Chapter 8. Potential Energy and Conservation of Energy


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1 Chapter 8 Potental Energy and Conservaton of Energy In ths chapter we wll ntroduce the followng concepts: Potental Energy Conservatve and nonconservatve forces Mechancal Energy Conservaton of Mechancal Energy The conservaton of energy theorem wll be used to solve a varety of problems As was done n Chapter 7 we use scalars such as work,knetc energy, and mechancal energy rather than vectors. Therefore the approach s mathematcally smpler. (81)
2 h v o B A v o g Work and Potental Energy: Consder the tomato of mass m shown n the fgure. The tomato s taken together wth the earth as the system we wsh to study. The tomato s thrown upwards wth ntal speed v o at pont A. Under the acton of the gravtatonal force t slows down and stops completely at pont B. Then the tomato falls back and by the tme t reaches pont A ts speed has reached the orgnal value v o. Below we analyze n detal what happens to the tomatoearth system. Durng the trp from A to B the gravtatonal force F g does negatve work W 1 = mgh. Energy s transferred by F g from the knetc energy of the tomato to the gravtatonal potental energy U of the tomatoearth system. Durng the trp from B to A the transfer s reversed. The work W 2 done by F g s postve ( W 2 = mgh ). The gravtatonal force transfers energy from the gravtatonal potental energy U of the tomatoearth system to the knetc energy of the tomato. The change n the potental energy U s defned as: U = W (82)
3 A A m B k B Consder the mass m attached to a sprng of sprng constant k as shown n the fgure. The mass s taken together wth the sprng as the system we wsh to study. The mass s gven an ntal speed v o at pont A. Under the acton of the sprng force t slows down and stops completely at pont B whch corresponds to a sprng compresson. Then the mass reverses the drecton of ts moton and by the tme t reaches pont A ts speed has reached the orgnal value v o. As n the prevous eample we analyze n detal what happens to the masssprng system. Durng the trp from A to B the sprng force F s does negatve work W 1 = k 2 /2. Energy s transferred by F s from the knetc energy of the mass to the potental energy U of the masssprng system. Durng the trp from B to A the transfer s reversed. The work W 2 done by F s s postve ( W 2 = k 2 /2 ). The sprng force transfers energy from the potental energy U of the masssprng system to the knetc energy of the mass. The change n the potental energy U s defned as: U = W (83)
4 f k A m v o f k d B m Conservatve and nonconservatve forces. The gravtatonal force as the sprng force are called conservatve because the can transfer energy from the knetc energy of part of the system to potental energy and vce versa. Frctonal and drag forces on the other hand are called nonconservatve for reasons that are eplaned below. Consder a system that conssts of a block of mass m and the floor on whch t rests. The block starts to move on a horzontal floor wth ntal speed v o at pont A. The coeffcent of knetc frcton between the floor and the block s μ k. The block wll slow down by the knetc frcton f k and wll stop at pont B after t has traveled a dstance d. Durng the trp from pont A to pont B the frctonal force has done work W f =  μ k mgd. The frctonal force transfers energy from the knetc energy of the block to a type of energy called thermal energy. Ths energy transfer cannot be reversed. Thermal energy cannot be transferred back to knetc energy of the block by the knetc frcton. Ths s the hallmark of nonconservatve forces. (84)
5 (85) Path Independence of Conservatve Forces In ths secton we wll gve a test that wll help us decde whether a force s conservatve or nonconservatve. A force s conservatve f the net work done on a partcle durng a round trp s always equal to zero (see fg.b). W net = Such a round trp along a closed path s shown n fg.b. In the eamples of the tomatoearth and masssprng system W net = W ab,1 + W ba,2 = 0 We shall prove that f a force s conservatve then the work done on a partcle between two ponts a and b does not depend on the path. From fg. b we have: W net = W ab,1 + W ba,2 = 0 W ab,1 =  W ba,2 (eqs.1) 0 From fg.a we have: W ab,2 =  W ba,2 (eqs.2) If we compare eqs.1 and eqs.2 we get: W = W ab,1 ab,2
6 ... O F() f Determnng Potental Energy Values: In ths secton we wll dscuss a method that can be used to determne the dfference n potental energy U of a conservatve force F between ponts f and on the as f we know F( ) A conservatve force F moves an object along the as from an ntal pont to a fnal pont. The work W that the force F does on the object s gven by : W f = F( ) d The correspondng change n potental energy U was defned as: f U = W Therefore the epresson for U becomes: U f = F( ) d (86)
7 dy mg Gravtatonal Potental Consder a partcle of mass m movng vertcally along the yas ( ) from pont y to pont y. At the same tme the gravtatonal force does work W energy: f on the partcle whch changes the potental energy of the partcleearth system. We use the result of the prevous secton to calculate U ( ) [ ] U = F( ) dy F = mg U = mg dy = mg dy = mg y U = mg y y = mg y We assgn the fnal pont y to be the "generc" ( ) pont y on the yas whose potental s U ( y). U ( y) U = mg y y Snce y y y f f f y y y f f only changes n the potental are physcally menangful, ths allows us to defne arbtrarly y and U The most convenent choce s: y y... y f m y y O = 0, U = 0 Ths partcular choce gves: y y f U ( y) = mgy (87)
8 O O O (a) (b) f (c) 2 f k U = k 2 = f k We assgn the fnal pont to be the "generc" 2 2 k k pont on the as whose potental s U( ). U( ) U = 2 2 Snce only changes n the potental are physcally menangful, ths allows us to defne arbtrarly and U The most convenent choce s: y = 0, U = 0 Ths partcular choce gves: Potental Energy of a sprng: Consder the blockmass system shown n the fgure. The block moves from pont to pont. At the same tme the sprng force does work W on the block whch change s the potental energy of the blocksprng system by an amount f f f W = F( ) d = kd = k d U = W U = f f k 2 2 (88)
