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Power Power is the time rate at which work W is done by a force •Average power Pavg = W/∆t (energy per time) •Instantaneous power P = dW/dt = (Fcosφ dx)/dt = F v cosφ= F . v Unit: watt 1 watt = 1 W = 1 J/s 1 horsepower = 1 hp = 550 ft lb/s = 746 W kilowatt-hour is a unit for energy or work: 1 kW h = 3.6 M J Problem 07-51 v Force F = (3.0î + 7.0 ĵ + 7.0k̂ ) Initial Displacement v d i = (3.0î − 2.0 ĵ + 5.0k̂ ) in ∆t = 4.0s Final Displacement v d f = (−5.0î + 4.0 ĵ + 7.0k̂ ) A) Work done on the particle B) Average power C) Angle between di and df Problem 07-51 v Force F = (3.0î + 7.0 ĵ + 7.0k̂ ) Initial Displacement v d i = (3.0î − 2.0 ĵ + 5.0k̂ ) in ∆t = 4.0s Final Displacement v d f = (−5.0î + 4.0 ĵ + 7.0k̂ ) A) Work done on the particle F is constant in this problem, so ( v v v W = F ⋅ df − di ) = (3.0î + 7.0 ĵ + 7.0k̂ ) ⋅ ((−5.0 − 3.0)î + ((4.0 + 2.0) ĵ + (7.0 − 5.0)k̂ ) = (3.0î + 7.0 ĵ + 7.0k̂ ) ⋅ (−8.0î + 6.0 ĵ + 2.0k̂ ) = −24.0 + 42.0 + 14.0 = 32.0J Problem 07-51 v Force F = (3.0î + 7.0 ĵ + 7.0k̂ ) Initial Displacement v d i = (3.0î − 2.0 ĵ + 5.0k̂ ) in ∆t = 4.0s Final Displacement v d f = (−5.0î + 4.0 ĵ + 7.0k̂ ) B) Average power Pave = W/∆t = 32.0J/4s = 8.0W Problem 07-51 v Force F = (3.0î + 7.0 ĵ + 7.0k̂ ) Initial Displacement v d i = (3.0î − 2.0 ĵ + 5.0k̂ ) v d i = 38 in ∆t = 4.0s Final Displacement v d f = (−5.0î + 4.0ˆj + 7.0k̂ ) v d f = 90 C) Angle between di and df di and df are two vectors, so take the dot product v v d f ⋅ d i = d f d i cos φ ⇒ cos φ = (−5.0î + 4.0 ĵ + 7.0k̂ ) ⋅ (3.0î − 2.0 ĵ + 5.0k̂ ) / ( d f d i ) (−15.0 − 8.0 + 35.0k̂ ) / ((9.49)(6.16) ) = 12.0 / 58.5 = 0.205 ⇒ φ = cos −1 0.205 = 78.2o Chapter 8: Potential Energy and Conservation of Energy Work and kinetic energy are energies of motion. rf v v 1 1 2 2 ∆K = K f − K i = mv f − mvi = Wnet = ∫ F ⋅ d r 2 2 ri Consider a vertical spring oscillating with mass m attached to one end. At the extreme ends of travel the kinetic energy is zero, but something caused it to accelerate back to the equilibrium point. We need to introduce an energy that depends on location or position. This energy is called potential energy. Chapter 8: Potential Energy and Conservation of Energy Work done by gravitation for a ball thrown upward then it fell back down b Wab+ Wba = − mgd + mg d = 0 The gravitational force is said to be a conservative force. A force is a conservative force if the net work it does on a particle moving around every closed path is zero. a Conservative forces Wab,1 + Wba,2 = 0 Wab,2 + Wba,2 = 0 therefore: Wab,1 = Wab,2 i.e. The work done by a conservative force on a particle moving between two points does not depend on the path taken by the particle. So, ... choose the easiest path!! Conservative & Non-conservative forces Conservative Forces: (path independent) gravitational spring electrostatic Non-conservative forces: (dependent on path) friction air resistance electrical resistance These forces will convert mechanical energy into heat and/or deformation. Gravitation Potential Energy Potential energy is associated with the configuration of a system in which a conservative force acts: ∆U = −W For a general conservative force F = F(x) xf v ∆U = − W = − ∫ F( x ) ⋅ dx̂ xi Gravitational potential energy: ∆U = − ∫ (−mg)dy = mg(y f − y i ) yf yi assume Ui = 0 at yi = 0 (reference point) U(y) = mgy (gravitational potential energy) only depends on vertical position Nature only considers changes in Potential energy important. Thus, we will always measure ∆U. So, . . . We need to set a reference point! The potential at that point could be defined to be zero (i.e., Uref. point = 0) in which case we drop the “∆” for convenience. BUT, it is always understood to be there! • Potential energy and reference point A 0.5 kg physics book falls from a table 1.0 m to the ground. What is U and ∆U if we take reference point y = 0 (assume U = 0 at y = 0) at the ground? y ∆U = Uf − Ui = −(mgh) Physics y=h 0 The book lost potential energy. Actually, it was converted to kinetic energy h y=0 • Potential energy and reference point A 0.5 kg physics book falls from a table 1.0 m to the ground. What is U and ∆U if we take reference point y = 0 (assume U = 0 at y = 0) at the table? y ∆U = Uf − Ui = mg(−h) Physics y=0 0 The book still the same lost potential energy. h y = −h Sample problem 8-1: A block of cheese, m = 2 kg, slides along frictionless track from a to b, the cheese traveled a total distance of 2 m along the track, and a net vertical distance of 0.8 m. How much is Wg? Easier (or another way)? b Wnet v v = ∫ Fg ⋅ d r a But, ... We don’t know the angle between F and dr Sample problem 8-1: A block of cheese, m = 2 kg, slides along frictionless track from a to b, the cheese