Chapter 8: Potential Energy and The Conservation of Total Energy

Transcription

1 Chapter 8: Potental Energy and The Conservaton o Total Energy Work and knetc energy are energes o moton. K K K mv r v v F dr Potental energy s an energy that depends on locaton. -Dmenson F x d U( x) dx v F v mv W net () r U() r v v v U() r U() r U() r 3-Dmensons v x î y r ĵ z kˆ

2 Force s the dervatve o Potental Energy: F(x) du(x)/dx Thus, the orce n the x-drecton s the negatve dervatve o the potental energy! The same holds true or y- and z-drectons. U(x, y, z) Fx (x, y, z) x U(x, y, z) Fy (x, y, z) y U(x, y, z) Fz (x, y, z) z v F U(x, y,z) x î U(x, y,z) y ĵ U(x, y,z) z kˆ

3 Work done by Sprng Force -- Summary Sprng orce s a conservatve orce Work done by the sprng orce: x v v x v W F dx ( kxxˆ ) dx k s x kx x kx v v F kx I x > x (urther away rom equlbrum poston); W s < 0 I x < x (closer to equlbrum poston); W s > 0 kx kx Let x 0, x x then W s ½ k x x x x dx (xˆ xˆ )

4 Elastc Potental Energy Sprng orce s a conservatve orce v v F kx U U U W kx kx Choose the ree end o the relaxed sprng as the reerence pont: U 0 at x 0 U (x) W(x) kx The work went nto potental energy, snce the speeds are zero beore and ater.

5 potental energy curve or sprngs U (x) W(x) kx U (parabola) E X 0 X E 0 K 0 X max 0 X max 0 X max U X

6 Daly Quz-, February 08, 006 F -kx A x0 A U F dx -½kx ) Horzontal sprng wth mass oscllatng wth maxmum ampltude x max A. At whch dsplacement(s) would the knetc energy equal the potental energy? ± A ± A x 0 ) x ± A 3) x 4) x 5) none o the above

7 A Quz Horzontal sprng wth mass oscllatng wth maxmum ampltude x max A. At whch dsplacement(s) would the knetc energy equal the potental energy? ) x 0 ) x none o the above ± A 3) x ± A A 4) x0 ± x A A

8 A Quz E tot ka ka kx K(x) U(x) x ± ( U(x) ) A kx Horzontal sprng wth mass oscllatng wth maxmum ampltude x max A. At whch dsplacement(s) would the knetc energy equal the potental energy? ) x 0 ) x ± A 5) none o the above 3) x ± A A 4) x0 ± x A A

9 consder a general potental energy curve-- U(x) Thus, what causes a orce s the varaton o the potental energy uncton,.e., the orce s the negatve dervatve o the potental energy! We know: U(x) W F(x) x Thereore: F(x) du(x)/dx

10 U A hydrogen atom wth knetc energy o 4 ev s approachng another hydrogen atom n ts ground state. The potental energy s shown to the rght. X

11 A hydrogen atom wth knetc energy o 4 ev s approachng another hydrogen atom n ts ground state. The potental energy s shown to the rght. Wll ths H atom be captured and thereby become a H molecule?

12 A hydrogen atom wth knetc energy o 4 ev s approachng another hydrogen atom n ts ground state. The potental energy s shown to the rght. Wll ths H atom be captured and thereby become a H molecule? ) yes ) no 3) maybe V(r) 4.5eV 4 ev 0 ev

13 Hydrogen Atom collsons Wll ths H atom be captured and thereby become a H molecule? V(r) 4 ev. yes. no 3. maybe 4.5eV 0 ev

14 A hydrogen atom wth knetc energy o 4 ev s approachng another hydrogen atom n ts ground state. The potental energy s shown to the rght. E ntal K ntal U ntal 4.0 ev 4.0 ev 0 ev

15 The H-atom hts the repulsve potental energy wall at about 0.04 nm and s relected back to nnty. Another object s needed to absorb the excess knetc energy. Note that the speed (knetc energy) ncreases as the potental well becomes more negatve, but the total energy s constant. Wll ths H atom be captured and thereby become a H molecule? ) yes ) no 3) maybe 4.5eV V(r) 4 ev 0 ev

16 Pendulum and/or Sprng Mechancal energy s conserved there are only conservatve orces actng on the system.

17 analyss o pendulum moton We can use mechancal energy to nd the speed o the mass, m, as a uncton o angle: Let the maxmum heght be y max and y can be any heght-- v g gl ( ymax y) g( L( cosθmax ) L( cosθ) ) ( cosθ cosθ ) max y y0 θ L mg E K 0 θ max 0 θ max 0 θ max θ U

18 wth conservaton o energy, t s always mportant to specy the system For an solated system wth only conservatve orces (e.g., F mg and F kx) actng on the system: E mec, E mec, E total (ntal) (nal) > K U K U E total ½ mv mgx ½ kx ½ mv mgx ½ kx E total Intal mechancal energy Fnal mechancal energy

19 Consder a vertcal sprng wth mass m: (we must consder both gravtatonal and elastc energes) Assume that the mass m was held at the nonextended sprng poston (y 0) and then slowly brought down to the equlbrum y 0 poston. Fnd ths equlbrum poston. y y 0 Equlbrum poston (F g F s 0): F g mg F s ky 0 > y 0 mg/k y A ky mg K e 0, U g,e mgy 0 mg( mg/k) (mg) /k, U s,e ½ ky 0 ½ k ( mg/k) ½ (mg) /k

