Chapter 7

1. Chapter 7 Work and Kinetic Energy

2. Introduction • The concept of energy is one of the most important topics in science andengineering. • In everyday life, we think of energy in terms of fuel for transportationand heating, electricity for lights and appliances, and foods for consumption.However, these ideas do not really define energy. They merely tell us that fuels areneeded to do a job and that those fuels provide us with something we call energy. Prof Dr Ahmet ATAÇ

3. Introduction • In this chapter, • we first introduce the concept of work. • Work is done by a forceacting on an object when the point of application of that force moves throughsome distance and the force has a component along the line of motion. • Next, wedefine kinetic energy, which is energy an object possesses because of its motion. • We shall see that the concepts of work and kinetic energy can be applied tothe dynamics of a mechanical system without resorting to Newton’s laws. Prof Dr Ahmet ATAÇ

4. Work • The work, W, done on a system by an agent exerting a constant force on the system is the product of the magnitude F of the force, the magnitude Drof the displacement of the point of application of the force, and cos q, where qis the angle between the force and the displacement vectors. • The meaning of the term workis distinctly different in physics than in everyday meaning. • Work is done by some part of the environment that is interacting directly with the system. • Work is done on the system. • Section 7.2 Prof Dr Ahmet ATAÇ

5. Work, cont. • W = FDrcos qor • W = F d cos q • The displacement is that of the point of application of the force. • A force does no work on the object if the force does not move through a displacement. • The work done by a force on a moving object is zero when the force applied is perpendicular to the displacement of its point of application. Prof Dr Ahmet ATAÇ • Section 7.2

6. Displacement in the Work Equation • The displacement is that of the point of application of the force. • If the force is applied to a rigid object that can be modeled as a particle, the displacement is the same as that of the particle. • For a deformable system, the displacement of the object generally is not the same as the displacement associated with the forces applied. • Section 7.2 Prof Dr Ahmet ATAÇ

7. Work Example • The normal force and the gravitational force do no work on the object. • cos q = cos 90° = 0 • The force is the only force that does work on the object. • If an applied forceFacts along the direction of the displacement, then q=0 • and cos 0=1. In this case • W = F d • Section 7.2 Prof Dr Ahmet ATAÇ

8. The sign of the work depends on the direction of the force relative to the displacement. Work is positive when projection of onto is in the same direction as the displacement. Work is negative when the projection is in the opposite direction. The work done by a force can be calculated, but that force is not necessarily the cause of the displacement. Work is a scalar quantity. The unit of work is a joule (J) 1 joule = 1 newton . 1 meter = kg ∙ m² / s² J = N · m More About Work • Section 7.2 Prof Dr Ahmet ATAÇ

9. Work Is An Energy Transfer • This is important for a system approach to solving a problem. • If the work is done on a system and it is positive, energy is transferred to the system. • If the work done on the system is negative, energy is transferred from the system. • If a system interacts with its environment, this interaction can be described as a transfer of energy across the system boundary. • This will result in a change in the amount of energy stored in the system. • Section 7.2 Prof Dr Ahmet ATAÇ

10. Example 7.1 • A man cleaning a floor pulls a vacuum cleaner with a force ofmagnitude F=50.0 N at an angle of 30.0° with the horizontal (Fig.) Calculate the work done by the force on thevacuum cleaner as the vacuum cleaner is displaced 3.00 m tothe right. Prof Dr Ahmet ATAÇ

11. Solutions 7.1 Prof Dr Ahmet ATAÇ

12. Scalar Product of Two Vectors Because of the way the force and displacement vectors are combined in Equation of work, it is helpful to use a convenient mathematical tool called the scalar product. • The scalar product of two vectors is written as . • It is also called the dot product. • q is the angle betweenA and B • Applied to work, this means • or • Section 7.3 Prof Dr Ahmet ATAÇ

13. The scalar product is commutative. The scalar product obeys the distributive law of multiplication. Scalar Product, cont • Section 7.3 Prof Dr Ahmet ATAÇ

14. Dot Products of Unit Vectors • Using component form with vectors: • In the special case where • Section 7.3 Prof Dr Ahmet ATAÇ

15. Example 7.3 • A particle moving in the x-yplane undergoes a displacementd=(2.0i-3.0j) m as a constant force F=(5.0i-2.0j)Nacts on the particle. (a) Calculate the magnitude of the displacement and that of the force. (b) Calculate the work done by F. Prof Dr Ahmet ATAÇ

16. Solutions • a) • b) Prof Dr Ahmet ATAÇ

17. Work Done by a Varying Force • To use W = F d cos θ, the force must be constant, so the equation cannot be used to calculate the work done by a varying force. • Assume that during a very small displacement, Dxor d, F is constant. • For that displacement, W ~ FDx • For all of the intervals, • Section 7.4 Prof Dr Ahmet ATAÇ

18. Work Done by a Varying Force, cont. • Let the size of the small displacements approach zero . • Since • Therefore, • The work done is equal to the area under the curve between xi and xf. • Section 7.4 Prof Dr Ahmet ATAÇ

