Chapter 8: Potential Energy and Conservation of Energy. Work and kinetic energy are energies of motion.


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1 Chapter 8: Potential Energy and Conservation of Energy Work and kinetic energy are energies of motion. Consider a vertical spring oscillating with mass m attached to one end. At the extreme ends of travel the kinetic energy is zero, but something caused it to accelerate back to the equilibrium point. We need to introduce an energy that depends on location or position. This energy is called potential energy.
2 Chapter 8: Potential Energy and Conservation of Energy Work done by gravitation for a ball thrown upward that then falls back down W ab + W ba = mgd + mg d = 0 The gravitational force is said to be a conservative force. b A force is a conservative force if the net work it does on a particle moving around every closed path is zero. a
3 W ab,1 + W ba,2 = 0 W ab,2 + W ba,2 = 0 therefore: W ab,1 = W ab,2 Conservative forces i.e. The work done by a conservative force on a particle moving between two points does not depend on the path taken by the particle. So,... choose the easiest path!!
4 Conservative & Nonconservative forces Conservative Forces: (path independent) gravitational spring electrostatic Nonconservative forces: (dependent on path) friction air resistance electrical resistance These forces will convert mechanical energy into heat and/or deformation.
5 Gravitation Potential Energy Potential energy is associated with the configuration of a system in which a conservative force acts: U = W For a general conservative force F = F(x) Gravitational potential energy: assume U i = 0 at y i = 0 (reference point) U(y) = mgy (gravitational potential energy) only depends on vertical position
6 Nature only considers changes in Potential energy important. Thus, we will always measure U. So,... We need to set a reference point! The potential at that point could be defined to be zero (i.e., U ref. point = 0) in which case we drop the for convenience. BUT, it is always understood to be there!
7 Potential energy and reference point (y = 0 at ground) A 0.5 kg physics book falls from a table 1.0 m to the ground. What is U and U if we take reference point y = 0 (assume U = 0 at y = 0) at the ground? y U = U f U i = (mgh) Physics y = h 0 The book lost potential energy. Actually, it was converted into kinetic energy h y = 0
8 Potential energy and reference point ( y= 0 at table) A 0.5 kg physics book falls from a table 1.0 m to the ground. What is U and U if we take reference point y = 0 (assume U = 0 at y = 0) at the table? y U = U f U i = mg( h) Physics y = 0 0 h The book lost the same potential energy. (independent of y axis assignment) y = h
9 Sample problem 81: A block of cheese, m = 2 kg, slides along frictionless track from a to b, the cheese traveled a total distance of 2 m along the track, and a net vertical distance of 0.8 m. How much is W g? W net = b a F g d r But,... We don t know the angle between F and dr Easier (or another) way?
10 Sample problem 81: A block of cheese, m = 2 kg, slides along frictionless track from a to b, the cheese traveled a total distance of 2 m along the track, and a net vertical distance of 0.8 m. How much is W g? Split the problem into two parts (two paths) since gravity is conservative c }=d 0 mg(b c) = mgd
11 Elastic Potential Energy Spring force is also a conservative force F = kx U f U i = ½ kx f 2 ½ kx i 2 Choose the free end of the relaxed spring as the reference point: that is: U i = 0 at x i = 0 U = ½ kx 2 Elastic potential energy
12 Conservation of Mechanical Energy Mechanical energy E mec = K + U For an isolated system (no external forces), if there are only conservative forces causing energy transfer within the system. We know: K = W (workkinetic energy theorem) Also: U = W (definition of potential energy) Therefore: K + U = 0 (K f K i ) + (U f U i ) = 0 therefore K 1 + U 1 = K 2 + U 2 (States 1 and 2 of system) E mec,1 = E mec,2 the mechanical energy is conserved
13 Conservation of mechanical energy For an object moved by a spring in the presence of a gravitational force. This is an isolated system with only conservative forces ( F = mg, F = kx ) acting inside the system E mec,1 = E mec,2 K 1 + U 1 = K 2 + U 2 ½ mv mgy 1 + ½ kx 1 2 = ½ mv mgy 2 + ½ kx 2 2
14 The bowling ball pendulum demonstration Mechanical energy is conserved if there are only conservative forces acting on the system.
