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Chapter 6: Work & Energy

Chapter 6: Work & Energy. THE COURSE THEME is NEWTON’S LAWS OF MOTION ! So Far : Motion analysis with forces . NOW : An alternative analysis using the concepts of Work & Energy . Easier? My opinion is yes! Conservation of Energy is NOT a new law!

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Chapter 6: Work & Energy

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  1. Chapter 6: Work & Energy

  2. THE COURSE THEME is NEWTON’S LAWS OF MOTION! • So Far: Motion analysis with forces. • NOW : An alternative analysis using the concepts of Work & Energy. • Easier? My opinion is yes! Conservation of Energy is NOT a new law! • We’ll see that this is just Newton’s Laws of Motion re-formulated or re-expressed (translated) from Force Language to Energy Language.

  3. Newton’s 2nd Law: • So far, we’ve expressed Newton’s Laws of Motion using the concepts of position, displacement, velocity, acceleration & force. • Newton’s Laws with Forces: Are quite general (macroscopic objects). In principle, they could be used to solve any dynamics problem, But, often, they are very difficult to apply, especially to very complicated systems. So, alternate formulations have been developed. These alternate formulations are often easier to apply. • Also, a practical problem is that, often, we may not even know all of the forces.

  4. One alternate approach uses Energy instead of Forceas the most basic physical quantity. • So, we’ll discuss Newton’s Lawsin Energylanguage. • Before we discuss these, we need to learn some vocabulary in • Energy Language. Newton’s 2nd Law:

  5. Newton’s 2nd Law: • Energy: A very common term in everyday usage. Everyday meanings might not coincide with the PHYSICS meaning! • Every physical process involves energy or energy transfer or transformations. Energy in physics can be somewhat abstract.

  6. Work • A connection between force & energy is work. Workdepends on the force, the displacement & the direction between them. • Work in physics has a more specific meaning than in everyday usage. • Work: Precisely defined in physics. Describes what is accomplishedby a force in moving an object through a distance.

  7. Work Done by Constant Force • Work: Precisely defined in physics. Describes what is accomplishedby a force in moving an object through a distance. For an object moving under the influence of a Constant Force, (as in the figure!) Work done (W) product of the magnitude of the displacement (d)  the component of force parallel to the displacement (F||). d W  F||d = Fd cosθ

  8. Work: W  FΔr • For 1-dimensional motion, W = F Δx • In 2- or 3-dimensions, the vector nature of the force F & displacement Δr must obviously be taken into into account: W = FΔrcos θ • θ is the angle between the force and the displacement

  9. Work and Directions • The component of F along the displacement direction is: Fcosθ F|| • When F|| is parallel to the displacement, the work is positive. • When F|| is anti- parallelto the displacement, the work is negative. • When the component of the force is perpendicular to the displacement, the work is zero.

  10. Work • Experiments have verified that although various forces produce different times & distances, the product of the force & the distance remains the same. • To accelerate an object to a specific velocity, you can exert a large force over a short distance or a small force over a long distance

  11. Work, Force and Displacement Work (W) is done bya force acting on an object. • W depends on the force F acting on the object and on the object’s displacement d. • The value of W depends on the direction of the force relative to d. • W may be positive, negative, or zero, depending on the angle θ between the force and the displacement. • If the displacement is zero (the object does not move), then W = 0, even though the force may be very large.

  12. What Does the Work? • When an “agent” (another mass from N’s 3rd Law!) applies a force to an object and does an amount of work W on that object, the object will do an amount of work equal to –W back on the agent • The forces form a Newton’s 3rd Law action-reaction pair. • Multiple agents (masses): • When multiple agents act on an object, you can calculate the work done by each separate agent

  13. Work and Amplifying Force • See figure. Suppose the person lifts his end of the rope through a distance L. • The pulley will move through a distance (½ )L • Work on the crate: Won crate = (2T)(L/2) = TL • Work on the rope: Won rope = TL • Work done on the rope is equal to the work done on the crate

  14. Work and Amplifying Force • Work on the crate: Won crate = (2T)(L/2) = TL • Work on the rope: Won rope = TL • Work done by the person is effectively “transferred” to the crate. • Forces can be amplified, but work can’t be increased this way. • The force is amplified, but not the work • The associated displacement is decreased • The work-energy theorem suggests work can be converted to energy, but since work cannot be amplified the exchange will not increase the amount of energy available • The result that work cannot be amplified is a consequence of the principle of conservation of energy

  15. Work Done by a Constant Force Work:W  F||d = Fd cosθ For a CONSTANT force!

  16. W = F||d = Fd cosθ • Consider a simple special case when F & d are parallel: θ = 0, cosθ = 1  W = Fd • Example: d = 50 m, F = 30 N  W = (30N)(50m) = 1500 N m SI Work Units Newton - meter = Joule 1 N m = 1 Joule = 1 J

  17. W  F||d = Fdcosθ • Its possible to exert a force & yet do no work! • Could have d = 0  W = 0 • Or could have F  d θ = 90º, cosθ = 0  W = 0 • Example,walking at constant speed v with a grocery bag.

