PHYSC 1114, Lecture 11 Contents:
PHYSC 1114, Lecture 11 Contents: 1◦ Collected Homework 5. Preassigned part of Homework 6: Chapter 4, Problem 54. Chapter 6, Problems 20, 30, 36. This is due coming Wednesday in class. 2◦ Discussed concepts of work and energy: a. Discussed parts of Chapter 6, Sections 6-1 through 6-8, showing one movie and transparencies of Figures 6-1, 6-3, 6-4, 6-11, 6-12, and 6-17 in the book. b. Defined work, kinetic energy, and potential energy due to gravity. c. Discussed conservative and nonconservative forces. d. Introduced Work-Energy Principle and Conservation of Energy. 3◦ Repeat of definition of Work: Work done on an object by a constant force F while the object is displaced over d is, (see Figure 6-1 in book), W = F d = F d cos θ Unit: newton × meter = N m = J = joule Here F = |F|, d = |d|, and θ = (F, d) is the angle between the force and displacement vectors. Figure 6-3 in the book i
shows the free-body diagram with the four forces acting on the box: gravity, normal force, pull and friction. Remember cos 0◦ = 1, cos 90◦ = 0, cos 180◦ = −1, so that the work is positive and maximal if the displacement and force are parallel (F d), zero if force and displacement are perpendicular (F ⊥ d), and maximally negative if force and displacement are antiparallel (F −d). 4◦ Example 6-2: Hiker (H) carrying backpack up a hill. Acceleration a = 0, or constant speed. θ d Forces on backpack: h = d cosθ FH by hiker and FG by gravity, which FG = mg cancel each other as F = ma = 0. The magnitude of FH satisfies FH = mg. FH
Work by hiker on backpack: WH = FH d cos θ = mgh. Work done by gravity: WG = FG d cos(180◦ − θ) = −mgh, using cos(180◦ − θ) = − cos θ, compare Appendix A-8. This was to be expected as FG = −FH , or FG = −FH , with indicating the component in the direction of d. 5◦ Kinetic Energy: In Chapter 2 we had the equation v 2 = vo2 + 2a(x − xo ), or v22 − v12 = 2ad, rewriting x − xo = d, vo = v1 , v = v2 . Multiplying this equation by 12 m on both sides we find 2 1 2 mv2
− 12 mv12 = mad = Fnet d = Wnet ii
using Newton’s Second Law and the definition of work. We define Kinetic Energy = KE = 12 mv 2 ,
SI Unit = J = joule
so that we obtain the Work-Energy Principle: Wnet = ∆KE = KE2 − KE1 The net work on an object equals the change in its kinetic energy. 5◦ Example 6-5: See book. 6◦ Gravitational Potential Energy: From the hiker example we see that the hiker does work W = mgh to carry mass m up to height h. We define Gravitational Potential Energy = PEgrav = mgy SI Unit = J = joule where y is the vertical coordinate and y = 0 corresponds to some reference level. Then the work done by the hiker is WH = mgh = mg(y2 − y1 ) = PE2 − PE1 = ∆PE The work done by gravity is WG = −mgh = −∆PE
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7◦ Example of Figure 6-11: The hand lifts a brick of mass m vertically from height y1 to y2 . [The book uses the funny subscript “ext” (exerted), but any other name like “hand” would do also.] So the work done by the hand is Wext = Fext d cos 0◦ = mg × (y2 − y1 ) × 1 = = mg(y2 − y1 ) = PE2 − PE1 = ∆PE The work done by gravity is WG = FG d cos 180◦ = mg × (y2 − y1 ) × (−1) = = −mg(y2 − y1 ) = −(PE2 − PE1 ) = −∆PE In short, WG = −∆PEgrav in full generality. 8◦ Example 6-7 (See Figure 6-12 in book): Roller-coaster car of mass m = 1000 kg moves from A (yA = 0), to B (yB = 10 m), to C (yC = −15 m). The potential energies: a. PEA = 0, 2 b. PEB = 1000 kg × 9.80 m/s × 10 m = 9.8 × 104 J, 2 c. PEC = 1000 kg × 9.80 m/s × (−15 m) = −1.5 × 105 J. Going from B to C: ∆PEB→C = PEC −PEB = −2.5×105 J.
