1 Physics for Scientists & Engineers, with Modern Physics, 4 th edition Giancoli Piri Reis University / Physics -I

2 Piri Reis University Fall Semester Physics -I Chapter 7 and 8 Work and Energy

3 Piri Reis University Lecture VII and VIII I. Work Done by a Constant Force II. Kinetic Energy, and the Work-Energy Principle III. Potential Energy IV. Conservative and Nonconservative Forces V. The Conservation of Mechanical Energy VI. Problem Solving Using Conservation of Mechanical Energy VII. Other Forms of Energy; Energy Transformations and the Law of Conservation of Energy VIII. Energy Conservation with Dissipative Forces: Solving Problems IX. Power

4 I. Work Done by a Constant Force The work done by a constant force is defined as the distance moved multiplied by the component of the force in the direction of displacement:

5 W = Fd cosθ Simple special case: F & d are parallel: θ = 0, cosθ = 1  W = Fd Example: d = 50 m, F = 30 N W = (30N)(50m) = 1500 N m Work units: Newton - meter = Joule 1 N m = 1 Joule = 1 J θ F d I. Work Done by a Constant Force

6 In the SI system, the units of work are joules: As long as this person does not lift or lower the bag of groceries, he is doing no work on it. The force he exerts has no component in the direction of motion. W = 0: Could have d = 0  W = 0 Could have F  d  θ = 90º, cosθ = 0  W = 0

7 Solving work problems: 1. Draw a free-body diagram. 2. Choose a coordinate system. 3. Apply Newton’s laws to determine any unknown forces. 4. Find the work done by a specific force. 5. To find the net work, either find the net force and then find the work it does, or find the work done by each force and add. I. Work Done by a Constant Force

8 W = F || d = Fd cosθ m = 50 kg, F p = 100 N, F fr = 50 N, θ = 37º W net = W G + W N +W P + W fr = 1200 J W net = (F net ) x x = (F P cos θ – F fr )x = 1200 J Example

9 ΣF =ma F H = mg W H = F H (d cosθ) = mgh F G = mg W G = F G (d cos(180-θ)) = -mgh W net = W H + W G = 0

10 I. Work Done by a Constant Force Work done by forces that oppose the direction of motion, such as friction, will be negative. Centripetal forces do no work, as they are always perpendicular to the direction of motion.

11 II. Kinetic Energy, and the Work-Energy Principle Energy was 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. Energy  The ability to do work Kinetic Energy  The energy of motion “Kinetic”  Greek word for motion An object in motion has the ability to do work

12 Consider an object moving in straight line. Starts at speed v 1. Due to presence of a net force F net, accelerates (uniform) to speed v 2, over distance d. II. Kinetic Energy, and the Work-Energy Principle

13 Newton’s 2nd Law: F net = ma (1) 1d motion, constant a  (v 2 ) 2 = (v 1 ) 2 + 2ad  a = [(v 2 ) 2 - (v 1 ) 2 ]/(2d) (2) Work done: W net = F net d (3) Combine (1), (2), (3):  W net = mad = md [(v 2 ) 2 - (v 1 ) 2 ]/(2d) OR W net = (½)m(v 2 ) 2 – (½)m(v 1 ) 2 II. Kinetic Energy, and the Work-Energy Principle

14 If we write the acceleration in terms of the velocity and the distance, we find that the work done here is We define the translational kinetic energy: II. Kinetic Energy, and the Work-Energy Principle

15 This means that the work done is equal to the change in the kinetic energy: If the net work is positive, the kinetic energy increases. If the net work is negative, the kinetic energy decreases. II. Kinetic Energy, and the Work-Energy Principle

16 Net work on an object = Change in KE. W net =  KE  The Work-Energy Principle Note: W net = work done by the net (total) force. W net is a scalar. W net can be positive or negative (because  KE can be both + & -) Units are Joules for both work & KE.