  18. W = F||d = Fdcosθ m = 50 kg, FP = 100 N, Ffr = 50 N, θ = 37º Example

  19. W = F||d = Fdcosθ m = 50 kg, FP = 100 N, Ffr = 50 N, θ = 37º Example Solution: WG = mgxcos(90) = 0, WN = FN xcos(90) = 0 WP = Fpxcos(37) = (100)(40)cos(37) = 3200 J WFfr= Ffrxcos(180) = - 2000J Wnet = 0 + 0 + 3200 -2000 = 1200 J

  20. Solving Work Problems • Sketch a free-body diagram. • Choosea coordinate system. • ApplyNewton’s Lawsto determine any unknown forces. • Findthe work done by a specific force. • Findthe net work by either • a. Find the net force & then find the work it does, or • b. Find the work done by each force & add.

  21. Typical Problem An object is displaced by a force Fon a frictionless, horizontal surface. The free body diagram is shown. The normal force FN & weightmg do no work in the process, since both are perpendicular to the displacement. W = F||d = Fd cosθ d Angles for forces in the figure: Normal Force: θ = 90°,cosθ = 0 Weight: θ = 270° (or - 90°), cosθ = 0

  22. Example: Work on a Backpack (a) Calculate the work a hiker must do on a backpack of mass m = 15 kg to carry it up a hill of height h = 10 m, as shown. (b) Calculate the work done by gravity on the backpack. (c) Calculate the net work done on the backpack. For simplicity, assume that the motion is smooth & at constant velocity (zero acceleration). For the hiker: ∑Fy = 0 = FH – mg  FH = mg WH = FHdcosθ = FHh

  23. Conceptual Example: Does Earth do Work on the Moon? The Moon revolves around the Earth in a nearly circular orbit, with approximately constant tangential speed, kept there by the gravitational force exerted by the Earth. Does gravity do (a) positive work (b) negative work, or (c) no work at all on the Moon?

  24. Example The force shown has magnitude FP = 20 N & makes an angle θ = 30° to the ground. Calculate the work done by this force when the wagon is dragged a displacement d = 100 m along the ground.

  25. Example The force shown has magnitude FP = 20 N & makes an angle θ = 30° to the ground. Calculate the work done by this force when the wagon is dragged a displacement d = 100 m along the ground. Solution: WP = FPdcos(30) = (20)(100)cos(30) = 1732 J

  26. A man cleaning a floor pulls a vacuum cleaner with a force F = 50 N at an angle θ = 30º with the horizontal. Calculate the work done by the force F as the vacuum cleaner is moved a distance Δr = 3.0 mto the right. Example Δr = 3.0 m

  27. A man cleaning a floor pulls a vacuum cleaner with a force F = 50 N at an angle θ = 30º with the horizontal. Calculate the work done by the force F as the vacuum cleaner is moved a distance Δr = 3.0 m to the right. Solution W = Fd cosθ = (50)(3)cos(30º) = 130 J Δr = 3.0 m

  28. Kinetic Energy & Work-Energy Theorem

  29. Energy: Traditionally defined as the ability to do work. We now know that not all forces are able to do work; however, we are dealing in these chapters with mechanical energy, which does follow this definition. • Kinetic Energy The energy of motion “Kinetic” Greek word for motion An object in motion has the ability to do work.

  30. Kinetic Energy • Find the work done on an object as a force F moves it from an initial position xito a final position xf Wnet = FnetΔx = (ma)Δx (1) • If it is a constant, the acceleration a can be expressed in terms of velocities (1 d kinematic equations): • Combining gives: • Wnet = (½)mvf² - (½)mvi² (2) • (½)mv²  kinetic energy  KE • KE is the energy due to the motion of the object. • (1) & (2) together give the Work-Energy Theorem: • Wnet= KE

  31. The Work-Energy Theorem Wnet = KE Physics of the Work-Energy Theorem: The kinetic energy of an object can be changed by doing work on the object

