英文版大学物理功和能共44页
大学物理第四章--功和能
a
l
xdx
2l
前已得出：
Af
mg(l a)2
2l
mg(l 2 a2 ) mg(l a)2 1 mv2
2l
2l
2
得v
g l
1
(l 2 a 2 ) (l a)2 2
§3 保守力的功与势能 一、 保守力
rB
B
两个质点之间的引力
B
第四章 功和能
§4.1 功 §4.2 动能定理 §4.3 保守力功与势能 §4.4 功能原理机械能守恒定律
§1 功和功率
一、恒力做功 直线运动
A=Fcos S
记作A F S F r
F
F
M
M
S
位移无限小时：
dA
F
dr
dA称为元功
功等于质点受的力和它的位移的点积(标积)
例１一水平放置的弹簧，其一端固定，另一端系一小球，求小
球的位置由A到B的过程中弹力对它所做的功。（在O处弹簧无 形变）
解：根据胡克定律 F F kx
W F dr
xB Fdx
xA
xB xA
kxdx
O
1 2
A
k xB2
B
xA2
1 2
k xA2
作用在质点
上．在该质点从坐标原点运动到(0，2R)位
置过程中，力
F
对它所作的功为多少？
y
b
b
A a F.dr a (Fxdx Fydy)
R
x O
例4 如图，水平桌面上有质点 m ，桌面的摩 擦系数为μ 求：两种情况下摩擦力作的功
英文版物理功和能
Example
x 4y
2
2. The trajectory is a straight line 4 y x 6
Work is a scalar, no direction. π positive work; 0 d W 0 ,
π 2 π 2 π 2 dW 0 , d W 0,
Constant Force (including its direction)
A
rA
o
s
(2) Work done by a varying force: When a particle moves from A to B along a curve path, the total work ds done by the force equals the integral of F from A to B: A
7-1 Work and Power (P147-)
1. Work
研究力和运动的空间过程关系。
The work done by a force F in displacing a s body s as the scalar product of F and s.
W F r cos F r
1
2
O
3 X
A1
( Fx dx Fy dy ) 2 ydx 4dy
大学物理《功和能》课件
L A L B
L
L
B
L
B
A
B f dr f dr 0
L
A
§4.3 保守力与势能
2.势能
A引 Gm 1 m 2 rB Gm 1 m 2 rA
A弹 1 2 ks
2 A
1 2
ks B
2
A引
Gm 1 m 2 r B r rA
B f dr k (r
A
r k ( r r0 ) A r 1 1 2 2 k ( r A r0 ) k ( rB r0 ) 2 2
O rA r r0 ) d r r r d r k ( r r0 ) d r r
第4章 功 和 能
Work and Energy
第4章
功和能
质点受力的作用时，如果持续一段时间，质点的动量会 改变；如果质点由空间位置的变化，则力对位移的累积（功） 会使质点的能量（动能和势能）发生变化。对功和能的研究， 是经典力学中重要的组成部分。 与机械运动相联系的能量守恒定律（机械能守恒定律）， 是普遍的能量守恒定律的一种特殊形式。
一般引力势能的零点取质点相距无穷远，E
r
一般弹性势能的零点取弹簧无伸缩状态，Ep
0 ， 0 C
s 0
A点势能可表为 E p ( A )
Ep 0 A
f保 dr
§4.4 引力势能与弹性势能
2.势能曲线
Ep
Ep
Gm1m2 r
Gm1m2 r0
引力势能曲线
引力势能是空间变量
动量动量角动量角动量能量能量守恒量对称性时空性质空间平移空间平移空间转动空间转动时间平移时间平移空间均匀性空间均匀性空间各向同性空间各向同性时间均匀性时间均匀性守称守恒守恒空间反演对称性空间反演对称性安保是指为了达到安全的目的而进行的对人或物的保护活动安保工作是指为集体或个人的安全而进行保卫的各种活动
大学物理课件第4章-功和能
如图,求船从离岸 x1处移到 x2 处的 过程中,力 F 对船所作的功.
