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Physics II Today’s Agenda

Physics II Today’s Agenda. Work & Energy. Discussion. Definition. Scalar Product. Work of a constant force. Work kinetic-energy theorem. Work of a sum of constant forces. Work for a sum of displacements with constant force. Comments.

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Physics II Today’s Agenda

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  1. Physics IIToday’s Agenda • Work & Energy. • Discussion. • Definition. • Scalar Product. • Work of a constant force. • Work kinetic-energy theorem. • Work of a sum of constant forces. • Work for a sum of displacements with constant force. • Comments. • Look at textbook problems Chp7 -7,11,12,17,20,23,24,27,29,36

  2. See text: 1-1 and 7-1 Work & Energy • One of the most important concepts in physics. • Alternative approach to mechanics. • Many applications beyond mechanics. • Thermodynamics (movement of heat). • Quantum mechanics... • Very useful tools. • You will learn new (sometimes much easier) ways to solve problems.

  3. See text: 1-1 and 7-1 Forms of Energy • Kinetic: Energy of motion. • A car on the highway has kinetic energy. • We have to remove this energy to stop it. • The breaks of a car get HOT! • This is an example of turning one form of energy into another. (More about this soon)...

  4. See text: 1-1 and 7-1 Forms of Energy • Potential: Stored, “potentially” ready to use. • Gravitational. • Hydro-electric dams etc... • Electromagnetic • Atomic (springs, chemical...) • Nuclear • Sun, power stations, bombs...

  5. e+ e- + 5,000,000,000 V - 5,000,000,000 V Mass = Energy (but not in Physics II) • Particle Physics: E = 1010 eV (a) (b) E = MC2 M ( poof ! ) (c)

  6. See text: 7-1 Energy Conservation • Energy cannot be destroyed or created. • Just changed from one form to another. • We say energy is conserved ! • True for any isolated system. • i.e when we put on the brakes, the kinetic energy of the car is turned into heat using friction in the brakes. The total energy of the “car-breaks-road-atmosphere” system is the same. • The energy of the car “alone” is not conserved... • It is reduced by the braking. • Doing “work” on a system will change it’s “energy”...

  7. See text: 7-1 and 7-2 Definition of Work: Ingredients: Force (F), displacement (S) Work, W, of a constant force F acting through a displacement S is: W = F.S = FScos(q) = FSS F S q FS displacement “Dot Product”

  8. a ba q b ab See text: 7-2 Aside: Scalar Product ( or Dot Product) Definition: a.b = abcos(q) = a[bcos(q)] = aba = b[acos(q)] = bab Some properties: a.b =b.a q(a.b) = (qb).a = b.(qa)(q is a scalar) a.(b + c) = (a.b)+ (a.c)(c is a vector) The dot product of perpendicular vectors is 0 !!

  9. See text: 7-2 Aside: Examples of dot products i .i = j .j = k .k = 1 i .j = j .k = k .i = 0 Then Suppose a.b = 1x4 + 2x(-5) + 3x6 = 12 a.a = 1x1 + 2x2 + 3x3 = 14 b.b = 4x4 + (-5)x(-5) + 6x6 = 77 a = 1 i + 2 j + 3 k b = 4 i - 5 j + 6 k

  10. a ay ax j i See text: 3-5 and 7-2 Aside: Properties of dot products • Magnitude: a2 = |a|2 = a .a =(axi + ay j ) .(axi + ayj ) = ax2( i .i ) + ay2( j .j ) + 2axay( i .j ) = ax2 + ay2 • Pythagoras Theorem !!

  11. See text: 7-2 Aside: Properties of dot products • Components: a = ax i + ay j + az k = (ax , ay , az ) = (a. i , a. j , a. k ) • Derivatives: • Apply to velocity • So if v is constant (like for UCM): since a and v are perpendicular

  12. See text: 7-2 Back to the definition of Work: Work, W, of a force F acting through a displacement S is: W = F.S F S

  13. See text: 7-1 and 7-2 Work: 1-D Example (constant force) • A force F= 10Npushes a box across a frictionless floor for a distance Dx = 5m. F Dx Work done byF on box : WF=F.Dx=FDx (since F is parallel to Dx) WF = (50 N)x(5m) = 50 N-m. See example 7.1

  14. mks cgs other BTU = 1054 J calorie = 4.184 J foot-lb = 1.356 J eV = 1.6x10-19 J Dyne-cm (erg) = 10-7 J N-m (Joule) See text: 7-1 Units: Force x Distance = Work Newton x [M][L] / [T]2 Meter = Joule [L] [M][L]2 / [T]2

