1 1. Why we introduce the concepts of work and energy? CWE theoremamF r r = total CWE theorem Conservation of energy (More universal) 2. About the systems that we discuss Ignore the size, internal structure, internal motion, deformations, and thermal effects. §8.1 Work done by variable force 2 SFSFW vv ?=??= θcos θ F v s v F v 3. The work done by a constant force sF rF rFW dcos cosd dd θ θ = ??= ?= v v v differential work a b o F v r v d sd r v r′ v θ F r 4. The work done by variable force §8.1 Work done by variable force §8.1 Work done by variable force Define the differential work done by any force is rFW r r dd ?= The work done by a particular force is F r ∫∫ ?== f i rFW r r ddW In Cartesian coordinate system: )ddd( ) ? d ? d ? d( ) ??? (W zFyFxF kzjyix kFjFiF zy r r x zy r r x f i f i ++= ++ ?++= ∫ ∫ r r r r a b o F v r v d sd r v r′ v θ F r 3 ∫∫∫ ∫ ++= ++= f i f i f i f i z z z y y y x x x zy r r x zFyFxF zFyFxF ddd )ddd(W r r Work done by a constant force: kzjyixr kzjyixr ffff iiii ??? ??? ++= ++= r r )()()( dddW ifzifyifx z z z y y y x x x zzFyyFxxF zFyFxF f i f i f i ?+?+?= ++= ∫∫∫ )( if rrFrFW rr r r r ??=?= ? §8.1 Work done by variable force The properties of the work: 1work is a scalar quantity; 2the work done by a force depends on the path followed by the system; §8.1 Work done by variable force sfisifrfrfW kkk B A k ?=??=?=?= ∫ ) ? () ? (d r r r r ? sfisifrfrfW kkk A B k ?=??=?=?= ∫ ) ? () ? (d r r r r ? A B s x y sfsfsfWWW kkk 2 pathclsd ?=??= ′ += 0,0,0 >=< AAA 4 3the work done by a force depends on the choice of the reference frame. Elevator:W=0, Earth: W≠0 y g W r v r 0=+ ′NN WW N r ′ c v N r v m c f r ′ s s ′ M 0<+ ′ff WW f r m §8.1 Work done by variable force 4the work done by a pair of forces is not always zero. 5. The work done by the total force The work done by the total force acting on the system is the algebraic, scalar sum of the work done by the individual forces. §8.1 Work done by variable force L L r r r r r L rr r r ++= +?+?= ?++=?= ∫∫ ∫∫ 21 21 21totaltotal dd d)(d WW rFrF rFFrFW f i f i f i 5 §8.1 Work done by variable force 6. The the geometric interpretation ∫∫∫ ++= f i f i f i z z z y y y x x x zFyFxF dddW The work done on the system by the force component as the system moves from to is the area under the curves of the graph of that force components versus the corresponding coordinate. i r r f r r Positive or negative work (area) depends on both the sign of the force and the direction of . r r ? §8.1 Work done by variable force 6 Example 1: A spring hangs vertically in its relaxed state. A block of mass m is attached to the spring, but the block is held in place so that the spring at first does not stretch. Now the hand holding block is slowly lowered, so that the block descends at constant speed until it reaches the point at which it hangs at equilibrium with the hand removed. At this point the spring is measured to have stretched a distance d from its previous relaxed length. Find the work done on the block in this process by (a)gravity, (b)the spring, (c) the hand. §8.1 Work done by variable force Solution: (a)gravity—a constant force mgdjdjmgrFW gg =???