9 Conservaton of Mechancal Energy: Mechancal energy of a system s defned as the sum of potental and knetc energes Emech = K + U We assume that the system s solated.e. no eternal forces change the energy of the system. We also assume that all the forces n the system are conservatve. When an nteral force does work W on an object of the system ths changes the knetc energy by K = W (eqs.1) Ths amount of work also changes the potental energy of the system by an amount If we compare equatons 1 and 2 we have: K ( ) 1 K = U U K + U = K + U K = U Ths U = W equaton s known as the prncple of conservaton of mechancal energy. It can be summarzed as: Emech = K + U = 0 For an solated system n whch the forces are a mture of conservatve and non conservatve forces the prncple takes the followng form E = W mech nc (89) (eqs.2) Here, s defned as the work of all the nonconsrvatve W nc force s of the system
10 (810) An eample of the prncple of conservaton of mechancal energy s gven n the fgure. It conssts of a pendulum bob of mass m movng under the acton of the gravtatonal force The total mechancal energy of the bobearth system remans constant. As the pendulum swngs, the total energy E s transferred back and forth between knetc energy K of the bob and potental energy U of the bobearth system We assume that U s zero at the lowest pont of the pendulum orbt. K s mamum n frame a, and e (U s mnmum there). U s mamum n frames c and g (K s mnmum there)
11 ... O A F B + Δ Fndng the Force F( ) analytcally from the potental energy U ( ) Consder an object that moves along the as under the nfluence of an unknown force F whose potental energy U() we know at all ponts of the as. The object moves from pont A (coordnate ) to a close by pont B (coordnate + ). The force does work W on the object gven by the equaton: W = F eqs.1 The work of the force changes the potental energy U of the system by the amount: U = W eqs.2 If we combne equatons 1 and 2 we get: U F = We take the lmt as 0 and we end up wth the equaton: F( ) = du ( ) d (811)
12 (812) The potental Energy Curve If we plot the potental energy U versus for a force F that acts along the as we can glean a wealth of nformaton about the moton of a partcle on whch F s actng. The frst parameter that we can determne s the force F() usng the equaton: F( ) = du ( ) d An eample s gven n the fgures below. In fg.a we plot U() versus. In fg.b we plot F() versus. For eample at 2, 3 and 4 the slope of the U() vs curve s zero, thus F = 0. The slope du/d between 3 and 4 s negatve; Thus F > 0 for the ths nterval. The slope du/d between 2 and 3 s postve; Thus F < 0 for the same nterval
13 (813) Turnng Ponts: The total mechancal energy s E = K( ) + U ( ) Ths energy s constant (equal to 5 J n the fgure) and s thus represented by a horzontal lne. We can slolve ths equaton for mec K( ) and get: K( ) = E U ( ) At any pont on the as mec we can read the value of the equaton above and determne U ( ). Then we can solve 2 mv From the defnton of K = the knetc energy cannot be negatve. 2 Ths property of K allows us to determne whch regons of the as moton s allowed. K ( ) = K ( ) = E U ( ) If K > 0 E U ( ) > 0 U ( ) < mech If K < 0 E U ( ) < 0 U ( ) > mech The ponts at whch: E mec for the moton. For eample 1 mec E mec mec = U ( ) are known as turnng ponts E s the turnng K Moton s allowed Moton s forbdden pont U versus plot above. At the turnng pont K = 0 for the
14 (814) Gven the U() versus curve the turnng ponts and the regons for whch moton s allowed depends on the value of the mechancal energy E mec In the pcture to the left consder the stuaton when E mec = 4 J (purple lne) The turnng ponts (E mec = U ) occur at 1 and > 5. Moton s allowed for > 1 If we reduce E mec to 3 J or 1 J the turnng ponts and regons of allowed moton change accordngly. Equlbrum Ponts: A poston at whch the slope du/d = 0 and thus F = 0 s called an equlbrum pont. A regon for whch F = 0 such as the regon > 5 s called a regon of neutral equlbrum. If we set E mec = 4 J the knetc energy K = 0 and any partcle movng under the nfluence of U wll be statonary at any pont wth > 5 Mnma n the U versus curve are postons of stable equlbrum Mama n the U versus curve are postons of unstable equlbrum
15 (815) Note: The blue arrows n the fgure ndcate the drecton of the force F as determned from the equaton: F( ) = du ( ) d Postons of Stable Equlbrum. An eample s pont 4 where U has a mnmum. If we arrange E mec = 1 J then K = 0 at pont 4. A partcle wth E mec = 1 J s statonary at 4. If we dsplace slghtly the partcle ether to the rght or to the left of 4 the force tends to brng t back to the equlbrum poston. Ths equlbrum s stable. Postons of Unstable Equlbrum. An eample s pont 3 where U has a mamum. If we arrange E mec = 3 J then K = 0 at pont 3. A partcle wth E mec = 3 J s statonary at 3. If we dsplace slghtly the partcle ether to the rght or to the left of 3 the force tends to take t further away from the equlbrum poston. Ths equlbrum s unstable
16 Work done on a System by an Eternal Force Up to ths pont we have consdered only solated systems n whch no eternal forces were present. We wll now consder a system n whch there are forces eternal to the system The system under study s a bowlng ball beng hurled by a player. The system conssts of the ball and the earth taken together. The force eerted on the ball by the player s an eternal force. In ths case the mechancal energy E mec of the system s not constant. Instead t changes by an amount equal to the work W done by the eternal force accordng to the equaton: W = Emec = K + U (816)
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