Chapter 7 Energy of a system

1. Chapter 7Energy of a system

2. General concept of energy • The concept of energy is one of the most important concepts in physics. • It is a scalar quantity. • All process occurring in the universe involve transformation/exchange of energy • Without transformation/exchange of energy the universe is a dead world devoid of activity. • Broadly speaking, any change of state in any system necessarily involves energy exchange/transformation.

3. Bull-leh • This Red Bull canned drink symbolises “energy”. • However, in physics, the concept of energy is very much different from that used in ordinary day language. • Physics requires that all quantities used must be defined precisely using the language of mathematics whenever necessary. This includes the term “energy”.

4. “Energy approach” vs. “force approach” • One can solve a simple dynamical problem using Newton’s second law (the “force approach”), but this usually leads to much mathematical inconvenience when the system involve more than a few particles. • One can use energy approach instead, which are much more convenient and mathematically simpler. • Energy approach (mostly used in “hybrid” mode with the force approach) is mostly preferred because (i) energy is a scalar quantity, and (ii) energy obeys conservation laws (that force doesn’t). • In calculations involve conservation of energy in simple dynamics problems we will see later, mostly we will be dealing with only algebraic equations instead of the more “inconvenient” second order differential equations. • Also energy is a more general approach than force approach of the Newton second law.

5. System and its environment • A system is a subset of the Universe. • The complimentary set not contained in the system is known as “the environment”, “the rest of the Universe”. • A hypothetical, closed “boundary” separates the two. • This closed boundary defines the chosen system, and generally the boundary is not “impenetrable” but allows interactions/exchange of energies to happen between the system and its environment across boundary. • The system’s boundary does not necessarily correspond to physical boundary. • In the process of solving a physical problem, identification of a suitable system (its boundary and constituents) is an critically essential step.

6. System System Environment System Boundary System Boundary Boundary Environment Environment Environment Boundary Choosing a system is a matter of convenience • In a particular physical problem, one can define/chose the system conveniently. There is a flexibility in defining a system. • Examples:

7. F F F r r system Work done by an external force on a system • Consider a system (e.g. a block) is being acted upon solely by a external force F, which “drags” the system along through a displacement r. • When the external force acts upon the block, Newton’s 2nd law asserts that the block shall accelerates at an acceleration a=F/m. • As a result, the system’s velocity changes according to vivf = vi + (F/m)t. • We define “ the work done by the force on the system along the infinitesimal path r ” as: W = Fr

8. system External force path sr = s F system Simplest example: F // r along a finite path s • The work done by the constant external force F on a block which is dragged along the x-direction through a finite displacement s in the +x direction: W = path sFr = Fs

9. path sr = s F system path (b) F system s1 s2 Force vs. displacement diagram path (a) • Work done by the external force on the system can be interpreted as the area under the force vs. displacement “curve” along the path s as defined by the red line in (a) • The work done along such a path is Ws1s2 = +Fs = +Fs • Vice versa, if the force traverses in the opposite direction instead, as in (b), the work done would be Ws2s1 = F(-s)= -Fs s2 s1 path -sr = -s

10. Real life example • In the following example, the work done by the man on the weight in (a) is W(a)=Fs • While in (c), W(c)= -Fs. • In (b), W(b)= 0. • Note: The system is the weight itself; the force and the lifter are ‘external’ to the system.

11. F  r Work done by a force on a system is path-dependent • Generally, the work done by an external force on a system is path-dependent (Wpath = pathFdr) • depends on the relative orientation between the vector F and the displacement dr along the path, W=Fdr = Fcosdr

12. Be very clear about your definition of “work done” • Consider the case of your hand catching a base ball. • (a) The direction of displacement of the baseball is the same as the direction of the displacement of your hand, s. • (b) The ball exerts a force FB on H on your hand in the same direction as s. • The work done by the force of the ball on your hand, WB on H= FB on Hs= +FB on Hs. • (c) Your hand exerts the force FH on B (= - FB on H) on the ball. The work done by the force of your hand on the ball is WH on B= FH on Bs= -FB on Hs = -FB on Hs. • In other words, WB on H = FB on Hs = - WH on B

13. Be very clear about your definition of “work done” (cont.) • From the previous example, you should realise that, the term “work done” is not well defined without clearly stating which is the force that is doing the work, and what is the system on which the work is being done onto. • For example, we must distinguished clearly between “the work done by the force of the ball on the hand”, and “the work done by the force of the hand on the ball”. These two “works done” refers to physically distinct quantities. • Work done by an external force on a system can be positive or negative.