=?= ? )0( ? r r ? (b)the spring—a variable force mgdkdjyjkyjyFW ss 2 1 2 1 ? d ?? d 2 -d 0 -d 0 ?=?=??=?= ∫∫ r jkyF s ? ?= r d mg kmgkd == (c)the hand —a variable force jmgjkyFFF FFF gsh ghs ?? 0 +=??= =++ rrr rrv gm r s F r h F ry §8.1 Work done by variable force 7 mgdgmdmgd ymgkyjyFW dd hh 2 1 2 1 )d( ? d 00 ?=?= +=?= ∫∫ ?? r 0=++ ghs WWW Can you explain? gm r s F r h F r y §8.1 Work done by variable force Example 2: A small object of mass m is suspended from a string of length L. the object is pulled sideways by a force F that is always horizontal, until the string finally makes an angle φ m with the vertical. The displacement is accomplished at a small constant speed. Find the work done by all forces that act on the object. §8.1 Work done by variable force 8 Solution: From the force diagram φtanmgF = Eliminate T, we find )cos-(1dsin )sind(tandtan dd m 0 φφφ φφφ φ mgLmgL Lmgxmg xFrFW m f i f i f i f i F == == =?= ∫ ∫∫ ∫∫ r r x component: y component: 0cos 0sin =? =? mgT TF φ φ §8.1 Work done by variable force )cos( d d ? 0d cos 0 m LL f i g T LLmg ymg rjmgW rTW m φ φ ??= ?= ??= =?= ∫ ∫ ∫ ? r r r 0=++= gTF WWWW §8.1 Work done by variable force r r d 9 M F k S Example 3: As shown in figure, M=2kg , k =200N·m -1 , S=0.2m , g ≈ 10m·s -2 . A student pull the string very slowly, ignore the friction between the pulley and the string , as well as the masses of the pulley and string . At initial time, the spring is in the relaxed state. Find the work done by the force . F r §8.1 Work done by variable force Solution: m2.0and m1.0 00 ==→= SxMgkxQ 0.1mbewilltheofheightthe M∴ When the string is pulled down 0.2 m, what will be the height of M? F= k x (0<x≤0.1m) a variable force k x 0 =Mg (0.1<x≤0.2m) a constant force ()J3|| dd 2.0 1.0 1.0 0 2 2 1 2.0 1.0 1.0 0 =+= += ∫∫ Mgxkx xMgxkxW F then §8.1 Work done by variable force 10 §8.2 Conservative force and the potential energy 1. What is conservative force If a work done by the force on any system as it moves between two points is independent of the path following by the system between the two points; or if the work done by the force on any system around any closed path is zero. Then the force is called conservative force. 0d =? ∫ rF r r ∫∫ ?=? b a b a rFrF )2()1( dd r r r r b 0d ≠? ∫ rF r r If , then the force is called nonconservative force. If , then the force is called zero-work force. 0d =? ∫ rF r r 2. Some examples of conservative forces 1local gravitational force rFW r r ??= jmgF ? ?