15 Pendulum (1D) vertically Find the maximum speed of the mass. Isolated system with only gravity (conservative force) acting on it. θ L y=0 Let 1 be at maximum height (y = y max ), Let 2 be at minimum height (y = 0), and 3 be somewhere between max and min. 3 y 1 mg 2 y = L(1 cosθ)
16 Pendulum (1D) vertically Where is the speed a max? L Isolated system with only gravity (conservative force) acting on it. 0 K 1 + U 1 = K 2 + U 2 = E total ½ mv mgy 1 = ½ mv mgy K max must be when U min (= 0) Lcosθ 3 y=0 y 2 L 1 mg In general, y = L(1 cosθ) Answer: 2
17 Pendulum (1D) vertically Find the maximum speed of the mass. Let 1 be at maximum height (y = y max ) and 2 be at minimum height (y = 0) E mech,1 = E mech,2 = E total y = L(1 cosθ) θ L K 1 + U 1 = K 2 + U 2 = E total 0 0 mgy max = ½ mv 2 max = E total y y=0 mg
18 Pendulum (1D) vertically Now find the speed of the mass as a function of angle. Let 1 be at maximum height (y = y max ) and 2 be at some height (y) E mech,1 = E mech,2 = E total y = L(1 cosθ) θ L K 1 + U 1 = K 2 + U 2 = E total mgy max = ½ mv 2 + mgy = E total y y=0 mg
19 Pendulum (1D) vertically Find the speed of the mass as a function of angle. Let 1 be at maximum height (y = y max ) and 2 be at some height (y) y θ L y=0 mg U K 0 θ max 0 +θ max 0 θ max θ
20 Pendulum (1D) vertically Find the speed of the mass as a function of angle. Let 1 be at maximum height (y = y max ) and 2 be at some height (y) y θ L y=0 mg E Sum of U + K is constant, =E K 0 θ max 0 +θ max 0 θ max θ U
21 Potential Energy Curve We know U(x) = W = F(x) x Therefore F(x) = du(x)/dx
22 Work done by external force When no friction acts within the system, the net work done by the external force equals to the change in mechanical energy W = E mec = K + U Friction is a nonconservative force When a kinetic friction force acts within the system, then the thermal energy of the system changes: E th = f k d Therefore W = E mec + E th
23 Work done by external force When there are nonconservative forces (like friction) acting on the system, the net work done by them equals to the change in mechanic energy W net = E mec = K + U W friction = f k d
24 Conservation of Energy The total energy E of a system can change only by amounts of energy that are transferred to or from the system W = E = E mec + E th + E int If there is no change in internal energy, but friction acts within the system: W = E mec + E th If there are only conservative forces acting within the system: W = E mec If we take only the single object as the system W = K
25 Law of Conservation of Energy For an isolated system (W = 0), the total energy E of the system cannot change E mec + E th + E int = 0 For an isolated system with only conservative forces, E th and E int are both zero. Therefore: E mec = 0
26 Chapter 8: Potential Energy and The Conservation of Total Energy Work and kinetic energy are energies of motion. Potential energy is an energy that depends on location. 1Dimension 3Dimensions
27 Force Potential Energy Relationship F(x) = du(x)/dx Thus, the force in the xdirection is the negative derivative of the potential energy! The same holds true for y and zdirections.
28 Potential Energy Curve Thus, what causes a force is the variation of the potential energy function, i.e., the force is the negative 3D derivative of the potential energy!
