106 PHYS - CH6 - Part2

Transcription

1 106 PHYS - CH6 - Part Conservative Forces (a) A force is conservative if work one by tat force acting on a particle moving between points is inepenent of te pat te particle takes between te two points (b) Te total work one by a conservative force is zero wen te particle moves aroun any close pat an returns to its initial position Conservative Forces To repeat te iea on te last slie: We ave seen tat te work one by a conservative force oes not epen on te pat taken. W W 1 = W Terefore te work one in a close pat is 0. W 1 W W NET = W 1 -W = W 1 -W 1 = 0 W 1 1

2 106 PHYS - CH6 - Part Work one by gravity W g = F. Δr = mg Δr cos θ= mg W g = mg (Depens only on!) Δr m θ mg W NET = W 1 + W W n m = Fi Δr 1 + Fi Δr Fi Δr n = Fi (Δr + Δr Δr ) n = Fi Δr = F W g = mg Δr 1 Δr Δr Δr 3 m mg Depens only on, not on pat taken! Δr n Non-conservative forces: A force is non-conservative if it causes a cange in mecanical energy; mecanical energy is te sum of kinetic an potential energy. Example: Frictional force. Tis energy cannot be converte back into oter forms of energy (irreversible). Work oes epen on pat. For straigt line W = -f For semi-circle pat W = - f (π /) Work varies epening on te pat. Energy is issipate Te presence of a non-conservative force reuces te ability of a system to o work (issipative force)

3 106 PHYS - CH6 - Part Energy issipation: e.g. sliing friction As te parts scrape by eac oter tey start small-scale vibrations, wic transfer energy into atomic motion Te atoms vibrations go back an forttey ave energy, but no average momentum. Te increase atomic vibrations appear to us as a rise in te temperature of te parts. Te temperature of an object is relate to te termal energy it as. Friction transfers some energy into termal energy Example : Hotweel A toy car slies on te frictionless track sown below. It starts at rest, rops a istance, moves orizontally at spee v 1, rises a istance, an ens up moving orizontally wit spee v. Fin v 1 an v. v 1 v 3

4 106 PHYS - CH6 - Part K+U energy is conserve, so ΔE = 0 ΔK = - ΔU Moving own a istance, 1 ΔU = -mg, Δ K = mv 1 Solving for te spee: v1 = g v 1 At te en, we are a istance - below our starting point. ΔU = -mg( - ), Solving for te spee: 1 Δ K = mv v = g ( ) - v 4

5 106 PHYS - CH6 - Part Example: Wit wat spee oes te weigt ave just before contact wit te nail? ΔK + ΔU = 0 U i = mg U f = 0 K i = 0 1 K f = mv i f v = g Wat is te force of resistance between te nail an te block? ΔK + ΔU = W = f nc U i = 0 U f = -mg K i = mg K f = 0 i f y i y f Solve you get f f + = mg 5

6 106 PHYS - CH6 - Part example Nose cruser? A bowling ball of mass m is suspene from te ceiling by a cor of lengt L. Te ball is release from rest wen te cor makes an angle θ A wit te vertical. (a) Fin te spee of te ball at te lowest point B. (b) Assume a cor lengt L = 5 m an an angle θ A = 0. (c) Te ball swings back. Will it crus te operator s nose? 6

7 106 PHYS - CH6 - Part Example; A block slies own a frictionless ramp. Suppose te orizontal (bottom) portion of te track is roug, suc tat te coefficient of kinetic friction between te block an te track is μ k. How far, x, oes te block go along te bottom portion of te track before stopping? μ k x Using W nc = ΔK + ΔU As before, ΔU = -mg W nc = work one by friction = -μ k mgx. ΔK = 0 since te block starts out an ens up at rest. W nc = ΔU -μ k mgx = -mg x = / μ k 7