= r --a constant force §8.2 Conservative force and the potential energy 11 y a 2 L 1 L f y i y r′ r r r r r d gm r b o z x yd )( d ) ? d ? d ? d() ? ( d f i b a b a if y y mgymgy ymg kzjyixjmg rFW ??= ?= ++??= ?= ∫ ∫ ∫ r r )( ] ? )( ? )( ? )[( ? if ififif mgymgy kzzjyyixxjmg rFW ??= ?+?+???= ??= r r or §8.2 Conservative force and the potential energy 2the gravitational force r r GMm F ? 2 ?= r )]()[( d cosd d ? d 2 2 2 if r r b a b a b a r mM G r mM G r r mM G r r mM G rr r mM GrFW f i ????= ?= ?= ??=?= ∫ ∫ ∫∫ θ r rr r r r rd r ′ r a b M L i r r f r r F r r r d or ∫∫ ?=??= f i r r b a r r mM Grrr r mM GW d ? d ? 22 §8.2 Conservative force and the potential energy 12 3spring force i x f x 0 ixr ikxF ? dd ? = ?= r r ) 22 (d ? d) ? ( 2 2 i f x x x x x k x kxxkixikxW f i f i ??=?=??= ∫∫ 3. Potential energy To exploit the position dependence of the work done by conservative force on a system, we introduce the concept of the potential energy. §8.2 Conservative force and the potential energy define PEPEPEW if ??=??= )( consrv 1gravitational potential energy near the earth 2gravitational potential energy §8.2 Conservative force and the potential energy )()( local ifif PEPEmgymgyW ??=??= mgyPE = local y=0, PE local =0. )]()[( if r mM G r mM GW ????= r mM GPE ?= grav r=∞, PE grav =0. 13 3 potential energy of spring force system )( 22 2 2 if i f PEPE x k x kW ??=+?= §8.2 Conservative force and the potential energy Note: 1Potential energy is the function of position; 2the choice of the position where the potential energy is zero is discretional. 3Potential energy is belong to the two particles of interaction. ④potential energy is the work done by conservative force from some position to the zero potential energy point. 2 2 spring x kPE = x=0, PE spring =0. §8.2 Conservative force and the potential energy 2 d ? d) ? ( 2 0 i x x x x kxxkixikxW i f i =?=??= ∫∫ Potential energy of spring force system: r mM Gr r mM G r r mM Grrr r mM GW i f i r r r b a ?=?= ?=??= ∫ ∫∫ ∞ d d ? d ? 2 22 Gravitational potential energy: i y y y mgyymgymgrFW =?=?=?= ∫∫∫ 0b a i f i ddd r r Gravitational potential energy near the earth: 14 Example : A uniform chain of mass m and length l is put on a frictionless horizontal surface.The length of hanging down part is 0.2 l ，pull the chain very slowly back to the surface. Find the work done by the pulling force. 0.8 l 0.2 l 0=μ m x F §8.2 Conservative force and the potential energy Solution 1: Work done by a variable force g l mx G = 50 d ? d ? d 0 2.0 mgl xx l mg ixi l mg rGW l G ?==?=?= ∫∫∫ r r 50 mgl WW GF =?= 0.8 l 0.2 l 0=μ m x F The mass of the hanging down part x l m §8.2 Conservative force and the potential energy 15 Solution 2: On the surface Initial state: Final state: Work done by mg: Work done by F: 0=PE ( ) 501055 1 mgl l mg c mg hPE ?=?== 0 2 =PE () 50 12 mgl PEPEPEW G ?=??=??= 50 mgl WW GF =?