19. If more than one force acts on a system and the system can be modeled as a particle, the total work done on the system is the work done by the net force. In the general case of a net force whose magnitude and direction may vary. The subscript “ext” indicates the work is done by an external agent on the system. Work Done By Multiple Forces • Section 7.4 Prof Dr Ahmet ATAÇ

20. Work Done by Multiple Forces, cont. • If the system cannot be modeled as a particle, then the total work is equal to the algebraic sum of the work done by the individual forces. • Remember work is a scalar, so this is the algebraic sum. • Section 7.4 Prof Dr Ahmet ATAÇ

21. Example 7.4 • A force acting on a particle varies with x,as shown in Figure. Calculate the work done by the force as the particlemoves from x=0 to x=6.0 m. Prof Dr Ahmet ATAÇ

22. Solution • The work done by the force is equal to the areaunder the curve from xA=0 to xC=6.0 m. This area isequal to the area of the rectangular section from Ato Bplus the area of the triangular section from B toC. • The area ofthe rectangle is (4.0)(5.0) Nm = 20J, and • the area of thetriangle is½ (2.0)(5.0) Nm = 50J. • Therefore, the total work done is 25 J. Prof Dr Ahmet ATAÇ

23. Work Done By A Spring • A model of a common physical system for which the force varies with position. • The block is on a horizontal, frictionless surface. • Observe the motion of the block with various values of the spring constant. Prof Dr Ahmet ATAÇ • Section 7.4

24. Spring Force (Hooke’s Law) • The force exerted by the spring is Fs = - kx • x is the position of the block with respect to the equilibrium position (x = 0). • k is called the spring constant or force constant and measures the stiffness of the spring. • k measures the stiffness of the spring. • This is called Hooke’s Law. Prof Dr Ahmet ATAÇ • Section 7.4

25. Hooke’s Law, cont. • The vector form of Hooke’s Law is • When x is positive (spring is stretched), F is negative • When x is 0 (at the equilibrium position), F is 0 • When x is negative (spring is compressed), F is positive • Section 7.4 Prof Dr Ahmet ATAÇ

26. Hooke’s Law, final • The force exerted by the spring is always directed opposite to the displacement from equilibrium. • The spring force is sometimes called the restoring force. • If the block is released it will oscillate back and forth between –x and x. • Section 7.4 Prof Dr Ahmet ATAÇ

27. Work Done by a Spring • Identify the block as the system. • Calculate the work as the block moves from xi = - xmax to xf = 0. • The net work done as the block moves from -xmax to xmax is zero • Section 7.4 Prof Dr Ahmet ATAÇ

28. Assume the block undergoes an arbitrary displacement from x = xi to x = xf. The work done by the spring on the block is If the motion ends where it begins, W = 0 Work Done by a Spring, cont. • Section 7.4 Prof Dr Ahmet ATAÇ

29. Spring with an Applied Force • Suppose an external agent, Fapp, stretches the spring. • The applied force is equal and opposite to the spring force. • Work done by Fapp as the block moves from –xmax to x = 0 is equal to -½ kx2max • For any displacement, the work done by the applied force is • Section 7.4

30. Example 7.6 • A common technique used to measure the force constant ofa spring is describedin Figure. The spring is hung vertically, and an object of mass mis attached to its lower end. Under the action of the “load” mg, the spring stretches a distance d from its equilibrium position. What is the force constant ? Prof Dr Ahmet ATAÇ

31. Solution Because the springforce is upward (opposite the displacement), it must balancethe downward force of gravity mgwhen the system is at rest. Prof Dr Ahmet ATAÇ

32. Energy Review • Kinetic Energy • Associated with movement of members of a system • Potential Energy • Determined by the configuration of the system • Gravitational and Elastic Potential Energies have been studied • Internal Energy • Related to the temperature of the system • Introduction

33. Types of Systems • Non-isolated systems • Energy can cross the system boundary in a variety of ways. • Total energy of the system changes • Isolated systems • Energy does not cross the boundary of the system • Total energy of the system is constant • Conservation of energy • Can be used if no non-conservative forces act within the isolated system • Applies to biological organisms, technological systems, engineering situations, etc • Introduction

34. Ways to Transfer Energy Into or Out of A System • In non-isolated systems, energy crosses the boundary of the system during some time interval due to an interaction with the environment. • Work – transfers energy by applying a force and causing a displacement of the point of application of the force. • Mechanical Wave – transfers energy by allowing a disturbance to propagate through a medium. • Heat – the mechanism of energy transfer that is driven by a temperature difference between two regions in space. • Matter Transfer – matter physically crosses the boundary of the system, carrying energy with it. • Electrical Transmission – energy transfer into or out of a system by electric current. • Electromagnetic Radiation – energy is transferred by electromagnetic waves. • Section 8.1

35. Examples of Ways to Transfer Energy • Section 8.1

36. Conservation of Energy • Energy is conserved • This means that energy cannot be created nor destroyed. • If the total amount of energy in a system changes, it can only be due to the fact that energy has crossed the boundary of the system by some method of energy transfer. • Section 8.1

