2 1 tdt10dv2a 0.2m / s 2dt10v F cte ma (2kg )( 0.2m / s 2 ) 0.4 NW F x ( 0.4 N )(t 0.1t 2 )( a ) t 1s v f 0.8m / sK f 0.5( 2kg )(0.8m / s ) 2 0.64 J(b ) t 5 s v f 0K f 0J(c) W K K f (5s ) K f (1s )W 0 0.64 0.64 J74. (a) Find the work done on the particle by the force represented inthe graph below as the particle moves from x 1 to x 3m. (b) The curveis given by F a/x2, with a 9Nm2. Calculate the work using integration( a ) W Area under curveW (11.5squares)(0.5m)(1N ) 5.75 J3(b) W 1391 1 dx 9 9( 1) 6 J23x x 173. An elevator has a mass of 4500kg and can carry a maximum load of1800kg. If the cab is moving upward at full load at 3.8m/s, what power isrequired of the force moving the cab to maintain that speed?Famgmtotal 4500kg 1800kg 6300kg Fa mg Fnet 0 Fa Fg 0 Fa mg (6300kg )(9.8m / s 2 ) 61.74kN P F v (61.74kN )(3.8m / s )P 234.61kW7

A single force acts on a body that moves along an x-axis. The figure below showsthe velocity component versus time for the body. For each of the intervals AB, BC,CD, and DE, give the sign (plus or minus) of the work done by the force, or statethat the work is zero.W K K f K 0 vBC()AB vB v A W 0DA1m v 2f v022tEBC vC vB W 0CD v D vC W 0DE vE 0, vD 0 W 050. A 250g block is dropped onto a relaxed vertical spring that has a springconstant of k 2.5N/cm. The block becomes attached to the spring andcompresses the spring 12 cm before momentarily stopping. While the spring isbeing compressed, what work is done on the block by (a) the gravitational force onit and (b) the spring force? (c) What is the speed of the block just before it hits thespring? (Friction negligible) (d) If the speed at impact is doubled, what is themaximum compression of the spring? ( a ) W Fg Fg d mgd (0.25kg )(9.8m / s 2 )(0.12 m ) 0.29 J1(b ) Ws kd 2 0.5 ( 250 N / m )( 0.12 m ) 2 1.8 J2mgdFs(c ) Wnet K 0.5mv 2f 0.5mvi2mgv f 0 K f 0 K K i 0.5mvi2 WFg Ws0.29 J 1.8 J 0.5 (0.25 kg )vi2 vi 3.47 m / s( d ) If vi ' 6.95 m / s Maximum spring compressio n ? v f 0Wnet mgd ' 0.5kd '2 K 0.5mvi '2d ' 0.23m8

62. In the figure below, a cord runs around two massless, frictionless pulleys; acanister with mass m 20kg hangs from one pulley; and you exert a force F on thefree end of the cord. (a) What must be the magnitude of F if you are to lift thecanister at a constant speed? (b) To lift the canister by 2cm, how far must youpull the free end of the cord? During that lift, what is the work done on thecanister by (c) your force (via the cord) and (d) the gravitational force on thecanister?( a ) Pulley 1 : v cte Fnet 0 2T mg 0 T 98 NHand cord : T F 0 F P2TTTmg 98 N2(b) To rise “m” 0.02m, two segments of the cord mustbe shorten by that amount. Thus, the amount of thestring pulled down at the left end is: 0.04mP1mg(c ) WF F d (98 N )(0.04 m ) 3.92 J( d ) WFg mgd ( 0.02 m )( 20 kg )(9.8m / s 2 ) 3.92 JW F W Fg 0There is no change in kinetic energy.9

50. A 250g block is dropped onto a relaxed vertical spring that has a spring constant of k 2.5N/cm. The block becomes attached to the spring and compresses the spring 12 cm before momentarily stopping. While the spring is being compressed, what work is done on the block by (a) the gravitational force on 1 i / File Size: 630KBPage Count: 9

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