= the potential energy 0.8 l 0.2 l 0=μ m x F §8.2 Conservative force and the potential energy §8.2 Conservative force and the potential energy 5. Conservative force and the potential energy Simple harmonic oscillation: x PE F PExFixiFrFW x xx d )d( )(dd ? d ? dd spring ?= ?==?=?= r r The gravitational force: r PE F PErFrrrFrFW r rr d )d( )(dd ? d ? dd ?= ?==?=?= r r Generalization: k z PE j y PE i x PE kFjFiFF zyx ? )( ? )( ? )( ??? ? ? ? ? ? ? ? ? ?=++= r 16 §8.3 The CWE theorem 1. Work and kinetic energy For a single particle: ∫∫∫ ?=?=?= === f i f i vvmr t v mrFW t r v t v mamF rrr r r r r r r r r dd d d d d d , d d totaltotal total 222 total 2 1 2 1 d 2 1 )(d 2 1 )(d 2 1 d d d 2 d d d d d )(d if f i v v mvmvvmvvmW vvvv v t v t v vv t v t vv f i ?==?=∴ ?=? ?=?+?= ? ∫∫ rr rrrr r rr rr rrr Q Kinetic energy is defined : 2 2 1 mvKE = 1Kinetic energy is a scalar; 2Kinetic energy never be negative; 3this definition holds true only at v<<c; 4 the SI unit for kinetic energy is J. Note: For a particle-group: ∑∑∑ ?== nn nn n n ninf vmvmWW 22 total,total 2 1 2 1 §8.3 The CWE theorem 17 2. The CWE theorem ∑∑∑ ?== nn nn n n ninf vmvmWW 22 total,total 2 1 2 1 22 total 2 1 2 1 if mvmvW ?= KEKEKEW if ?=?= total The effect of the work done by all forces acting on a particle or a particle-group is that to change the value of the kinetic energy property of the particle or the particle-group. §8.3 The CWE theorem Conclusions: 1work is the measure of the energy transfer. If nonzero total work is done on a system, the kinetic energy of the system changes, it means that the energy is transferred to the system. 2The CWE theorem of a particle is the scalar equivalent of the second law on the particle. 3. The total mechanical energy and the conservation of mechanical energy forces Conservative force Noncoservative force §8.3 The CWE theorem 18 EPEKEPEKE PEPEKEKEPEKEW PEW KEWW iiff ifif ?=+?+= ?+?=?+?= ??= ?=+ )()( )()( other consrv consrvother 4. Conservation law of mechanical energy PEKEE += mechanical energy: a. If the forces acting on the system are all conservative forces; b. If the work done by all “others” forces acting on the system is zero. ttancons=+==+= fffiii PEKEEPEKEE §8.3 The CWE theorem （1）The kinetic energy of the satellite； （2）The gravitational potential energy； （3）The total mechanical energy。 O r F 2R R M m §8.3 The CWE theorem Example 1: A man-made satellite of mass m moves around a circular path above the surface of the earth 2R height (v<<c). In terms of m, R, G, and M, present 19 Solution： ,cv << nonrelative () R GmM mvKE R v m R mM G 62 1 3 3 2 2 2 == =① ② R mM Gr r mM GPE R 3 d 3 2 ?=?= ∫ ∞ ③ R GMm PEKEE 6 ?=+= Confined in the gravitational field of the Earth O r F 2R R M m §8.3 The CWE theorem 思考:卫星对接问题 a b a R b R 设飞船a 、b 圆轨道在同一平面内，飞船a 要追 上b并与之对接，能否直接加速？ R GMm PEKEE 6 ?