8 106 PHYS - CH6 - Part Example Two blocks, A an B (m A =50 kg an m B =100 kg), are connecte by a string as sown. If te blocks begin at rest, wat will teir spees be after A as sli a istance = 0.5 m? Assume te pulley an incline are frictionless. ANS: 1.51 m/s Example A rescue elicopter lifts a 79-kg person straigt up by means of a cable. Te person as an upwar acceleration of 0.70 m/s an is lifte from rest troug a istance of 11 m. (a) Wat is te tension in te cable? (b) How muc work is one by te tension in te cable (c) How muc work is one by te person s weigt? () Use te work-energy teorem an fin te final spee of te person. a) T-mg = ma, T= N b) W T = T = J g T mg c) W w = -mg = J ) W T + W W = 1/ mv f -1/ mv o v f = 3.9 m/s 8

9 106 PHYS - CH6 - Part Example: A skier (m=58 kg) is traveling own a 5 egree slope. His skies against te snow exert a frictional force of 70 N. He starts out wit a velocity of 3.6 m/s. Wat velocity oes e en up wit after traveling 57 m ownill? Wat is te net force along te irection of te isplacement? F = s mgsin θ + ( 70N) 1 1 W = Fs s = mv mv0 1 1 [ mg sin θ + ( 70N )] s = mv mv0 From tis, we can solve for v! example (similar to examples 6.5&6.6) A skier starts from rest at te top of a frictionless incline of eigt 0.0 m, as sown. At te bottom of te incline, se encounters a orizontal surface were te coefficient of kinetic friction between te skis an te snow is (a)how far oes se travel on te orizontal stretc. 9

10 106 PHYS - CH6 - Part Δ E = 0 = Δ K + ΔU 1 mg = mv v = g v = = 19.8 m / s Δ K = Kf Ki = fk Since K f = 0 K = f; f = K k k k i f = μ n = μ mg k k i 1 mv 19.8 ( ) = K i v = = μ mg μ mg μ g k k k 95. m Power Power is te rate at wic work is one by a force P AVG = W/Δt Average Power P = W/t Instantaneous Power Te unit of power is a Joule/secon (J/s) wic we efine as a Watt (W) 1 W = 1 J/s W r r P = = F x t t r r x r r P = F = F v t ( ) 10

11 106 PHYS - CH6 - Part v θ F ΔW = FΔx cosθ Δx = vδt ΔW FΔx cosθ P = = = Fvcosθ Δt Δt Note: power time = work, so work can be measure in units of kw. (1 KW = (103 Watt) (3600 s) = W s = J =3.6 MJ.) Britis units are p (orse power) 1 p = 550 ft lb/s = 746 W ( a) F x = 0 F F R mg sinθ = 0 P = Fv ( b) Now, θ = 0 F x = ma F F R = 0 v = v 0 + at P = Fv Example : Power Nees of Car Calculate te power nee far te car (a) to climb a ill. (b) to pass anoter car. 11

12 106 PHYS - CH6 - Part Problem: etermine te power after an elapse time of 3.0 s mg 50 kg 9.8 m/s 0. m P= = = 330W Δt 3.0s.0 m 50 kg g Energy Loss in Automobile Automobile uses only at 13% of its fuel to propel te veicle. Wy? 67% in te engine: 1. Incomplete burning. Heat 3. Soun 16% in friction in mecanical parts 4% in operating oter crucial parts suc as oil an fuel pumps, etc 13% use for balancing energy loss relate to moving veicle, like air resistance an roa friction to tire, etc Two frictional forces involve in moving veicles m Coefficient of Rolling Friction; μ=0.016 μn= μmg = 7N 1 1 Air Drag f a = DρAv = v = 0.647v Total Resistance f t = fr + fa Total power to keep spee v=6.8m/s=60mi/ P = ftv = ( 691N) 6.8 = 18.5kW Power to overcome eac component of resistance P r = = 6.08kW car P = = 1450kg Weigt = mg = 1400N frv =( ) f v = ( 464.7) 6.8 = 1. kw a a 5 1