17 Because work and kinetic energy can be equated, they must have the same units: kinetic energy is measured in joules. Work done on hammer: W h =  KE h = -Fd = 0 – (½)m h (v h ) 2 Work done on nail: W n =  KE n = Fd = (½)m n (v n ) II. Kinetic Energy, and the Work-Energy Principle

18 An object can have potential energy by virtue of its surroundings. Familiar examples of potential energy: A wound-up spring A stretched elastic band An object at some height above the ground III. Potential Energy

19 In raising a mass m to a height h, the work done by the external force is We therefore define the gravitational potential energy: W G = -mg(y 2 –y 1 ) = -(PE 2 – PE 1 ) = -  PE III. Potential Energy

20 W ext =  PE The Work done by an external force to move the object of mass m from point 1 to point 2 (without acceleration) is equal to the change in potential energy between positions 1 and 2 W G = -  PE III. Potential Energy

21 This potential energy can become kinetic energy if the object is dropped. Potential energy is a property of a system as a whole, not just of the object (because it depends on external forces). If, where do we measure y from? It turns out not to matter, as long as we are consistent about where we choose y = 0. Only changes in potential energy can be measured. III. Potential Energy

22 Potential energy can also be stored in a spring when it is compressed; the figure below shows potential energy yielding kinetic energy. III. Potential Energy

23 Restoring force “Spring equation or Hook’s law” where k is called the spring constant, and needs to be measured for each spring. (6-8) The force required to compress or stretch a spring is : F P = kx III. Potential Energy

24 The force increases as the spring is stretched or compressed further (elastic) (not constant). We find that the potential energy of the compressed or stretched spring, measured from its equilibrium position, can be written: (6-9) III. Potential Energy

25 Conservative Force  The work done by that force depends only on initial & final conditions & not on path taken between the initial & final positions of the mass.  A PE CAN be defined for conservative forces Non-Conservative Force  The work done by that force depends on the path taken between the initial & final positions of the mass.  A PE CAN’T be defined for non-conservative forces The most common example of a non-conservative force is FRICTION IV. Conservative and Nonconservative Forces

26 If friction is present, the work done depends not only on the starting and ending points, but also on the path taken. Friction is called a nonconservative force. IV. Conservative and Nonconservative Forces

27 Potential energy can only be defined for conservative forces. IV. Conservative and Nonconservative Forces

28 Therefore, we distinguish between the work done by conservative forces and the work done by nonconservative forces. We find that the work done by nonconservative forces is equal to the total change in kinetic and potential energies: W net = W c + W nc W net = ΔKE = W c + W nc W nc = ΔKE - W c = ΔKE – (– ΔPE) IV. Conservative and Nonconservative Forces

29 If there are no nonconservative forces, the sum of the changes in the kinetic energy and in the potential energy is zero – the kinetic and potential energy changes are equal but opposite in sign.  KE +  PE = 0 This allows us to define the total mechanical energy: And its conservation: (6-12b) V. The Conservation of Mechanical Energy

30  PE 2 = 0 KE 2 = (½)mv 2 KE 1 + PE 1 = KE 2 + PE mgh = (½)mv v 2 = 2gh  KE + PE = same as at points 1 & 2 The sum remains constant  PE 1 = mgh KE 1 = 0

31 In the image on the left, the total mechanical energy is: The energy buckets (right) show how the energy moves from all potential to all kinetic. VI. Problem Solving Using Conservation of Mechanical Energy

32 If there is no friction, the speed of a roller coaster will depend only on its height compared to its starting point. VI. Problem Solving Using Conservation of Mechanical Energy

33 For an elastic force, conservation of energy tells us: VI. Problem Solving Using Conservation of Mechanical Energy

34 Some other forms of energy: Electric energy, nuclear energy, thermal energy, chemical energy. Work is done when energy is transferred from one object to another. Accounting for all forms of energy, we find that the total energy neither increases nor decreases. Energy as a whole is conserved. VII. Other Forms of Energy; Energy Transformations and the Law of Conservation of Energy

35 If there is a nonconservative force such as friction, where do the kinetic and potential energies go? They become heat; the actual temperature rise of the materials involved can be calculated. VIII. Energy Conservation with Dissipative Forces: Solving Problems

36 We had, in general: W NC =  KE +  PE W NC = Work done by non-conservative forces  KE = Change in KE  PE = Change in PE (conservative forces) Friction is a non-conservative force! So, if friction is present, we have (W NC  W fr ) W fr = Work done by friction In moving through a distance d, force of friction F fr does work W fr = - F fr d VIII. Energy Conservation with Dissipative Forces: Solving Problems