=+= 加速，发动机做功，ΔE＞0, 轨道半径R增大,不能对接; 方法: a 减速 ΔE<0 R减小 R C 轨道 加速 R b 轨道 §8.3 The CWE theorem 20 a b a R b R c c R 方法: a 减速 ΔE<0 R减小 R C 轨道 加速 R b 轨道 §8.3 The CWE theorem §8.3 The CWE theorem Example 2: P 346 8.18 The mass of the system m, the height h 0 , the spring constant k are given. Find the maximum extent to which the spring is compressed. y x 0 h o 21 Solution: 1The process Descent down Encounter with the spring 1 N r W r W r s F r 2 N r )()( other iiff PEKEPEKEEW +?+=?= According to the CWE theorem: 2a single application of CWE theorem §8.3 The CWE theorem 0)()( other =+?+=?= iiff PEKEPEKEEW Initial state: 0 0 mghPEKE ii +=+ Final state: 2 2 1 0 kxPEKE ff +=+ 0 2 1 0 2 =?mghkx Therefore k mgh x k mgh x 00 22 ?=?±= The solution is §8.3 The CWE theorem 22 Example 3: As shown in figure, on a thin staff of mass m 1 and length L, there is a small ring of mass m 2 which is connected with a light spring, the spring constant is k. At initial time, the system is static, under the application of an external force the system rotates at constant angular speed , the m 2 slide very slowly to the end of the staff A, at this instant, the kinetic energy of the staff is ω §8.3 The CWE theorem k m 2 m 1 L A o 22 1 3 1 ωLm Find the work done by the external force . KEWW ?=+ consvother 2 0 consv )( 2 1 d xkxkxWW x s ??=?== ∫ ? §8.3 The CWE theorem k m 2 m 1 L A o () 22 2 22 1 2 2 1 other 2 1 6 1 ωω LmLmxkW +=?? Lmxk 2 2 ω=? Solution: Solve the equations! 23 1. Energy diagram x PE 0 mgh h = 0 2 2 1 kx x = 0 r mM G? r = ∞ h PE 0 r PE 0 force Local gravity spring gravity PE PE=0 Energy diagram §8.4 energy diagram 2. Bound (confined) state of a system (a) Simple harmonic oscillator: )(sin 2 1 )( 2 1 )( )(cos 2 1 )( 2 1 )( )sin()( )cos()( 2222 222 φωω φω φωω φω +== +== +?= += tmAtmvtKE tkAtkxtPE tAtv tAtx x x 2 2 2 1 )()( kAtKEtPEE m k =+= =ω §8.4 energy diagram 24 )()( tKEtPE + )(tPE )(tKE )()( tKEtPE + )(tPE )(tKE A+A? PEE PEE m v PEmvE ≥ ≥?= += 0)]( 2 [ 2 1 2 2 x PE 0 PE KE A+A? §8.4 energy diagram x >A, v 2 <0, forbidden; x <-A, v 2 <0, forbidden; -A≤x ≤A, KE PE r PE 0 0<E 0>E No turning point ? Turning point x PE 0 PE KE A+A? ?? Turning point Turning point are called unbounded particles 0≥E , the motion is bounded 0<E 0>?= PEEKE (b)gravitational field §8.4 energy diagram 25 Example : A particle of mass m is in the one-dimension potential field. m7 ms2 kg2 0 1 0 = ?= = ? x iv m v find： ①the range of motion of the particle ②in which area F>0 ③in which area v=v max =? x (m) 2 PE (J) 4 -4 01 4 79 mv 0 r §8.4 energy diagram Solution： ①Initial state ()J4 2 02 1 == += mv PEKEE J4=E Draw the line x (m) 2 PE(J) 4 -4 01 4 79 mv 0 r J4=E ? §8.4 energy diagram 1≥x The intersection point at x=1, We have the range of the motion : 26 ③At x = 4 m , the potential energy is least and the kinetic energy is the most large. () EmvPE =+ 2 maxmin 2 1 ②0 d d >?= x PE FIf one wish 0 d d < x PE then That means the tangential of the potential curve must be negative. 