29 Potential Energy Curve We know: Therefore: U(x) = W = F(x) x F(x) = du(x)/dx Now integrate along the displacement: = Rearrange terms:
30 Conservation of Mechanical Energy Holds only for an isolated system (no external forces) and if only conservative forces are causing energy transfer between kinetic and potential energies within the system. Mechanical energy: E mec = K + U We know: K = W (workkinetic energy theorem) U = W (definition of potential energy) Therefore: K + U = 0 (K f K i ) + (U f U i ) = 0 Rearranging terms: K f + U f = K i + U i = K 2 + U 2 = E mec (a constant)
31 A x=0 A Horizontal spring with mass oscillating with maximum amplitude x max = A. At which displacement(s) would the kinetic energy equal the potential energy? 5) none of the above
32 U f U i = ½ kx f 2 ½ kx i 2 U i = 0 at x i = 0; U max = ½ kx max 2 Since K = U Horizontal spring with mass oscillating with maximum amplitude x max = A. At which displacement(s) would the kinetic energy equal the potential energy? A x=0 A 5) none of the above
33 Conservation of mechanical energy For an isolated system with only conservative forces (e.g., F = mg and F = kx) acting on the system: E mec,1 = E mec,2 = E total => K 1 + U 1 = K 2 + U 2 = E total ½ mv mgx 1 + ½ kx 1 2 = ½ mv mgx 2 + ½ kx 2 2 = E total Initial mechanical energy Final mechanical energy
34 Horizontal Spring x=0 Isolated system with only conservative forces acting on it. (e.g., ) E mech,1 = E mech,2 = E total K 1 + U 1 = K 2 + U 2 = E total ½ mv ½ kx 1 2 = ½ mv ½ kx 2 2 = E total
35 Work done by Spring Force Spring force is a conservative force Work done by the spring force: If x f > x i (further away from equilibrium position); W s < 0 My hand did positive work, while the spring did negative work so the total work on the object = 0
36 Work done by Spring Force Spring force is a conservative force Work done by the spring force: If x f < x i (closer to equilibrium position); W s > 0 My hand did negative work, while the spring did positive work so the total work on the object = 0
37 Work done by Spring Force  Summary Spring force is a conservative force Work done by the spring force: If x f > x i (further away from equilibrium position); W s < 0 If x f < x i (closer to equilibrium position); W s > 0 Let x i = 0, x f = x then W s = ½ k x 2
38 Elastic Potential Energy Spring force is a conservative force Choose the free end of the relaxed spring as the reference point: U i = 0 at x i = 0 The work went into potential energy, since the speeds are zero before and after.
39 Vertical Spring with mass m Consider that mass m was held so the spring was relaxed (y 1 =0) and then slowly + let down to the equilibrium position. y 1 =0 Find this equilibrium position. y= y Equilibrium position (F g + F s = 0): 0 F g = mg F s = ky 0 (now the spring is not relaxed!) => y 0 = mg/k (substitute for y o ) y 2 = A ky mg K e = 0, U g,e = mgy 0 = mg( mg/k) = (mg) 2 /k, U s,e = ½ ky 0 2 = ½ k ( mg/k) 2 = ½ (mg) 2 /k
40 Vertical Spring with mass m Now consider that the mass m was dropped from y 1 =0 + Find maximum amplitude A. y 1 =0 Initial position (y 1 = 0): ky y= y K 1 = 0, U g,1 = 0, U s,1 = ½ ky 2 = 0 0 => E mech,1 = 0 Choose = 0 Since y 1 = 0 mg y Final position (y 2 = A) : 2 = A K 2 = 0, U g,2 = mgy 2 = mg( A) = mga, U s,2 = ½ ky 2 2 = ½ ka 2 E mech,1 =E mech,2 = 0 = U g,2 +U s,2 = mga + ½ ka 2 => A = 2mg/k (= 2y 0 )
41 Vertical Spring with mass m Now consider that the mass m was dropped from y 1 =0 Where does the max speed occur? Maximum speed occurs when the potential energy is a minimum. For an Arbitrary position: y 2 = A E mech = 0 = K + U g +U s =½ mv 2 + mgy + ½ ky 2 + y 1 =0 y= y 0 ky mg du/dy = d(mgy + ½ ky 2 )/dy = 0 => mg + ky = 0 => y = mg/k which is y o (from 2 slides ago) => maximum speed is at equilibrium point (as expected)