  15. v1 v2 F Dx See text: 7-1 and 7-2 Work & Kinetic Energy: • A force F= 10Npushes a box across a frictionlessfloor for a distance Dx = 5m. The speed of the box is v1 before the push, and v2 after the push. m i

  16. v1 v2 F Dx See text: 7-1, 7-2, 7-5 Work & Kinetic Energy... • Since the force F is constant, acceleration a will be constant. We have shown that for constant a: • v22 - v12 = 2a(x2-x1 ) = 2aDx. • multiply by 1/2m: 1/2mv22 - 1/2mv12 = maDx • But F = ma1/2mv22 - 1/2mv12 = FDx m a i

  17. v1 v2 F Dx Work & Kinetic Energy... • So we find that • 1/2mv22 - 1/2mv12 = FDx = WF • Define Kinetic Energy K: K = 1/2mv2 • K2 - K1 = WF • WF = DK (Work kinetic-energy theorem) m a i

  18. Fnet K2 K1 dS See text: 7-3, 7-4, 7-5 Work Kinetic-Energy Theorem: {NetWork done on object} = {change in kinetic energy of object} • This is true in general:

  19. F(x) x1 x2 dx See text: 7-3 Work done by Variable Force: (1D) • When the force was constant, we wrote W = FDx • area under F vs x plot: • For variable force, we find the areaby integrating: • dW = F(x) dx. F Wg x Dx

  20. See text: 7-3, 7-4, 7-5 A simple application:Work done by gravity on a falling object • What is the speed of an object after falling a distance H, assuming it starts at rest ? • Wg = F.S = mgScos(0) = mgH Wg = mgH Work Kinetic-Energy Theorem: Wg = mgH= 1/2mv2 v0 = 0 mg j S H v

  21. See text: 7-3, 7-4, 7-5 What about a sum of forces? Suppose FTOT = F1 + F2 and the displacement is S. The work done by each force is: W1 = F1.S W2 = F2.S WTOT= W1 + W2 = F1.S + F2.S = (F1+ F2).S WTOT= FTOT .SIt’s the total force that mattters !! FTOT F1 S F2

  22. See text: 7-3, 7-4, 7-5 Work by sum of displacementswith constant force. W = W1 + W2 = F.S1 + F.S2 = F.( S1 + S2 ) W= F.S Work depends onlyon total displacement, not on the “path”. S2 S S1 F

  23. See text: 7-3, 7-4, 7-5 Work by sum of displacementswith constant force. W = W1 + W2 +. . .+ Wn = F.S1 + F.S2 + . . . + F.Sn = F.( S1 + S2 + . . .+ Sn ) W= F.S Same result as simple case. Sn S S3 S2 S1 F

  24. See text: 7-3, 7-4, 7-5 Comments: • Time interval not relevant. • Run up the stairs quickly or slowly...same W. Since W = F.S • No work is done if: • F = 0 or • S = 0 or • q= 90o

  25. See text: 7-3, 7-4, 7-5 Comments... W = F.S • No work done if q= 90o. • No work done by T. • No work done by N. T v v N

  26. Recap of today’s lecture • Work & Energy. • Discussion. • Definition. • Scalar Product. • Work of a constant force. • Work kinetic-energy theorem. • Properties (units, time independence etc). • Work of a combination of forces. • Comments. • Look at textbook problems Serway Chp7 -7,11,12,17,20,23,24,27,29,36

  27. Physics II: Lecture Todays Agenda • Review of Work. Last lecture • Work done by gravity near the earths surface. • Examples: • pendulum, inclined plane, free-fall. • Work done by variable force. • Spring • Problem involving spring & friction- Serway 7- 37,39,40,46,51

  28. v1 v2 F Dx See text: 7-3, 7-4, 7-5 Work Kinetic-Energy Theorem: • {NetWork done on object} • = • {change in kinetic energy of object} • WF = DK = 1/2mv22 - 1/2mv12 WF = FDx m

  29. See text: 7-1 Work done by gravity: • Wg = F.S1= mgS1cos(q1) = -mgS1cos(f1) = -mgDy Wg = -mgDy Depends only on Dy ! S1 Dy j f1 q1 m mg

  30. See text: 7-1 Work done by gravity... • Wg = S(F.s)= F.STOT Wg = -mgDy Depends only on Dy ! Dy j m mg See example 7-2 (easy) and example 7-9 (harder)

  31. See text: 7-1 Example: Falling Objects v=0 v=0 v=0 H vf vf vf Free Fall Frictionless inclinePendulum

  32. N mg See text: 7-1 and 7-2 Example: Falling Objects... W = F.S = FScos(q) • No work done if q= 90o. • No work done by T. • Only mg does work ! • No work done by N. • Only mg does work ! T v mg v

  33. See text: 7-3 and 7-4 Example: Falling Objects v=0 v=0 v=0 H vf vf vf Free Fall Frictionless inclinePendulum Only gravity will do work: Wg = mgH = 1/2 mvf2 does not depend on path !!

