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Announcements 1. Physics colloquium (Professor Brian Matthews) cancelled. 2. Scholarship opportunities for undergraduate students 3. Schedule -- First exam – Tuesday, Sept. 30th Review Chapters 1-8 – Thursday, Sept. 25th Extra practice problems on line Extra problem solving sessions??? 4. Some comments on friction and HW 6 5. The notion of energy and work 9/18/2003 PHY 113 -- Lecture 7 1 9/18/2003 PHY 113 -- Lecture 7 2 Surface friction force models ¾Static friction f = −Fapplied if | f | < µs N ¾Kinetic friction | f | = µs N 9/18/2003 PHY 113 -- Lecture 7 3 N f mg F sin θ N−mg−F sin θ = 0 f = µk N F cos θ − f = ma 9/18/2003 PHY 113 -- Lecture 7 4 without friction : mg sin θ = ma D = 12 at 2 with friction : mg sin θ − µ k mg cos θ = ma f t f = 2t 9/18/2003 ⇒ 4a f = a D = 12 a f t 2f ⇒ 4( g sin θ − µ k g cos θ ) = g sin θ PHY 113 -- Lecture 7 5 Energy Î work, kinetic energy Force Î effects acceleration A related quantity is Work rf Wi → f = ∫ F ⋅ dr ri A·B = AB cosθ F dr ri 9/18/2003 rj PHY 113 -- Lecture 7 6 Units of work: work = force · displacement = (N · m) = (joule) •Only the component of force in the direction of the displacement contributes to work. •Work is a scalar quantity. •If the force is not constant, the integral form must be used. •Work can be defined for a specific force or for a combination of forces rf W1 = ∫ F1 ⋅ dr ri 9/18/2003 rf rf W2 = ∫ F2 ⋅ dr W1+ 2 = ∫ (F1 + F2 ) ⋅ dr = W1 + W2 ri PHY 113 -- Lecture 7 ri 7 Peer instruction question A ball with a weight of 5 N follows the trajectory shown. What is the work done by gravity from the initial ri to final displacement rf? 2.5m rf ri 1m 1m (a) 0 J 9/18/2003 10 m (b) 7.5 J (c) 12.5 J PHY 113 -- Lecture 7 (d) 50 J 8 Gravity does negative work: rf Gravity does positive work: ri mg mg rf ri W=−mg(rf−ri)<0 9/18/2003 W=−mg(rf−ri)>0 PHY 113 -- Lecture 7 9 F Positive work x Negative work 9/18/2003 PHY 113 -- Lecture 7 10 More examples: Suppose a rope lifts a weight of 1000N by 0.5m at a constant upward velocity of 2m/s. How much work is done by the rope? W=500 J Suppose a rope lifts a weight of 1000N by 0.5m at a constant upward acceleration of 2m/s2. How much work is done by the rope? W=602 J 9/18/2003 PHY 113 -- Lecture 7 11 Why is work a useful concept? Consider Newton’s second law: Ftotal = m a Î Ftotal · dr= m a · dr rf rf rf rf rf dv d v dr dv ∫ Ftotal ⋅ dr = ∫ ma ⋅ dr = ∫ m ⋅ dr = ∫ m ⋅ dt = ∫ m ⋅ vdt dt dt dt dt ri ri ri ri ri Wtotal = ½ m vf2 - ½ m vi2 9/18/2003 Kinetic energy (joules) PHY 113 -- Lecture 7 12 Introduction of the notion of Kinetic energy Some more details: Consider Newton’s second law: Ftotal = m a Î Ftotal · dr= m a · dr rf rf rf tf tf dv d v dr dv F ⋅ d r = m a ⋅ d r = m ⋅ d r = m ⋅ dt = m ⋅ vdt ∫ total ∫ ∫ ∫ ∫ dt dt dt dt ri ri ri ti ti tf vf f dv 1 1 2 1 2 ⎛ ⎞ ∫ m ⋅ vdt = ∫ mdv ⋅ v = ∫ d ⎜ mv ⋅ v ⎟ = mv f − mvi 2 dt ⎠ 2 ti i ⎝2 vi ÎWtotal = ½ m vf2 - ½ m vi2 9/18/2003 Kinetic energy (joules) PHY 113 -- Lecture 7 13 Kinetic energy: K = ½ m v2 units: (kg) (m/s)2 = (kg m/s2) m N m = joules Work – kinetic energy relation: Wtotal = Kf – Ki 9/18/2003 PHY 113 -- Lecture 7 14 9/18/2003 PHY 113 -- Lecture 7 15 Examples of the work—kinetic energy relation: Suppose Ftotal = constant= F0 Wtotal = Kf – Ki Î F0(xf-xi) = ½ mvf2 - ½mvi2 In this case, we also know that F0 = ma0 so that, F0(xf-xi) = ma0 (xf-xi) = ½ mvf2 - ½mvi2 Îvf2 =vi2 +2a0 (xf-xi) 9/18/2003 PHY 113 -- Lecture 7 16 More examples of work—kinetic energy relation without friction: gsinθ Assume block of mass m is sliding down a g h frictionless surface at an angle of θ L θ kinematic analysis: vf2 =vi2 +2a0 (xf-xi) = 0 +2(g sinθ)L = 2gh energy analysis: Wtotal = mgh = ½ mvf2 9/18/2003 PHY 113 -- Lecture 7 17 More examples of work—kinetic energy relation with friction: gsinθ g Assume block of mass m is sliding down a fk surface with a kinetic friction coefficient of µk at an angle of θ h L θ kinematic analysis: vf2 =vi2 +2a0 (xf-xi) = 2(g sinθ−µkg cosθ)L energy analysis: Wtotal = mgh −µkmg cosθL = ½ mvf2 9/18/2003 PHY 113 -- Lecture 7 18 -Ftrack-mg=-mvt2/R If Ftrack=0, then mg(h-2R)= ½ mvt2 Îh=5/2R h v 2R 9/18/2003 PHY 113 -- Lecture 7 19 Suppose that the coefficient of friction between the skis and the snow is µk=0.2. What is the stopping distance d? L h mgh- µkmgcosθL- µkmgd=0 d=h/µk -Lcosθ 9/18/2003 PHY 113 -- Lecture 7 20 θ L 1 2 A ball attached to a rope is initially at an angle θ. After being released from rest, what is its velocity at the lowest point 2? 2 gL(1 − cosθ ) 9/18/2003 PHY 113 -- Lecture 7 21 Hooke’s “law” (model force) F = −kx xf Wi → f = ∫ − kx dx = − 12k (x 2f − xi2 ) xi xi 9/18/2003 PHY 113 -- Lecture 7 xf 22 Example problem: A mass m with initial velocity v compresses a spring with Hooke’s law constant k by a distance xf. What is xf when the mass momentarily comes to rest? K i = 12 mv 2 Kf = 0 Wi → f = − 12 kx f Wi → f = K f − K i xf = 9/18/2003 2 ⇒ 12 kx f = 12 mv 2 2 m v k PHY 113 -- Lecture 7 23