42 Vertical Spring with mass m Now consider that the mass m was dropped from y 1 =0 What is the max speed? Maximum speed position: 0 = K max + U g + U s = ½ mv 2 + mg( mg/k) + ½ k( mg/k) 2 + y 1 =0 y= y 0 y 2 = A ky mg = ½ mv 2 (mg) 2 /k + ½ k(mg/k) 2 = ½ mv 2 ½ (mg) 2 /k => v 2 = mg 2 /k
43 Vertical Spring with mass m Now consider that the mass m was dropped from y 1 =0 Find the mechanical energy of the mass at the equilibrium point. Kinetic energy at y = y 0 : K y0 = ½ (mg) 2 /k + y 1 =0 y= y 0 y 2 = A Potential energy at y = y 0 : (U g +U s ) y = y0 = mg( y 0 ) + ½ k( y 0 ) 2 = mgy 0 + ½k(y 0 ) 2 = mg(mg/k) + ½k(mg/k) 2 = ½(mg) 2 /k Mechanical energy at y = y 0 : E mech = K y0 + (U g +U s ) y = y0 = ½ (mg) 2 /k ½(mg) 2 /k = 0 ky mg
44 Vertical Spring with mass m When we slowly let the mass relax to y 0, the mechanical energy is: E mech = K y0 + (U g +U s ) y = y0 0 => E mech = ½(mg) 2 /k (mg) 2 /k ½ (mg) 2 /k But when we let the mass drop, the mechanical energy at y = y 0 is E mech = 0 + y 1 =0 y= y 0 y 2 = A ky mg Where did the missing energy (½(mg) 2 /k) go? Answer: Our hand provided a nonconservative force on m!
45 Work done by external force When no friction acts within the system, the net work done by an external force from outside the system equals the change in mechanical energy of the system W net = E mec = K + U Friction is a nonconservative force that opposes motion Work done by friction is: W friction = f k d When a kinetic friction force acts within the system, then the thermal energy of the system changes: E th = f k d Therefore W = E mec + E th
46 Conservation of Total Energy The total energy E of a system can change only by an amount of energy that is transferred to or from the system. W = E = E mec + E th + E int Here, E int are energy changes due to other nonconservative internal forces. If there is no internal energy change, but friction acts within the system: W = E mec + E th If there are only conservative forces acting within the system: W = E mec
47 Isolated Systems For an isolated system (no work done on the system, W = 0), the total energy E of the system cannot change E mec + E th + E int = 0 For an isolated system with only conservative forces, E th and E int are both zero. Therefore: E mec = 0
48 Law of Conservation of Total Energy in an Isolated System Rearrange terms.
49 Law of Conservation of Energy Count up the initial energy in all of its forms. Count up the final energy in all of its forms. These two must be equal (if nothing is added form outside the system).
50 Sample Problem 86 A wooden crate of m = 14kg is pushed along a horizontal floor with a constant force of F = 40N for a total distance of d = 0.5m, during which the crate s speed decreased from v o = 0.60 m/s to v = 0.20m/s. A) Find the work done by F. W = Fd cosθ = (40N)(0.50m)cos0 o = 20J B) Find the increase in thermal energy. (Consider the work done on the block/spring system) W = E mec + E thermal = 20J = ½mv 2 ½mv o 2 + E thermal => E thermal = W E mec =20J (½mv 2 ½mv o2 ) = 22.2J
51 Sample Problem 87 In the figure, a 2.0kg package slides along a floor with speed v 1 = 4.0m/s. It then runs into and compresses a spring, until the package momentarily stops. Its path to the initially relaxed spring is frictionless, but as it compresses the spring, a kinetic friction force from the floor, of magnitude 15 N, acts on it. The spring constant is 10,000 N/m. By what distance d is the spring compressed when the package stops?
52 The net change in energy in the system must equal zero. Initial mechanical energy, (The spring is relaxed) Final mechanical energy, (The mass is stopped)
53 The change in mechanical energy must equal the energy converted to thermal energy. Thus, a quadratic equation in d with: m = 2.0 kg, v 1 = 4.0 m/s, f k = 15 N, and k = 10,000 N/m
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