14. Unit for work done • Unit for work done: joule, J (same unit as for energy) • 1 joule = 1 Nm • 1 joule is the amount of work done by a 1N force that is displaced through 1m in the same direction.

15. Work done is energy transfer • If work W is done by an external force on a system, that means the same amount of energy is being transferred across the system boundary into the system. • The definition of the term “energy” shall be defined more clearly later. • Work done by an external force on a system is a mechanism where energy can be transferred across the system boundary in both direction. • If W > 0, we say “the force does positive work on the system”. In this case there is a net flow of energy into the system. • If W < 0, we say “the force does negative work on the system”. In this case there is a net flow of energy out of the system. • A non-zero work done on a system shall render a change in the total energy contained in the system.

16. Work done by a varying force • Consider a varying force in 1D Fx = Fx(x) • In a sufficiently small interval, x0, Fx(x) is approximately constant. • Work done by the force in this interval is hence W ≈ Fxx • This is represented by the shaded area in the column with width x. • Assuming Fx ≥ 0, the sum of all the area of such columns for all interval xk gives the area under the curve.

17. Work done by a varying force (cont.) • Hence the work done by the varying force Fx(x) when it traverses from the point xi to xfalong the specified path Ais path A

18. Example • A force acting upon a particle varies in position according the graph shown. Calculate the work done by the force between x =0 =6.0 m. • Solution: • Work done by the force along the part A  B  C is the area under the “curve” = [(4 x 5) + ½(6-4)x5] joule = 25 J

19. Example • The force acting on a inter-planetary satellite is given by in S.I unit, where r is the distance between them. • Calculate the work done by the Sun on the satellite when its distance from the Sun changes from 1.5 x 1011 m to 2.3 x 1011 m.

20. r2= r1+dr dr=dr r1 Fsun on sat Solution • Want to calculate work done by the force Fsun on sat when the satellite changes from r1=1.5 x 1011 m to r2 = 2.3 x 1011 m . • This is the case F // -dr, hence we expect that the numerical answer of W, the work done by the force Fsun on sat long the path from r1 to r2 along the positive radial direction is negative. • Note that the displacement dr points in the direction of r2 – r1.

21. Solution (cont.) • By definition, The work done by the force of the sun on the satellite is negative, as expected.

22. Hook’s law Fs x=xi +i x=0 xx’ x=x0 • The force exerted by the spring on the block, Fs, when the block is displaced by x from the equilibrium position (at x=x0) is given by Fs = Fsx= -k(x-x0) = -k(x-x0)i -kxi -kx’= -kx’i • k is the spring constant (also known as the force constant). It measures the stiffness of the spring. • x0 = x0i the equilibrium position where the spring is at the relaxed state. • The force by the spring on the block is always in opposite direction to the block’s displacement.

23. Hooke’s law (cont.) Fs points in -i; x points in +i; • If x > x0, the spring is stretched. By the Hooke’s law, Fs = - kxi, Fs points in the –idirection • If x < x0, the spring is compressed. By the Hooke’s law, Fs = -kxi, Fs points in the +idirection • When x = 0, the spring is neither compressed nor stretched. Fs = 0. This is the equilibrium position. x’ i x=x0 Fs points in +i; x points in -i x’ Active figure 7.10

24. Hooke’s law (cont.) Fs points in -i; x points in +i; • Fs is known as the restoration force, as it always tend to “restore” the block back to the equilibrium position, but usually the block overshoots due to its inertia. • As a result, the block is set into an oscillation motion about the equilibrium position. Such motion is a simple harmonic motion due to the simple fact that the restoration force is always proportional to the negative of the block’s displacement. x’ i x=x0 Fs points in +i; x points in -i x’

25. Work done by the spring force • We wish to calculate the work done by the spring force acting on the block when the block is displaced from xi to xf. • The spring force on the block is given by the Hooke’s as Fs = - k(x-x0). • The block is displaced from xi = xii to xf =xfi • The work done by the srping force Fs, by definition is i x=x0 x=0 xi Fs x’i xf Fs x’f