37  When friction is present, we have: W fr = -F fr d =  KE +  PE = KE 2 – KE 1 + PE 2 – PE 1 –Also now, KE + PE  Constant! –Instead, KE 1 + PE 1 + W fr = KE 2 + PE 2 OR: KE 1 + PE 1 - F fr d = KE 2 + PE 2 For gravitational PE: ( ½) m(v 1 ) 2 + mgy 1 = ( ½) m(v 2 ) 2 + mgy 2 + F fr d For elastic or spring PE: (½)m(v 1 ) 2 + (½)k(x 1 ) 2 = (½)m(v 2 ) 2 + (½)k(x 2 ) 2 + F fr d

38 Problem Solving: 1. Draw a picture. 2. Determine the system for which energy will be conserved. 3. Figure out what you are looking for, and decide on the initial and final positions. 4. Choose a logical reference frame. 5. Apply conservation of energy. 6. Solve. VIII. Energy Conservation with Dissipative Forces: Solving Problems

39 Power is the rate at which work is done – The difference between walking and running up these stairs is power – the change in gravitational potential energy is the same. (6-17) In the SI system, the units of power are watts: British units: Horsepower (hp). 1hp = 746 W IX. Power

40 Power is also needed for acceleration and for moving against the force of gravity. The average power can be written in terms of the force and the average velocity: IX. Power

41 Summary of Chapter VII and VIII Work: Kinetic energy is energy of motion: Potential energy is energy associated with forces that depend on the position or configuration of objects. The net work done on an object equals the change in its kinetic energy. If only conservative forces are acting, mechanical energy is conserved. Power is the rate at which work is done.

42 Example 1 Pulling a Suitcase-on-Wheels Find the work done if the force is 45.0-N, the angle is 50.0 degrees, and the displacement is 75.0 m.

43 Example 2

44 Example 3 Accelerating a Crate The truck is accelerating at a rate of m/s 2. The mass of the crate is 120-kg and it does not slip. The magnitude of the displacement is 65 m. What is the total work done on the crate by all of the forces acting on it?

45 The angle between the displacement and the friction force is 0 degrees. Example 3 Accelerating a Crate

46 Example 4 Deep Space 1 The mass of the space probe is 474-kg and its initial velocity is 275 m/s. If the.056 N force acts on the probe through a displacement of 2.42×10 9 m, what is its final speed?

47 Example 4 Deep Space 1

48 Example 4 Deep Space 1

49 In this case the net force is Example 5

50 Example 6

51 Example 7

52 Example 8 A Gymnast on a Trampoline The gymnast leaves the trampoline at an initial height of 1.20 m and reaches a maximum height of 4.80 m before falling back down. What was the initial speed of the gymnast?

53 Example 8 A Gymnast on a Trampoline

54 Example 9 A Daredevil Motorcyclist A motorcyclist is trying to leap across the canyon by driving horizontally off a cliff 38.0 m/s. Ignoring air resistance, find the speed with which the cycle strikes the ground on the other side.

55 Example 9 A Daredevil Motorcyclist

56 Example 9 A Daredevil Motorcyclist

57 Example 10 Fireworks Assuming that the nonconservative force generated by the burning propellant does 425 J of work, what is the final speed of the rocket. Ignore air resistance.

58 Example 10 Fireworks

59 Part a) constant speed a = 0 ∑F x = 0 F – F R – mg sinθ = 0 F = F R + mg sinθ F = 3100 N P = Fv = 91 hp Part b) constant acceleration Now, θ = 0 ∑F x = ma F – F R = ma v = v 0 + at a = 0.93 m/s 2 F = ma + F R = 2000 N P = Fv = 82 hp Example 11

60 HOMEWORK Giancoli, Chapter 7 3, 7, 8, 11, 14, 25, 32, 40, 45, 47 Giancoli, Chapter 8 7, 9, 10, 17, 20, 21, 23, 28, 36, 38 References o “Physics For Scientists &Engineers with Modern Physics” Giancoli 4 th edition, Pearson International Edition