Power. Power is the time rate at which work W is done by a force Average power. (energy per time) P = dw/dt = (Fcosφ dx)/dt = F v cosφ= F.

Transcription

1 Power Power is the time rate at which work W is done by a force Aerage power P ag = W/ t Instantaneous power (energy per time) P = dw/dt = (Fcosφ dx)/dt = F cosφ= F. 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

2 Force F = (3.0î + 7.0ĵ+ 7.0kˆ) Problem in t = 4.0s Initial Displacement d i = (3.0î 2.0ĵ+ 5.0kˆ) Final Displacement d f = ( 5.0î + 4.0ĵ+ 7.0kˆ) A) Work done on the particle B) Aerage power C) Angle between d i and d f

3 Force F = (3.0î + 7.0ĵ+ 7.0kˆ) Problem in t = 4.0s Initial Displacement d i = (3.0î 2.0ĵ+ 5.0kˆ) Final Displacement d f = ( 5.0î + 4.0ĵ+ 7.0kˆ) A) Work done on the particle W = F is constant in this problem, so F ( d d ) f i = (3.0î + 7.0ĵ+ 7.0kˆ) (( )î + ((4.0 + = (3.0î + 7.0ĵ+ 7.0kˆ) ( 8.0î + 6.0ĵ+ 2.0kˆ) = = 32.0J 2.0)ĵ+ ( )kˆ)

4 Force F = (3.0î + 7.0ĵ+ 7.0kˆ) Problem in t = 4.0s Initial Displacement d i = (3.0î 2.0ĵ+ 5.0kˆ) Final Displacement d f = ( 5.0î + 4.0ĵ+ 7.0kˆ) B) Aerage power P ae = W/ t = 32.0J/4s = 8.0W

5 d i F = (3.0î + 7.0ĵ+ 7.0kˆ) Force Initial Displacement = (3.0î 2.0ĵ+ 5.0kˆ) d = 38 i Problem in t = 4.0s Final Displacement d f = d f ( 5.0î + 4.0ĵ+ 7.0kˆ) = 90 C) Angle between d i and d f d and d are two ectors, so take the dot product i f d f d i = d cosφ = ( 5.0î + f d ( kˆ) / φ = cos i cosφ 1 4.0ĵ+ 7.0kˆ) (3.0î 2.0ĵ+ 5.0kˆ) / = ((9.49)(6.16)) 78.2 o ( d d ) = 12.0 / 58.5 = f i

6 Chapter 8: Potential Energy and Conseration of Energy Work and kinetic energy are energies of motion. K = K f K i = 1 2 m 2 f rf = F dr Consider a ertical spring oscillating with mass m attached to one end. At the extreme ends of trael 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. 1 2 m 2 i = W net r i

7 Chapter 8: Potential Energy and Conseration of Energy Work done by graitation for a ball thrown upward then it fell back down W ab + W ba = mgd + mg d = 0 The graitational force is said to be a conseratie force. b A force is a conseratie force if the net work it does on a particle moing around eery closed path is zero. a

8 W ab,1 + W ba,2 = 0 W ab,2 + W ba,2 = 0 therefore: W ab,1 = W ab,2 Conseratie forces i.e. The work done by a conseratie force on a particle moing between two points does not depend on the path taken by the particle. So,... choose the easiest path!!

9 Conseratie & Non-conseratie forces Conseratie Forces: (path independent) graitational spring electrostatic Non-conseratie forces: (dependent on path) friction air resistance electrical resistance These forces will conert mechanical energy into heat and/or deformation.

10 Graitation Potential Energy Potential energy is associated with the configuration of a system in which a conseratie force acts: U = W For a general conseratie force F = F(x) = = xf U W F(x) dxˆ Graitational potential energy: U = y f y i x ( mg)dy = i mg ( ) y f y i assume U i = 0 at y i = 0 (reference point) U(y) = mgy (graitational potential energy) only depends on ertical position

11 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., U ref. point = 0) in which case we drop the for conenience. BUT, it is always understood to be there!

12 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 = U f U i = (mgh) 0 Physics y = h The book lost potential energy. h Actually, it was conerted to kinetic energy y = 0

13 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 = U f U i = mg( h) 0 Physics y = 0 The book still the same lost potential energy. h y = h

14 Sample problem 8-1: A block of cheese, m = 2 kg, slides along frictionless track from a to b, the cheese traeled a total distance of 2 m along the track, and a net ertical distance of 0.8 m. How much is W g? Easier (or another way)? W net b = F a g dr But,... We don t know the angle between F and dr

15 Sample problem 8-1: A block of cheese, m = 2 kg, slides along frictionless track from a to b, the cheese traeled a total distance of 2 m along the track, and a net ertical distance of 0.8 m. How much is W g? Split the problem into two parts since graity is conseratie c }=d W net = c a F g dxˆ + b c F g dŷ 0 mg(b c) = mgd