traveled a total distance of 2 m along the track, and a net vertical distance of 0.8 m. How much is Wg? c Split the problem into two parts since gravity is conservative }=d c Wnet b v v = ∫ Fg ⋅ dx̂ + ∫ Fg ⋅ dŷ a 0 c mg(b−c) = mgd Elastic Potential Energy Spring force is also a conservative force F = −kx xf ∆U = − W = ∫ (−kx )dx xi Uf – Ui = ½ kxf2 – ½ kxi2 Choose the free end of the relaxed spring as the reference point: that is: Ui = 0 at xi = 0 U = ½ kx2 Elastic potential energy Conservation of Mechanical Energy • Mechanical energy Emec = K + U • For an isolated system (no external forces), if there are only conservative forces causing energy transfer within the system…. We know: ∆K = W (work-kinetic energy theorem) Also: ∆U = −W (definition of potential energy) Therefore: ∆K + ∆U = 0 (Kf – Ki) + (Uf – Ui) = 0 therefore K1 + U1 = K2 + U2 Emec,1 = Emec,2 the mechanical energy is conserved Conservation of mechanical energy For an isolated system with only conservative forces ( F = mg, F = −kx ) act inside Emec,1 = Emec,2 K1 + U1 = K2 + U2 ½ mv12 + mgy1 + ½ kx12 = ½ mv22 + mgy2 + ½ kx22 The bowling ball pendulum demonstration Mechanical energy is conserved if there are only conservative forces acting on the system. Pendulum (1-D) vertically Find the maximum speed of the mass. θ Isolated system with only gravity (conservative force) acting on it. L 1 3 y mg y=0 Let 1 be at maximum height (y = ymax), 2 2 be at minimum height (y = 0), and 3 y = L(1−cosθ) be somewhere inbetween max and min. Pendulum (1-D) vertically Where does it have the max speed? θ L 1. 1: max. height 2. 2: min. height 3. 3: somewhere in between max and min 4. none of the above 3 y=0 1 y 2 mg θ Lcosθ Find the maximum speed of the mass. L Pendulum (1-D) vertically L 1 Isolated system with only gravity (conservative force) acting on it. 0 K1 + U1 = K2 + U2 = Etotal ½ mv12 0 + mgy1 = ½ mv22 + mgy2 0 Kmax must be when Umin (= 0) 3 y 2 mg y=0 In general, y = L(1−cosθ) Answer: 2 Pendulum (1-D) vertically Find the maximum speed of the mass. Let 1 be at maximum height (y = ymax) and 2 be at minimum height (y = 0) Emech,1 = Emech,2 = Etotal θ L y = L(1−cosθ) K1 + U1 = K2 + U2 = Etotal y 0 0 mgymax = ½ mvmax2 = Etotal y=0 ⇒ v max = 2gy max = 2gL(1 − cos θ max ) mg Pendulum (1-D) vertically Find the speed of the mass as a function of angle. Let 1 be at maximum height (y = ymax) and 2 be at some height (y) Emech,1 = Emech,2 = Etotal θ L y = L(1−cosθ) K1 + U1 = K2 + U2 = Etotal mgymax = ½ mv2 + mgy = Etotal y y=0 mg v = 2g (y max − y ) = 2g(L(1 − cos θ max ) − L(1 − cos θ)) = 2gL(cos θ − cos θ max ) Pendulum (1-D) vertically Find the speed of the mass as a function of angle. Let 1 be at maximum height (y = ymax) and 2 be at some height (y) θ y y=0 v = 2g (y max − y ) = 2g(L(1 − cos θ max ) − L(1 − cos θ)) = 2gL(cos θ − cos θ max ) U K 0 −θmax 0 +θmax 0 −θmax L θ mg Pendulum (1-D) vertically Find the speed of the mass as a function of angle. Let 1 be at maximum height (y = ymax) and 2 be at some height (y) θ y y=0 U E K 0 −θmax 0 +θmax 0 −θmax L θ mg Potential Energy Curve • We know ∆U(x) = −W = −F(x) ∆x Therefore F(x) = −dU(x)/dx Work done by external force • When no friction acts within the system, the net work done by the external force equals to the change in mechanical energy W = ∆Emec = ∆K + ∆U • Friction is a non-conservative force • When a kinetic friction force acts within the system, then the thermal energy of the system changes: ∆Eth = fkd Therefore W = ∆Emec + ∆Eth Work done by external force • When there are non-conservative forces (like friction) acting on the system, the net work done by them equals to the change in mechanic energy Wnet = ∆Emec = ∆K + ∆U Wfriction = −fkd Conservation of Energy The total energy E of a system can change only by amounts of energy that are transferred to or from the system W = ∆E = ∆Emec + ∆Eth + ∆Eint If there is no change in internal energy, but friction acts within the system: W = ∆Emec + ∆Eth If there are only conservative forces act within the system: W = ∆Emec If we take only the single object as the system W = ∆K Law of Conservation of Energy • For an isolated system (W = 0), the total energy E of the system cannot change ∆Emec + ∆Eth + ∆Eint = 0 • For an isolated system with only conservative forces, ∆Eth and ∆Eint are both zero. Therefore: ∆Emec = 0 Sample 8-7. In the figure, a 2.0kg package slides along a floor with speed v1 = 4.0m/s. It then runs into and compresses a spring, until the package momentarily stops. Its path to the initially relaxed spring is frictionless, but as it compresses the spring, a kinetic friction force from the floor, of magnitude 15 N , act on it. The spring constant is 10,000 N/m. By what distance d is the spring compressed when the package stops?