20 vertcal sprng... Assume that the mass m was held at the non-extended sprng poston (y 0) and then let drop. Fnd the lowest pont the mass reaches. Choose 0 y 0 Intal poston: y y 0 K 0, U g, 0, U s, ½ ky 0 > E mech, 0 0 y A Fnal poston: K 0, U g, mgy mg( A) mga, U s, ½ ky ½ ka ky mg E mech, E mech, 0 U g, U s, mga ½ ka

21 vertcal sprng... Assume that the mass m was held at the non-extended sprng poston (y 0) and then let drop. Fnd the lowest pont the y 0 mass reaches. y y 0 Fnal poston: y A ky mg E mech, E mech, 0 U g, U s, mga ½ ka > A mg/k ( y 0 )

22 vertcal sprng... Assume that the mass m was held at the non-extended sprng poston (y 0) and then let drop. Fnd the maxmum speed o the mass. (Maxmum speed occurs when the potental energy s mnmum.) Arbtrary poston: y A E mech 0 K U g U s ½ mv mgy ½ ky y 0 y y 0 ky mg du/dy d(mgy ½ ky )/dy 0 > mg ky 0 > y mg/k y 0 > maxmum speed occurs at the equlbrum pont

23 vertcal sprng... Assume that the mass m was held at the non-extended sprng poston (y 0) and then let drop. Fnd the maxmum speed o the mass. (Maxmum speed occurs when the potental energy s mnmum.) Maxmum speed poston: y A 0 K max U g U s ½ mv mg( mg/k) ½ k( mg/k) y 0 y y 0 ky mg 0 K max U g U s ½ mv (mg) /k ½(mg) /k ½ mv ½ (mg) /k > v mg /k

24 vertcal sprng... Assume that the mass m was held at the non-extended sprng poston (y 0) and then let drop. Fnd the mechancal energy o the mass at the equlbrum pont. Knetc energy at y y 0 : K y0 ½ (mg) /k y 0 y y 0 y A Potental energy at y y 0 : (U g U s ) y y0 mg( y 0 ) ½ k( y 0 ) mgy 0 ½k(y 0 ) mg(mg/k) ½k(mg/k) ½(mg) /k ky mg Mechancal energy at y y 0 : E mech K y0 (U g U s ) y y0 ½ (mg) /k ½(mg) /k 0

25 Work done by external orce When no rcton acts wthn the system, the net work done by the external orce equals the change n mechancal energy W net E mec K U Frcton s a non-conservatve orce that opposes moton Work done by rcton s: W rcton k d When a knetc rcton orce acts wthn the system, then the thermal energy o the system changes: E th k d Thereore W E mec E th

26 Work done by external orces When there are non-conservatve orces (lke rcton) actng on the system, the net work done by them equals the change n mechancal energy W net E mec K U

27 Conservaton o Total Energy The total energy E o a system can change only by an amount o energy that s transerred to or rom the system. W E E mec E th E nt I there s no nternal energy change, but rcton acts wthn the system: W E mec E th I there are only conservatve orces actng wthn the system: W E mec

28 Isolated Systems For an solated system (W 0), the total energy E o the system cannot change E mec E th E nt 0 For an solated system wth only conservatve orces, E th and E nt are both zero. Thereore: E mec 0

29 Law o Conservaton o Total Energy E E K K U U Rearrange terms. ( E E ) E E thermal thermal E nternal E nternal ( K K ) ( U U ) ( E E ) ( E E ) 0 thermal thermal nternal nternal E ( K U) E thermal E nternal E mechancal E thermal E nternal 0

30 Law o Conservaton o Energy Count up the ntal energy n all o ts orms. E E E K U E K K U U E E thermal E thermal nternal Count up the nal energy n all o ts orms. E K U E These two must be equal. thermal thermal nternal E nternal E E nternal

31 Sample Problem 8-6 A wooden crate o m 4kg s pushed along a horzontal loor wth a constant orce o F 40N or a total dstance o d 0.5m, durng whch the crate s speed decreased rom v o 0.60 m/s to v 0.0m/s. A) Fnd the work done by F. W Fd cosφ (40N)(0.50m)cos0 o 0J B) Fnd the ncrease n thermal energy. W E mec E thermal 0J ½mv ½mv o E thermal > E thermal W E mec 0J (½mv ½mv o ).J

32 Sample Problem 8-7 In the gure, a.0kg package sldes along a loor wth speed v 4.0m/s. It then runs nto and compresses a sprng, untl the package momentarly stops. Its path to the ntally relaxed sprng s rctonless, but as t compresses the sprng, a knetc rcton orce rom the loor, o magntude 5 N, act on t. The sprng constant s 0,000 N/m. By what dstance d s the sprng compressed when the package stops?

33 The change n mechancal energy must equal the energy converted to thermal energy. E mechancal E ( E E ) E 0 mec, mec, thermal 0 thermal E K U mv U mv 0 mec, Intal mechancal energy, mv E K U mv U 0 kd mec, Fnal mechancal energy, kd

34 The change n mechancal energy must equal the energy converted to thermal energy. E kd thermal ( ) ( ) E E E kd mv d 0 mec, mec, thermal k Thus, a quadratc equaton n d wth: m.0 kg, v 4.0 m/s, k 5 N, and k 0,000 N/m kd d k d 0.055m mv 0

35 Potental Energy Curve We know: Thereore: U(x) W F(x) x F(x) du(x)/dx Now ntegrate along the dsplacement: v v F dx v v F dx W v v du F dx dx dx K K du dx dx Rearrange terms: du dx K dx K ( U U ) U U U K U K U U

36 Horzontal Sprng x0 Isolated system wth only conservatve orces actng on t. v v (e.g., F kx) E mech, E mech, E total K U K U E total ½ mv ½ kx ½ mv ½ kx E total