941 ><< xx Then we have x (m) 2 PE (J) 4 -4 01 4 79 mv 0 r §8.4 energy diagram ( ) 1 max ms82.222 ? ≈=v () min 2 max 2 1 PEEmv ?= ( ) ( )J844 =??= §8.4 energy diagram 27 0>?= PEEKE 0)]( 2 [ 2 1 21 2 ≥?= += PEE m v PEmvE EPE, EPE, PE J0.5=E 2 at J0.5 xKE = 5 atJ0.1 xxKE >= J0.1=E J0.3=E J0.4=E §8.4 energy diagram §8.5 The escape speed and black holes 1. The escape speed The escape speed is the minimum speed needed to project a mass m upward from the surface of another mass M so that m never returns. R M m escape v r 0)()( other =+?+=?= if PEKEPEKEEW 00 2 1 2 escape =?== ?== f ff ii r GMm PEKE R GMm PEmvKE R GM v 2 escape = 28 §8.5 The escape speed and black holes 2. The relative escape speed R GM v 2 escape = ⊕ ⊕ ⊕ = R GM v 2 The escape speed from the Earth 21 escape ) ratioradius ratiomass ( / / == ⊕ ⊕ ⊕ RR MM v v §8.5 The escape speed and black holes 3. Black holes If the mass M is compressed to a small size that the escape speed exceeds the speed of light, the mass M form a black hole. The critical radius (Schwarzschild radius): 2 escape 2 2 lightofspeed c GM R c R GM cv s s = =∴ ==Q 29 恒星的起源及其演化 1、恒星分类 主序星——恒星一生 的主要阶段。其光度和 表面温度与质量成正 比；存在时间与质量成 反比；燃烧氢生成氦。 巨星——恒星演化的晚 期阶段。当主序星中心 随氢燃烧成氦，温度变 低而形成。 超巨星——最亮的恒 星。年轻的、演化迅速 的大质量恒星。 白矮星——低光度、高密度、高温 度的恒星。恒星的老年阶段，核能 接近枯竭，内温极高，爆发后余一 高密度核而成。 2、恒星的起源 星际空间稀薄物质由于涨落分裂成团块，并在引力作 用下凝聚成弥漫星云，经快收缩和慢收缩阶段成为“星胎 （原始恒星）”。再经过几千万到几亿年的收缩中心温度 达1000万K，氢燃烧的核反应提供足够的能量是内压与引 力处于相对平衡状态。 恒星的起源及其演化 30 3、恒星的演化和归宿 高温高压核聚变辐射能量、消耗核燃料(氢--氦)，燃料 用完，内压减小，引力引起塌缩，外部聚变成碳、氧、 氖、…、铁（结合能最大的核子），内部引力压缩将原子压 碎，变成中子核，同时引力使外围物质向内坍缩，与内核区 相撞，产生极强的巨大激波，使恒星增亮为太阳的100亿 倍——超新星爆发。 1054年 超新星 爆发后 的蟹状 星云 超新星爆发带来积极的一面:恒星 外层加热，进一步核聚变产生 金、铅、铀等元素，连同早期产 生的碳、氧等抛出，与其它碎片 混合，产生下一代恒星和行星。 恒星的起源及其演化 4、白矮星、中子星、黑洞 白矮星——原子态（超固态）的物质组成。 一般物质宏观状态下是致密的，但微观状态下 却是空虚的，犹如一个足球场上只有一两个足球。 在万有引力的作用下恒星要收缩使物质原子一个挨 一个成为原子态或超固态——白矮星。 什么样的恒星会演化成白矮星？ M=0.4~1M 日 R=1/40~1/100R 日 ρ =10 5 g/cm 3 M< 1.2M 日 的恒星将演化成白矮星。 恒星的起源及其演化 31 星际星云→原始恒星→主序星→红巨星→超新星爆发 （或白矮星）→黑洞（或中子星、星际物质） 恒星的起源及其演化 中子星——当恒星质量1.2M 日 <M< 3.2M 日 时，由于引力较大，恒星中的原子态物质将 进一步被压缩，电子和核内质子结合形成中 子，恒星处于中子态。 中子星（脉冲星） ρ =10 15 g/cm 3 中子星密度极高： 1967年发现中子星， 由于其强大的磁场和 高速旋转，地球上接 收到的信号为脉冲式 ——脉冲星。 恒星的起源及其演化 32 黑洞——当恒星质量M> 3.2M 日 时,中子星状态也无 法维持恒星的平衡，恒星 将继续收缩，直至形成致 密的、引力极大的、连光 也无法逃脱的天体。 1968年韦伯声称接 受到银河系中央的引力 波信号；2000年10月26 日,美麻省理工学院由美 国宇航局“钱德拉”X射 线望远镜观测到X射线 爆持续了3个小时，期 间有10分钟减弱消失， 是射线穿过黑洞区域的 证明,该区域不超过1.5 亿公里，质量260亿万 个太阳。X射线由黑洞 吞噬彗星形成。 恒星的起源及其演化 黑洞吸积盘 恒星的起源及其演化 33 黑洞的性质： ①黑洞无其他天体的性质，无磁场、无射线 爆、无物质结构、…… ②黑洞无毛定理 黑洞的全部性质可以用质量、角动量和电荷这 三个量（三根毛）描述。 ③黑洞可以辐射（蒸发） 黑洞辐射是一种量子效应。 恒星的起源及其演化 黑 洞 视 界 34 主要参考书目： 《相对论天体和宇宙》王永久著 《宇宙与人》忻迎一著 《时间简史插图本》史蒂芬?霍金著 《果壳中的宇宙》史蒂芬?霍金著 §8.6 The power 1. Average power t W P ? ? = av (J/s) 2. Instantaneous power vF t rF t W t W P t r r r r ?= ? == ? ? = →? d d d d lim 0 vF t rF t W t W P t r r r r ?= ? == ? ? = →? total total 0 total d d d d lim 3. The power of the total force acting on a system