16 Elastic Potential Energy Spring force is also a conseratie force F = kx U = W x = f ( x i kx)dx U f U i = ½ kx f2 ½ kx i 2 Choose the free end of the relaxed spring as the reference point: that is: U i = 0 at x i = 0 U = ½ kx 2 Elastic potential energy

17 Conseration of Mechanical Energy Mechanical energy E mec = K + U For an isolated system (no external forces), if there are only conseratie forces causing energy transfer within the system. We know: K = W Also: U = W (work-kinetic energy theorem) (definition of potential energy) Therefore: K + U = 0 (K f K i ) + (U f U i ) = 0 therefore K 1 + U 1 = K 2 + U 2 E mec,1 = E mec,2 the mechanical energy is consered

18 Conseration of mechanical energy For an isolated system with only conseratie forces ( F = mg, F = kx ) act inside E mec,1 = E mec,2 K 1 + U 1 = K 2 + U 2 ½ m 12 + mgy 1 + ½ kx 12 = ½ m 22 + mgy 2 + ½ kx 2 2

19 The bowling ball pendulum demonstration Mechanical energy is consered if there are only conseratie forces acting on the system.

20 Pendulum (1-D) ertically Find the maximum speed of the mass. Isolated system with only graity (conseratie force) acting on it. θ L y=0 Let 1 be at maximum height (y = y max ), 2 be at minimum height (y = 0), and 3 be somewhere inbetween max and min. 3 y 1 mg 2 y = L(1 cosθ)

21 Pendulum (1-D) ertically Where does it hae the max speed? θ L : max. height 2. 2: min. height 3. 3: somewhere in between max and min 3 y=0 y 2 mg 4. none of the aboe

22 Pendulum (1-D) ertically Find the maximum speed of the mass. Isolated system with only graity (conseratie force) acting on it. 0 K 1 + U 1 = K 2 + U 2 = E total L ½ m 12 + mgy 1 = ½ m 22 + mgy K max must be when U min (= 0) Lcosθ 3 y=0 y 2 θ L 1 mg In general, y = L(1 cosθ) Answer: 2

23 Pendulum (1-D) ertically Find the maximum speed of the mass. Let 1 be at maximum height (y = y max ) and 2 be at minimum height (y = 0) E mech,1 = E mech,2 = E total y = L(1 cosθ) θ L K 1 + U 1 = K 2 + U 2 = E total 0 0 mgy max = ½ m max2 = E total y y=0 mg max = 2gy max = 2gL 1 ( cosθ ) max

24 Pendulum (1-D) ertically Find the speed of the mass as a function of angle. Let 1 be at maximum height (y = y max ) and 2 be at some height (y) E mech,1 = E mech,2 = E total y = L(1 cosθ) θ L K 1 + U 1 = K 2 + U 2 = E total mgy max = ½ m 2 + mgy = E total y y=0 mg = 2g = 2gL ( y y) = 2g( L( 1 cosθ ) L( 1 cosθ) ) max ( cosθ cosθ ) max max

25 Pendulum (1-D) ertically Find the speed of the mass as a function of angle. Let 1 be at maximum height (y = y max ) and 2 be at some height (y) y θ L = 2g y=0 ( y y) = 2g( L( 1 cosθ ) L( 1 cosθ) ) max max mg = 2gL ( cosθ cosθ ) max U K 0 θ max 0 +θ max 0 θ max θ

26 Pendulum (1-D) ertically Find the speed of the mass as a function of angle. Let 1 be at maximum height (y = y max ) and 2 be at some height (y) y θ L y=0 mg E K 0 θ max 0 +θ max 0 θ max θ U

27 Potential Energy Cure We know U(x) = W = F(x) x Therefore F(x) = du(x)/dx

28 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 = E mec = K + U Friction is a non-conseratie force When a kinetic friction force acts within the system, then the thermal energy of the system changes: E th = f k d Therefore W = E mec + E th

29 Work done by external force When there are non-conseratie forces (like friction) acting on the system, the net work done by them equals to the change in mechanic energy W net = E mec = K + U W friction = f k d

30 Conseration 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 = E mec + E th + E int If there is no change in internal energy, but friction acts within the system: W = E mec + E th If there are only conseratie forces act within the system: W = E mec If we take only the single object as the system W = K

31 Law of Conseration of Energy For an isolated system (W = 0), the total energy E of the system cannot change E mec + E th + E int = 0 For an isolated system with only conseratie forces, E th and E int are both zero. Therefore: E mec = 0

32 Sample 8-7. In the figure, a 2.0kg package slides along a floor with speed 1 = 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?