  34. See text: 7-1 to 7-5 Lifting a book with your hand:What is the total work done on the book ?? • First calculate the work done by gravity: Wg = mg.S = -mgS • Now find the work done bythe hand: WHAND = FHAND.S = FHAND S FHAND S v = const a = 0 mg

  35. See text: 7-1 to 7-5 Example: Lifting a book... Wg = -mgS WHAND = FHAND S WTOT = WHAND + Wg = FHAND S -mgS = (FHAND -mg)S = 0 sincea = 0 • So WTOT= 0 !! FHAND S v = const a = 0 mg

  36. See text: 7-5 Example: Lifting a book... • Work Kinetic-Energy Theorem says: WF = DK {NetWork done on object}={change in kinetic energy of object} In this case, v is constant so DK = 0and so WF must be 0, as we found. FHAND S v = const a = 0 mg

  37. F(x) x1 x2 dx See text: 7-3 Work done by Variable Force: (1D) • When the force was constant, we wrote W = FDx • area under F vs x plot: • For variable force, we find the areaby integrating: • dW = F(x) dx. F Wg x Dx

  38. See text: 7-3 1-D Variable Force Example: Spring • For a spring we know that Fx = -kx. F(x) x1 x2 x equilibrium -kx F= - k x1 F= - k x2

  39. See text: 7-3 Spring... • The work done by the spring Wsduring a displacement from x1to x2 is the area under the F(x) vs x plot between x1and x2. F(x) x1 x2 x Ws equilibrium -kx

  40. See text: 7-3 Spring... • The work done by the spring Wsduring a displacement from x1to x2 is the area under the F(x) vs x plot between x1and x2. F(x) x1 x2 x Ws -kx

  41. See text: 7-3 Problem: Spring pulls on mass. • A spring (constant k) is stretched a distance d, and a mass m is hooked to its end. The mass is released (from rest). What is the speed of the mass when it returns to the equilibrium position if it slides without friction? m equilibrium position stretched position (at rest) m d after release m v back at equilibrium position m vE

  42. See text: 7-3 Problem: Spring pulls on mass. • First find the net work done on the mass during the motion from x=d to x=0 (only due to the spring): stretched position (at rest) m d equilibrium position m i ve

  43. See text: 7-3 Problem: Spring pulls on mass. • Now find the change in kinetic energy of the mass: stretched position (at rest) m d equilibrium position m i ve

  44. See text: 7-3 Problem: Spring pulls on mass. • Now use work kinetic-energy theorem: Wnet = WS = DK. stretched position (at rest) m d equilibrium position m i ve

  45. See text: 7-3 Problem: Spring pulls on mass. • Now suppose there is a coefficient of friction m between the block and the floor? • The total work done on the block is now the sum of the work done by the spring WS (same as before) and the work done by friction Wf.Wf = f.S = -mmg d S stretched position (at rest) m d equilibrium position f= mmg m i ve

  46. See text: 7-3 Problem: Spring pulls on mass. • Again use Wnet = WS + Wf = DKWf = -mmg d S stretched position (at rest) m d equilibrium position f= mmg m i ve

  47. Recap of today’s lecture • Review • Work done by gravity near the earths surface. • Examples: • pendulum, inclined plane, free-fall. • Work done by variable force. • Spring • Problems involving spring & friction. Serway 7- 37,39,40,46,51

  48. Physics II: Lecture Todays Agenda • Review. • Work done by variable force in 3-D. • Newtons gravitational force. • Conservative Forces & Potential energy. • Conservation of “Total Mechanical Energy” • Example: Pendulum. • Nonconservative force • friction • General work-energy theorem • Example problem. • Serway Chp 8- 1,2,7,9,11,12,14,20,22,25,27,32,35

  49. F dS See text: 7-4 Work by variable force in 3-D: • Work dWF of a force F acting • through an infinitesmal • displacement dS is: • dW = F.dS • The work of a big displacement through a variable force will be the integral of a set of infinitesmal displacements: • WTOT = F.dS

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