37. Conservation of Energy, cont. • Mathematically, DEsystem = ST • Esystem is the total energy of the system • T is the energy transferred across the system boundary by some mechanism • Established symbols: Twork = W and Theat = Q • Others just use subscripts • The primarily mathematical representation of the energy version of the analysis model of the non-isolated system is given by the full expansion of the above equation. • D K + D U + DEint = W + Q + TMW + TMT + TET + TER • TMW – transfer by mechanical waves • TMT – by matter transfer • TET – by electrical transmission • TER – by electromagnetic transmission • Section 8.1

38. Isolated System • For an isolated system, DEmech = 0 • Remember Emech = K + U • This is conservation of energy for an isolated system with no non-conservative forces acting. • If non-conservative forces are acting, some energy is transformed into internal energy. • Conservation of Energy becomes DEsystem = 0 • Esystem is all kinetic, potential, and internal energies • This is the most general statement of the isolated system model. • Section 8.2

39. Isolated System, cont. • The changes in energy can be written out and rearranged. • Kf + Uf = Ki + Ui • Remember, this applies only to a system in which conservative forces act. • Section 8.2

40. Kinetic Energy • One possible result of work acting as an influence on a system is that the system changes its speed. • The system could possess kinetic energy. • Kinetic Energy is the energy of a particle due to its motion. • K = ½ mv2 • K is the kinetic energy • m is the mass of the particle • v is the speed of the particle • A change in kinetic energy is one possible result of doing work to transfer energy into a system. • Section 7.5 Prof Dr Ahmet ATAÇ

41. Kinetic Energy, cont • A particle undergoing a displacement d • and a changein velocity under the action of aconstant net forceF. • A particle of mass mmoving to the right under the action ofa constant net forceF. • Because the force is constant, we know from Newton’s second law that the particle moves with a constant acceleration a. • If the particle is displaced a distance d, the net work done by the total force F is • Section 7.5 Prof Dr Ahmet ATAÇ

42. Kinetic Energy, cont • We found that the following relationships are valid when a particleundergoes constant acceleration: • where viis the speed at t=0 and vfis the speed at time t. Substituting these expressions into Equation Prof Dr Ahmet ATAÇ

43. Kinetic Energy, cont • The quantity ½ m v2represents the energy associated with the motion of theparticle. This quantity is so important that it has been given a special name —kinetic energy.The net work done on a particle by a constant net force Factingon it equals the change in kinetic energy of the particle. • In general, the kinetic energy Kof a particle of mass mmoving with a speed vis defined as Prof Dr Ahmet ATAÇ

44. Work-Kinetic Energy Theorem • The Work-Kinetic Energy Theorem states Wext = Kf – Ki = ΔK • When work is done on a system and the only change in the system is in its speed, the net work done on the system equals the change in kinetic energy of the system. • The speed of the system increases if the work done on it is positive. • The speed of the system decreases if the net work is negative. • Also valid for changes in rotational speed • The work-kinetic energy theorem is not valid if other changes (besides its speed) occur in the system or if there are other interactions with the environment besides work. • The work-kinetic energy theorem applies to the speed of the system, not its velocity. • Section 7.5 Prof Dr Ahmet ATAÇ

45. Work-Kinetic Energy Theorem, cont • We can apply Newton’s second law, Fx=max andthe net work done as • Section 7.5 Prof Dr Ahmet ATAÇ

46. Situations Involving Kinetic Friction • One way to include frictional forces in analyzing the motion of an object slidingon a horizontalsurface is to describe the kinetic energy lost because of friction.Suppose a book moving on a horizontal surface is given an initial horizontalvelocity viand slides a distance dbefore reaching a final velocity vf as shown in Figure. Prof Dr Ahmet ATAÇ

47. Situations Involving Kinetic Friction, cont., • The external force that causes the book to undergo an acceleration in thenegative xdirection is the force of kinetic friction fkacting to the left, opposite themotion.initial kinetic energy of the book is ½ mvi2and its final kinetic energyis½ mvf2.Newton’s second law gives -fk =max. • When friction — as well as other forces — acts on an object, the work – kineticenergy theorem reads Prof Dr Ahmet ATAÇ

48. Example 7.7 • A 6.0-kg block initially at rest is pulled to the right along ahorizontal, frictionless surface by a constant horizontal forceof 12 N. Find the speed of the block after it has moved 3.0 m. Prof Dr Ahmet ATAÇ

49. Solution • Because there is no friction, the net external force acting onthe block is the 12-N force. The work done by this force is; • Using the work – kinetic energy theorem and noting thatthe initial kinetic energy is zero, we obtain; Prof Dr Ahmet ATAÇ

50. Example 7.8 • A 6.0-kg block initially at rest is pulled to the right along ahorizontal, not frictionless surface by a constant horizontal forceof 12 N and has a coefficient ofkinetic friction of 0.15. Find the speed of the block after it has moved 3.0 m. Prof Dr Ahmet ATAÇ