iv
9◦ Conservative and Nonconservative Forces: a. Conservative Force: Work is independent of path, only depending on endpoints. Then potential energy is well defined as the work done by the force along a path from the point considered to a special reference point. Example: In the case of gravity the work only depends on the height difference of original and final positions. b. Nonconservative Force: Work done by such a force depends on the path and how this path is taken. No potential energy can be defined and the mechanical energy is not conserved. Example: In the case of friction, mechanical energy is partly converted into heat. Dragging a box to OU and back is different from dragging it 1 m and back. 10◦ Work-Energy Theorem: We can write Wnet = WC + WNC , where WC be the work done by conservative forces and WNC the work done by nonconservative forces. We had earlier in this lecture Wnet = ∆KE = KE2 − KE1 and WC = −∆PE = PE1 − PE2 . Hence, Wnet = WNC + WC = WNC − ∆PE = ∆KE or WNC = ∆KE + ∆PE
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11◦ Conservation of Mechanical Energy: If all nonconservative forces are perpendicular to the path, then WNC = 0 and mechanical energy is conserved: ∆KE + ∆PE = 0
=⇒
(KE2 − KE1 ) + (PE2 − PE1 ) = 0
=⇒
KE2 + PE2 = KE1 + PE1 = constant ≡ E Here we define the mechanical energy E = KE + PE . This principle allows us to do calculations easily that are inaccessible with the theory of the first four chapters. 13◦ Example of Figure 6-17: When dropping a rock, there is a conversion of potential energy to kinetic energy: • The moment the rock is dropped the rock’s energy is all potential energy E = mgh. • Just before the rock hits the floor its energy is all kinetic energy E = 12 mv 2 . • The potential energy mgy is proportional to height y. Therefore, if the rock has dropped half the distance, half of its energy is potential energy and half kinetic energy. • At any time during its fall the rock’s energy satisfies E = 12 mv 2 + mgy = constant, as long as air-friction can be ignored.
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12◦ Movies shown: 1. Conservation of Energy: Minimal Critical Velocity on a Vertical Loop a. To calculate from which height h the ball needs to be released to reach this critical speed on top of the loop, we use the Principle of Conservation of Energy. Originally the ball has zero speed and zero kinetic energy (KE = 0). All its energy is potential energy PE = mgh. As the ball rolls down the incline its kinetic energy KE = 12 mv 2 increases and its potential energy decreases. At the bottom the energy is all kinetic energy. Then, as the ball goes up the loop-theloop, kinetic energy is converted into potential energy.
2r
mg
h
b. At the top of the loop the height is h2 = 2r = 0.40 m. The original height is h1 = h. Hence, the principle of conservation of energy says PEinitial = PEfinal +KEfinal 2 or mgh1 = mgh2 + 12 mvcrit . Dividing this by mg gives 2 vcrit (1.4 m/s)2 = 0.40 m+ h = h 1 = h2 + 2 = 0.50 m. 2g 2 × 9.8 m/s
Indeed, released from this height the ball completes the loop, but released from a lower position it falls out of the loop. vii
shows the free-body diagram with the four forces acting on the box: gravity, normal force, pull and friction. Remember cos 0◦ = 1, cos 90◦ = 0, cos 180◦ = −1, so that the work is positive and maximal if the displacement and force are parallel (F d), zero if force and displacement are perpendicular (F ⊥ d), and maximally negative if force and displacement are antiparallel (F −d). 4◦ Example 6-2: Hiker (H) carrying backpack up a hill. Acceleration a = 0, or constant speed. θ d Forces on backpack: h = d cosθ FH by hiker and FG by gravity, which FG = mg cancel each other as F = ma = 0. The magnitude of FH satisfies FH = mg. FH
Work by hiker on backpack: WH = FH d cos θ = mgh. Work done by gravity: WG = FG d cos(180◦ − θ) = −mgh, using cos(180◦ − θ) = − cos θ, compare Appendix A-8. This was to be expected as FG = −FH , or FG = −FH , with indicating the component in the direction of d. 5◦ Kinetic Energy: In Chapter 2 we had the equation v 2 = vo2 + 2a(x − xo ), or v22 − v12 = 2ad, rewriting x − xo = d, vo = v1 , v = v2 . Multiplying this equation by 12 m on both sides we find 2 1 2 mv2
− 12 mv12 = mad = Fnet d = Wnet ii
using Newton’s Second Law and the definition of work. We define Kinetic Energy = KE = 12 mv 2 ,
SI Unit = J = joule