  32. Consider an object moving in straight line. It starts at speed v1. Due to the presence of a net forceFnet, (≡∑F),it accelerates (uniformly) to speed v2, over a distance d. Newton’s 2nd Law: Fnet= ma (1) 1d motion, constant a  (v2)2 = (v1)2 + 2ad  a = [(v2)2 - (v1)2]/(2d) (2) Work done: Wnet = Fnet d (3) Combine (1), (2), (3):

  33. Fnet= ma (1) a = [(v2)2 - (v1)2]/(2d) (2) Wnet = Fnet d (3) • Combine (1), (2) & (3):  Wnet = mad = md [(v2)2 - (v1)2]/(2d) OR Wnet = (½)m(v2)2 – (½)m(v1)2

  34. DEFINITION:Kinetic Energy(KE) (Kinetic = “motion”, units are Joules, J) • Summary: The net work done by a constant force in accelerating an object of mass m from v1to v2is: KE • We have: The WORK-ENERGY Theorem • Wnet = KE • ( = “change in”) • We’ve shown this for a 1d constant force. However, it is valid in general!

  35. The net work done on an object = the change in it’s KE. Wnet = KE (1)  The Work-Energy Theorem Note:(1) is Newton’s 2nd Lawin Work & Energy language!

  36. The net work done on an object = the change in it’s KE: Wnet = KE (1)  The Work-Energy Theorem Note: • Wnet = work done by the net (total) force. • Wnet is a scalar & can be positive or negative • This is true because because KE can be both + & -. • If Wnetis positive, the KE increases. If Wnetis negative, the KE decreases. • Units are Joules for both work & kinetic energy. Note: (1) is Newton’s 2nd Law in Work & Energy language!

  37. A moving hammer can do work on a nail! • For the hammer: Wh = KEh = -Fd = 0 – (½)mh(vh)2 • For the nail: Wn = KEn = Fd = (½)mn(vn)2 - 0

  38. Example: Kinetic Energy & Work Done on a Baseball • A baseball, mass m = 145 g (0.145 kg) is • thrown so that it has a speed v = 25 m/s. • a. What is its kinetic energy? • b. What was the net work done on the ball • to make it reach this speed, starting from rest?

  39. Example: Kinetic Energy & Work Done on a Baseball • A baseball, mass m = 145 g (0.145 kg) is • thrown so that it has a speed v = 25 m/s. • a. What is its kinetic energy? • b. What was the net work done on the ball • to make it reach this speed, starting from rest? Solution: a. KE = (½)mv2 = (½)(0.145)(25)2 = 45.3 J b. Wnet = KE = (45.3) – 0 = 45.3 J

  40. Work on a car to increase its kinetic energy Calculate the net work required to accelerate a car, mass m = 1000-kg car from v1 = 20 m/s to v2 = 30 m/s.

  41. Work on a car to increase its kinetic energy Calculate the net work required to accelerate a car, mass m = 1000-kg car from v1 = 20 m/s to v2 = 30 m/s. Solution: Wnet = KE = (½)m(v2)2 – (½)m(v1)2 Wnet = (½)(1000)[(30)2 – (20)2] Wnet = 2.5  104 J

  42. Conceptual Example: Work to Stop a Car A car traveling at speed v1 = 60 km/h can brake to a stop within a distance d = 20 m. If the car is going twice as fast, 120 km/h, what is its stopping distance? Assume that the maximum braking force is approximately independent of speed.

  43. Definition of Work: Wnet = Fdcos (180º) = -Fd • Work-Energy Theorem: • Wnet = KE = (½)m(v2)2 – (½)m(v1)2 • (v2)2 = 0(car stopped) so-Fd = KE= 0 - (½)m(v1)2or • d  (v1)2 • So the stopping distance is proportional to the square of the initial speed! If the initial speed is doubled, the stopping distance quadruples!Note:KE  (½)mv2  0 must be positive, since m & v2 are always positive (real v).

  44. Example A block, mass m = 6 kg, is pulled from rest (v0 = 0) to the right by a constant horizontal force F = 12 N. After it has been pulled for Δx = 3 m, find it’s final speed v.

  45. Example A block, mass m = 6 kg, is pulled from rest (v0 = 0) to the right by a constant horizontal force F = 12 N. After it has been pulled for Δx = 3 m, find it’s final speed v. • Solution:Work-Energy Theorem • Wnet = KE  (½)[m(v)2 - m(v)2] (1) • If F = 12 N is the only horizontal force, we have • Wnet = FΔx (2) • Combine (1) & (2): FΔx = (½)[m(v)2 - 0] • Solve for v: (v)2 = [2Δx/m] • v = [2Δx/m]½ = 3.5 m/s

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