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Physics 111: Lecture 9 Today’s Agenda

Physics 111: Lecture 9 Today’s Agenda. Work & Energy Discussion Definition Dot Product Work of a constant force Work/kinetic energy theorem Work of multiple constant forces Comments. Work & Energy. One of the most important concepts in physics Alternative approach to mechanics

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Physics 111: Lecture 9 Today’s Agenda

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  1. Physics 111: Lecture 9Today’s Agenda • Work & Energy • Discussion • Definition • Dot Product • Work of a constant force • Work/kinetic energy theorem • Work of multiple constant forces • Comments

  2. 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. 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 (thermal energy).

  4. e+ e- + 5,000,000,000 V - 5,000,000,000 V Mass = Energy (but not in Physics 111) • Particle Physics: E = 1010 eV (a) (b) E = MC2 M ( poof ! ) (c)

  5. Energy Conservation Wilberforce Returning Can • 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-brakes-road-atmosphere” system is the same. • The energy of the car “alone” is not conserved... • It is reduced by the braking. • Doing “work” on an isolated system will change its “energy”...

  6. Definition of Work: Ingredients: Force (F), displacement (r) Work, W, of a constant force F acting through a displacement r is: W = F r = F rcos  = Frr F r  Fr displacement “Dot Product”

  7. Definition of Work... Hairdryer • Only the component of F along the displacement is doing work. • Example: Train on a track. F r  Fcos 

  8. a ba  b a  b ab Aside: Dot Product (or Scalar Product) Definition: a.b = ab cos  = a[b cos ] = aba = b[a cos ] = 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. y j x i k z Aside: Examples of dot products i .i = j .j = k .k = 1 i .j = j .k = k .i = 0 Suppose Then a = 1 i + 2 j + 3 k b = 4 i - 5 j + 6 k a.b = 1x4 + 2x(-5) + 3x6 = 12 a.a = 1x1 + 2x2 + 3x3 = 14 b.b = 4x4 + (-5)x(-5) + 6x6 = 77

  10. a ay ax j i 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 • Pythagorean Theorem!!

  11. 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):

  12. Back to the definition of Work: Skateboard Work, W, of a force Facting through a displacement ris: W = F r F r

  13. Lecture 9, Act 1Work & Energy • A box is pulled up a rough (m > 0) incline by a rope-pulley-weight arrangement as shown below. • How many forces are doing work on the box? (a)2 (b)3 (c)4

  14. v • Consider direction ofmotion of the box 3 forcesdo work mg does negative work Lecture 9, Act 1Solution N • Draw FBD of box: T • Any force not perpendicularto the motion will do work: f N does no work (perp. to v) T does positive work mg f does negative work

  15. Work: 1-D Example (constant force) • A force F= 10 Npushes a box across a frictionlessfloor for a distance x = 5 m. F x Work done byF on box : WF=Fx=Fx (since F is parallel to x) WF = (10 N) x (5 m) = 50 Joules (J)

  16. 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) Units: Force x Distance = Work Newton x [M][L] / [T]2 Meter = Joule [L] [M][L]2 / [T]2

  17. Work & Kinetic Energy: • A force F= 10 Npushes a box across a frictionlessfloor for a distance x = 5 m. The speed of the box is v1 before the push and v2 after the push. v1 v2 F m i x

  18. v1 v2 F x Work & Kinetic Energy... • Since the force F is constant, acceleration awill be constant. We have shown that for constant a: • v22 - v12 = 2a(x2-x1) = 2ax. • multiply by 1/2m: 1/2mv22 - 1/2mv12 = max • But F = ma1/2mv22 - 1/2mv12 = Fx m a i

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

  20. Work/Kinetic Energy Theorem: {NetWork done on object} = {change in kinetic energy of object} • We’ll prove this for a variable force later.

  21. Lecture 9, Act 2Work & Energy • Two blocks have masses m1 and m2, where m1 > m2. They are sliding on a frictionless floor and have the same kinetic energy when they encounter a long rough stretch (i.e. m > 0) which slows them down to a stop.Which one will go farther before stopping? (a)m1(b)m2(c)they will go the same distance m1 m2

  22. Lecture 9, Act 2Solution • The work-energy theorem says that for any object WNET = DK • In this example the only force that does work is friction (since both N and mg are perpendicular to the block’s motion). N f m mg

  23. Lecture 9, Act 2Solution • The work-energy theorem says that for any object WNET = DK • In this example the only force that does work is friction (since both N and mg are perpendicular to the blocks motion). • The net work done to stop the box is - fD = -mmgD. • This work “removes” the kinetic energy that the box had: • WNET = K2 - K1 = 0 - K1 m D

  24. mm2gD2 = mm1gD1 m2D2 = m1D1 Since m1>m2 we can see that D2>D1 m1 m2 D1 D2 Lecture 9, Act 2Solution • The net work done to stop a box is - fD = -mmgD. • This work “removes” the kinetic energy that the box had: • WNET = K2 - K1 = 0 - K1 • This is the same for both boxes (same starting kinetic energy).

  25. 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 r = mg rcos(0) = mgH Wg = mgH Work/Kinetic Energy Theorem: Wg = mgH= 1/2mv2 v0 = 0 mg j r H v

  26. What about multiple forces? Suppose FNET = F1 + F2 and the displacement is r. The work done by each force is: W1 = F1r W2 = F2 r WTOT= W1 + W2 = F1 r + F2 r = (F1+ F2) r WTOT= FTOT rIt’s the total force that matters!! FNET F1 r F2

  27. Comments: • Time interval not relevant • Run up the stairs quickly or slowly...same W Since W = F r • No work is done if: • F = 0 or • r = 0 or • = 90o

  28. Comments... W = F r • No work done if = 90o. • No work done by T. No work done by N. T v v N

  29. a Lecture 9, Act 3Work & Energy • An inclined plane is accelerating with constant acceleration a. A box resting on the plane is held in place by static friction. How many forces are doing work on the block? (a)1(b)2(c) 3

  30. FS a N mg Lecture 9, Act 3Solution • First, draw all the forces in the system:

  31. Lecture 9, Act 3Solution • Recall that W = F Δrso only forces that have a component along the direction of the displacement are doing work. FS a N mg • The answer is (b) 2.

  32. Recap of today’s lecture • Work & Energy (Text: 6-1 and 7-4) • Discussion • Definition (Text: 6-1) • Dot Product (Text: 6-2) • Work of a constant force (Text: 7-1 and 7-2) • Work/kinetic energy theorem (Text: 6-1) • Properties (units, time independence, etc.) • Work of a multiple forces • Comments • Look at textbook problems Chapter 6: # 1, 50, 65

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