so that we obtain the Work-Energy Principle: Wnet = ∆KE = KE2 − KE1 The net work on an object equals the change in its kinetic energy. 5◦ Example 6-5: See book. 6◦ Gravitational Potential Energy: From the hiker example we see that the hiker does work W = mgh to carry mass m up to height h. We define Gravitational Potential Energy = PEgrav = mgy SI Unit = J = joule where y is the vertical coordinate and y = 0 corresponds to some reference level. Then the work done by the hiker is WH = mgh = mg(y2 − y1 ) = PE2 − PE1 = ∆PE The work done by gravity is WG = −mgh = −∆PE
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7◦ Example of Figure 6-11: The hand lifts a brick of mass m vertically from height y1 to y2 . [The book uses the funny subscript “ext” (exerted), but any other name like “hand” would do also.] So the work done by the hand is Wext = Fext d cos 0◦ = mg × (y2 − y1 ) × 1 = = mg(y2 − y1 ) = PE2 − PE1 = ∆PE The work done by gravity is WG = FG d cos 180◦ = mg × (y2 − y1 ) × (−1) = = −mg(y2 − y1 ) = −(PE2 − PE1 ) = −∆PE In short, WG = −∆PEgrav in full generality. 8◦ Example 6-7 (See Figure 6-12 in book): Roller-coaster car of mass m = 1000 kg moves from A (yA = 0), to B (yB = 10 m), to C (yC = −15 m). The potential energies: a. PEA = 0, 2 b. PEB = 1000 kg × 9.80 m/s × 10 m = 9.8 × 104 J, 2 c. PEC = 1000 kg × 9.80 m/s × (−15 m) = −1.5 × 105 J. Going from B to C: ∆PEB→C = PEC −PEB = −2.5×105 J.
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9◦ Conservative and Nonconservative Forces: a. Conservative Force: Work is independent of path, only depending on endpoints. Then potential energy is well defined as the work done by the force along a path from the point considered to a special reference point. Example: In the case of gravity the work only depends on the height difference of original and final positions. b. Nonconservative Force: Work done by such a force depends on the path and how this path is taken. No potential energy can be defined and the mechanical energy is not conserved. Example: In the case of friction, mechanical energy is partly converted into heat. Dragging a box to OU and back is different from dragging it 1 m and back. 10◦ Work-Energy Theorem: We can write Wnet = WC + WNC , where WC be the work done by conservative forces and WNC the work done by nonconservative forces. We had earlier in this lecture Wnet = ∆KE = KE2 − KE1 and WC = −∆PE = PE1 − PE2 . Hence, Wnet = WNC + WC = WNC − ∆PE = ∆KE or WNC = ∆KE + ∆PE
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11◦ Conservation of Mechanical Energy: If all nonconservative forces are perpendicular to the path, then WNC = 0 and mechanical energy is conserved: ∆KE + ∆PE = 0
=⇒
(KE2 − KE1 ) + (PE2 − PE1 ) = 0
=⇒
KE2 + PE2 = KE1 + PE1 = constant ≡ E Here we define the mechanical energy E = KE + PE . This principle allows us to do calculations easily that are inaccessible with the theory of the first four chapters. 13◦ Example of Figure 6-17: When dropping a rock, there is a conversion of potential energy to kinetic energy: • The moment the rock is dropped the rock’s energy is all potential energy E = mgh. • Just before the rock hits the floor its energy is all kinetic energy E = 12 mv 2 . • The potential energy mgy is proportional to height y. Therefore, if the rock has dropped half the distance, half of its energy is potential energy and half kinetic energy. • At any time during its fall the rock’s energy satisfies E = 12 mv 2 + mgy = constant, as long as air-friction can be ignored.
vi
12◦ Movies shown: 1. Conservation of Energy: Minimal Critical Velocity on a Vertical Loop a. To calculate from which height h the ball needs to be released to reach this critical speed on top of the loop, we use the Principle of Conservation of Energy. Originally the ball has zero speed and zero kinetic energy (KE = 0). All its energy is potential energy PE = mgh. As the ball rolls down the incline its kinetic energy KE = 12 mv 2 increases and its potential energy decreases. At the bottom the energy is all kinetic energy. Then, as the ball goes up the loop-theloop, kinetic energy is converted into potential energy.
2r
mg
h
b. At the top of the loop the height is h2 = 2r = 0.40 m. The original height is h1 = h. Hence, the principle of conservation of energy says PEinitial = PEfinal +KEfinal 2 or mgh1 = mgh2 + 12 mvcrit . Dividing this by mg gives 2 vcrit (1.4 m/s)2 = 0.40 m+ h = h 1 = h2 + 2 = 0.50 m. 2g 2 × 9.8 m/s
Indeed, released from this height the ball completes the loop, but released from a lower position it falls out of the loop. vii