26. a a b b Work done by the spring force along two opposite paths • Say the spring force traverses from xaxb along the path Pab does work Ws,ab = X, infers that when the spring traverses in the opposite path Pba from xbxa, will do a work Ws,ba = -Ws,ab = -X Ws,ba= -X Ws,ab= X path Pba path Pab

27. Work done by the spring force in a sequential process • Work done by the spring force on the block when it is displaced from x’i = -x’maxx’f =0: • Work done by the spring force on the block when it is displaced from x’i = 0 x’f =+x’max • Hence the total work done by the spring on the block from x’=–x’max ke x’ = x’max is 0

28. i Work done by an external force on the spring • Consider an external force Fapp is stretching a spring, as shown. • The external force is such that it is equal in magnitude but opposite in direction to the spring force Fs acting on the hand that stretch the spring: Fapp=-Fs • What is the work done by the external force on the spring when the spring is stretched from x’=x’i to x’=x’f? xf xi i x=x0 x’i Fapp Fs x’f

29. Solution • Since Fapp = -Fs = -(-kx’) = kx’, the work done by the external force W ’, is the same as that of the spring force, W, except for a sign difference. Note the distinction between the definition of Wapp, Ws’; these are two different quantities, which refer to works done by different forces on different objects.

30. Example: Measuring k • A spring loaded with a mass m is stretched by an amount of d when the system is in an EB state, as shown. What is the spring constant k?

31. Solution • At EB (equilibrium), the force on the mass is • Fnet = Fs + mg = 0; • Fs=-kx’ = -kdj; mg= mgj. • Projects out the vertical component: • jFnet =-kd + mg = 0 • i.e., k = mg / d x’=0 +j x’=x’max=dj

32. Work done by the spring force +j • What is the work done by the spring force on the mass when it is stretched? Solution: x’=0 x’=x’max=dj Note that in this solution the direction +j is purposely defined in a direction different from the previous case. This is to illustrate the point that the definition of direction is a choice of convenience. The final answer should not depend on the choice of the direction’s definition.

33. +j Stretch spring in an accelerating elevator • If the experiment is repeated in an accelerating elevator with acceleration ay = -ay jupwards. What will the equilibrium displacement be in this case? a=-ayj Fs’ d’

34. +j Solution • Analyse the problem in the Earth frame. • In this frame the ball is observed to accelerate upwards with an acceleration a=ayj (ay is a negative value if accelerates upwards) • The displacement while at EB is d' (d). • In this case the ball will feel a different spring force F’sFs since the spring is stretched to a different displacement at EB. • Write down the Newton’s law for the ball: F’net = F’s + mg = ma = mayj • Projects out the vertical component: • jF’net = -kd' + mg = may d' = m(g - ay) / k = d(1- ay /g) • An observer in the elevator can determine the acceleration ay by measuring the EB stretch d ’. ay can be inferred from ay = g[1-(d‘ / d)] • The spring-ball system can be used as an “accelerometer”. a=ayj Fs’ d’

35. r| Work done by a force on a block causes change in speed • Let’s calculate the work done by a net force Fnet=F on a block when it is displaced through a displacement ralong . • The block is our chosen system. The external force is not considered as part of the box’s system. • Work done by Fnetcauses the velocity of the particle to change from vivf. • In particular, W equals the change in the function K(v) = ½mv2: W= K(vf) - K(vi)

36. r| Definition of KE • We define the function K(v)=½mv2 to be the kinetic energy (KE) of the particle with instantaneous velocity v. • Kinetic energy of a particle is the energy that is associated with the motion of the particle in space. Kf= ½mvf2 Ki= ½mvi2

37. KE change is a result of external work done on a system • The work done by the external force causes the KE to change from Ki = ½mvi2 to Kf =½mvi2. W = Kf – KiK • We can also interpret the equation as: The work done by the external force on the block system is transferred into the system, and as a result, the kinetic energy of the block system is changed by the same amount. • In other words, the work done by the external force is transformed into the kinetic energy of the system.

38. KE change is a result of external work done on a system (cont.) • The change in KE (K) of the block’ system is only one of many possible effects caused by the work done by a net external force that transfers energy into or out from a system. • In general, when a force does work on a system, K is not the only energy change that could happen. There could be also changes in energies of other kind (e.g., potential energy) in the system too. • It so happens that in our discussion here, the work done by the external force only causes a change in KE of the system which comprise of a assumed structure-less block. There is no change of energy of other form is taking place within the block other than KE.

