Chapter 7: Work and EnergyForms of Mechanical EnergyPowerPoint PresentationSlide 4Work by a Baseball PitcherSlide 6Work Done by a Constant Force (I)Work Done by a Constant Force (II)Example 1AExample 1A (cont’d)Slide 11Slide 12Work-Energy TheoremExample 2Example 2 (cont’d)Slide 16Slide 17Slide 18Slide 19Slide 20Spring Force (Hooke’s Law)Work Done to Stretch a SpringSlide 23Work Done by a Varying ForceSlide 25Slide 26Example 1BExample 1B (cont’d)Slide 29Example 2 (cont’d)Potential Energy and Energy ConservationWork Done by the Gravitational Force (I)Work Done by the Gravitational Force (II)Work Done by the Gravitational Force (III)Work Done by the Gravitational Force (IV)Slide 36Work Done by the Gravitational Force (V)Work Done by Ff (I)Work Done by Ff (II)Example 1Wg(AC) = Ug(yA) – Ug(yC)Climbing the Sear towerSlide 43Slide 44Work and Energy1. Work Energy Work done by a constant force(scalar product) Work done by a varying force(scalar product & integrals)2. Kinetic EnergyChapter 7: Work and EnergyWork-Energy TheoremForms of Mechanical EnergyWork and EnergyWork and EnergyCONSERVATION OF ENERGYWork and EnergyWork and EnergyWork by a Baseball PitcherA baseball pitcher is doing work on the ball as he exerts the force overa displacement.v1 = 0v2 = 44 m/sWork done by several forcesWork and Energy Work Done by a Constant Force (I)Work (W) How effective is the force in moving a body ?W [Joule] = ( F cos  ) d Both magnitude (F) and directions () must be taken into account.Work and Energy Work Done bya Constant Force (II)Example: Work done on the bag by the person.. Special case: W = 0 Ja) WP = FP d cos ( 90o )b) Wg = m g d cos ( 90o ) Nothing to do with the motionWork and EnergyExample 1AA 50.0-kg crate is pulled 40.0 m by a constant force exerted (FP = 100 N and  = 37.0o) by a person. A friction force Ff =50.0 N is exerted to the crate. Determinethe work done by each force acting on thecrate.Work and EnergyExample 1A (cont’d)WP = FP d cos ( 37o )Wf = Ff d cos ( 180o )Wg = m g d cos ( 90o )WN = FN d cos ( 90o )180o90odF.B.D.Work and EnergyExample 1A (cont’d)WP = 3195 [J]Wf = -2000 [J] (< 0)Wg = 0 [J]WN = 0 [J]180oWork and EnergyExample 1A (cont’d)Wnet = Wi = 1195 [J] (> 0)The body’s speed increases.Work and Energy Work-Energy TheoremWnet = Fnet d = ( m a ) d = m [ (v2 2 – v1 2 ) / 2d ] d = (1/2) m v2 2 – (1/2) m v1 2 = K2 – K1Work and EnergyExample 2A car traveling 60.0 km/h to can brake toa stop within a distance of 20.0 m. If the caris going twice as fast, 120 km/h, what is itsstopping distance ?(a)(b)Work and Energy Example 2 (cont’d) (1) Wnet = F d(a) cos 180o = - F d(a) = 0 – m v(a)2 / 2  - F x (20.0 m) = - m (16.7 m/s)2 / 2(2) Wnet = F d(b) cos 180o = - F d(b) = 0 – m v(b)2 / 2  - F x (? m) = - m (33.3 m/s)2 / 2(3) F & m are common. Thus, ? = 80.0 mWork and EnergyWork and EnergyDoes the Earth do work on the satellite?Satellite in a circular orbitWork and Energy2 BWork and EnergyForcesForces on a hammerheadWork and EnergyS S23 FnWork and EnergySpring Force (Hooke’s Law)FS(x) = - k xFPFSNatural Lengthx > 0x < 0Spring Force(Restoring Force):The spring exerts its force in thedirection opposite the displacement.Work and EnergyWork Done to Stretch a Spring x2W = FP(x) dxx1FS(x) = - k xWNatural LengthFPFSWork and EnergyWork and Energy lbW = F|| dl la Work Done bya Varying Forcel  0Work and EnergyExample 1AA person pulls on the spring, stretching it3.0 cm, which requires a maximum force of 75 N. How much work does the person do ? If, instead, theperson compressesthe spring 3.0 cm,how much workdoes the person do ?Work and Energy(a) Find the spring constant k k = Fmax / xmax = (75 N) / (0.030 m) = 2.5 x 103 N/m(b) Then, the work done by the person is WP = (1/2) k xmax2 = 1.1 J(c) x2 = 0.030 mWP = FP(x) d x = 1.1 J x1 = 0Example 1A (cont’d)Work and EnergyExample 1BA person pulls on the spring, stretching it3.0 cm, which requires a maximum force of 75 N. How much work does the spring do ? If, instead, theperson compressesthe spring 3.0 cm,how much workdoes the spring do ?Work and Energy(a) Find the spring constant k k = Fmax / xmax = (75 N) / (0.030 m) = 2.5 x 103 N/m(b) Then, the work done by the spring is(c) x2 = -0.030 m  WS = -1.1 J x2 = -0.030 mWS = FS(x) d x = -1.1 J x1 = 0Example 1B (cont’d)Work and EnergyExample 2A 1.50-kg block is pushed against a spring(k = 250 N/m), compressing it 0.200 m, andreleased. What will be the speed of theblock when it separates from the spring atx = 0? Assume k =0.300. (i) F.B.D. first !(ii) x < 0FS = - k xWork and Energy(a) The work done by the spring is(b) Wf = - kFN (x2 – x1) = -4.41 (0 + 0.200)(c) Wnet = WS + Wf = 5.00 - 4.41 x 0.200(d) Work-Energy Theorem: Wnet = K2 – K1  4.12 = (1/2) m v2 – 0  v = 2.34 m/s x2 = 0 mWS = FS(x) d x = +5.00 J x1 = -0.200 mExample 2 (cont’d)Energy Conservation1. Conservative/Nonconservative Forces Work along a path(Path integral) Work around any closed path(Path integral)2. Potential EnergyPotential Energy and Energy ConservationMechanical Energy ConservationEnergy ConservationWork Done bythe Gravitational Force (I) 21)(ˆd ˆ d mgymgymgyy-mg2yy1y2y1l2l1--)jj lFW ()(ly(Path integral)Near the Earth’s surfaceEnergy ConservationNear the Earth’s surface Work Done bythe Gravitational Force (II) 21)(ˆd ˆd ˆ d mgymgymgyy x-mg2yy1y2y1l2l1--)jij lFW ()(dly(Path integral)Energy Conservation Work Done bythe Gravitational Force (III)Wg < 0 if y2 > y1Wg > 0 if y2 < y1The work done by the gravitationalforce depends only on the initial andfinal positions..Energy Conservation Work Done bythe Gravitational Force (IV)Wg(ABCA) = Wg(AB) + Wg(BC) + Wg(CA) = mg(y1 – y2) + 0 + mg(y2- y1) = 0 dlABCEnergy ConservationEnergy Conservation Work Done bythe Gravitational Force (V)Wg = 0 for a closed path The gravitational force is a conservative force.Energy ConservationWork Done by Ff (I) -)(d