39. Work-kinetic energy theorem • More formally the states that: “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.”

40. When work is done, KE is changed. • From the work-KE theorem W = Kf– KiK • If an external force does positive work on the system, K= Kf– Ki is positive Kf> Ki, i.e., the KE of the system increases. • If an external force does negative work on the system, K= Kf– Ki is negative Kf< Ki, i.e., the KE of the system decreases.

41. Example (a) • A 6.0 kg block initially at rest is dragged through a straight horizontal path by a horizontal force F of 12 N. Obtain the block’s velocity after it has been displaced for 3.0 m. Solution In this case, F // x. The work done by the force is WF = Fx = Fx = 12N3.0m=36 J. By the work-KE theorem, WF = K, the change in the KE is K=Kf - Ki = ½mvf2 – 0 = ½mvf2 Equating WF to K, vf = (2WF / m)½ = (236/6) ½ m/s = 3.46 m/s.

42. Example (b) • If the external force’s magnitude is doubled, FF’= 2 F, and the block is accelerated to the same vf after traversing a distance x’ . How is x’ compared to x? • Solution WF = Fx= K =½mvf2 WF’ = F’x’= K’ =½mv’f2 (Fx / F’x’)=(½mvf2/½mv’f2) (Fx / 2Fx’)=(vf2/v’f2) = 1 • (½)(x /x’)= 1 • x’= ½x

43. Example (c) • How is the time interval t’ in (b) compared to that of tin (a)? Solution: Since the force in (b) is twice that of (a), so is the acceleration: a’=2a. To accelerate the block from rest to a common speed, the time taken in (b) is only half to that in (a), i.e. t’ = ½ t. • Note that in this case, one could simply know the answer even without performing the calculation quantitatively. Only qualitative argument is sufficient.

44. Potential Energy

45. system with only a single ‘particle’ external force system with two ‘particles’ internal forces -mg Earth (the ‘particle’ always at rest) System with internal forces • In the previous discussion, our system comprises of only one single particle. The particle in the system interacts only with the external force. The work done by the external force causes a change in the system’s kinetic energy. • Now, we would like to consider a system containing two particles, such that there exists some internal forces between the two particles inside the system. Furthermore, we also assume one of the particles is always at rest.

46. Potential energy is the energy associated with spatial relative configuration • Potential energy (PE) of a system is the energy that is associated not with the motion of the system but the relative configuration (how the constituents are placed relative to each other in space) of the system, in which the constituents interacts through some internal forces. • A change in the relative configuration of the constituents in the system could cause a change in potential energy of the system (but this is not always so). • Examples of potential energies that should be familiar to us include • Gravitational PE • Electromagnetic PE • Elastic energy stored in a stretched/strained spring.

47. Example of a change in relative configuration in a system • Consider the book-Earth system: • The initial position of the book wrp to the Earth is a configuration, specified by the vertical coordinate yb. • The final position of the book wrp to the Earth after the book falls through a vertical displacement is another configuration, specified by the vertical coordinate ya. • Comparing these two configurations, we say that there occurs a relative configuration change, with y = yb- ya 0 system Earth

48. system with two ‘particles’ internal forces -mg Earth (the ‘particle’ always at rest) Example of a change in relative configuration in a system • The change in the configuration of the book-Earth system comes hand-in-hand with a change in the PE of the system. • Such an association (i.e., change in configuration  change in PE) is only possible in the presence of internal force within the system. • In other words, if there is no internal forces in the system, there shall be no potential energy which can be associated with the system. • In a system with no internal forces acting among the constituents, a relative configuration change does not correspond to any change in potential energy of that system (e.g. in an ideal gas system).

49. Boundary System The spring force is a pair of internal forces Environment PE is ‘force-specific’ • Potential energies are specifically related to the forces that “produce” them. In other word, a given type of force will produce a specific type of potential energy. • The change in potential energy of a given kind is due to the work done by a specific force related to that potential energy. • E.g., gravitational PE change in the book-Earth system is associated with the work done by gravitational force, which is an internal force to the Earth-book system. • E.g., Elastic potential energy in a spring-mass system is associated with the spring force, which is also an internal force in the spring-mass system.

50. Note • Given a defined system, such as a “block” or a “book-Earth” or a “block-spring” system, and also the forces in operation within the system, can you differentiate which forces are internal and which are not?