F
解:判别F 是否为变力作功(大小不变,方
向变元),功属于dW变力作F功.建dx立坐F标,取dx元过co程sa
h
o x2
a
dx x x1 x
cosa x
x2 h2 x
dW F dx
x2 h2
功在数值上等于示功图
F
曲线下的面积。
3. 功率
0 x1
x2 x
dx
平均功率: P =ΔΔWt
瞬时功率:
P
= lim
Δt 0
Δ Δ
Wt =
dW dt
=
F
. dr
dt
= F .v
[ 例1 ] 有一单摆，用一水平力作用于m
使其极其缓慢上升。当θ 由 0 增大到 0 时，
求: 此力的功。
{
F T sinθ T cosθ mg
两边平方
v 2 v12
由动量守恒
2v1
v2
v 22
v
v1
v2
由机械能守 恒（ 势能无变化）
v2
v12
v
2 2
v1 v2 0 两球速度总互相垂直
例8：已知半圆柱形光滑木凹槽，放在光滑桌面上，
如图，求质点下滑至最低点时给木块的压力.
解：
mv MV 0
•2.碰撞分类
正碰 斜碰
（从碰撞前后两球中心连线角度分类 ）
弹性碰撞 非弹性碰撞 完全非弹性碰撞
一般非弹性碰撞
（从碰撞能量损失角度分类）
例7：在平面上两相同的球做完全弹性碰撞，其中一球开始时处于
英文版大学物理 第三章
Bead a Wire
θ
F
b
x
Signs for work θ < 90o, W > 0; θ > 90o, W < 0; θ = 90o, W = 0.
r r ∆r = ∆xi
Units for Work 1 N·m= 1 J
Work done by a variable constant force The work done on the particle in moving through r the small displacement dr due to the force is
Chapter 3 Work and Energy
3-1 Work and Power 3-2 Work−Kinetic Energy Theorem − 3-3 Conservative Force 3-4 Potential Energy 3-5 Calculating the Conservative Force from the Potential 3-6 Conservation of Mechanical Energy 3-7 Conservation of Energy
3-1 Work and Power Work done by a constant force Suppose that a bead can slide along a frictionless wire. r A constant force F directed at an angle θ to the wire, r r causes the bead to move a displacement ∆r = ∆xi The work done by the force is defined to be the product of the magnitude of the displacement and the r component of force that is parallel to the displacement.
(英文版)大学物理-力学
Dep. physics
1.NEWTON’S LAW
Newton’s First Law
An object at rest remains at rest, and an object in motion continues in motion with constant velocity (that is, constant speed in a straight line) unless it experiences a net external force.
v v u
t t
•Acceleration commutation
a a
y S u y S
r
P
r
O
R uti
O
x x
t t z
z
Dep. physics
v
x
vx
u
v
y
vy
v
z
vz
t t
ax a x ay a y az az
t t
S
F
ma
S
F
ma
动画
22
Summarizing:
Dep. physics
7
a
dv
dt
d 2r dt 2
a
dvx dt
i
dv y dt
j
dvz dt
k
d2 dt
x
2
i
d2y dt 2
j
d 2z dt 2
k
axi ay j azk
a a
ax2
a
2 y
az2
Dep. physics
cos ax a
cos ay a
大学物理第4章 功与能work and energy(1)
2.动能的性质: 瞬时性;相对性
问题：
铅直下落的冰雹，质量为 m，某时刻的速率为v，试问从地 面上以速率v水平运动的车上观察，该冰雹的动能是多少？
（答案:mv2）
3.动能与动量的关系:
v v
v v
u
二 、动能定理 1. 质点的动能定理
合外力对质点所作的功, 等于质点动能的增量 .
质点运动的动能定理 合外力对质点所做的功等于质点动能的增量
典型的保守力： 重力、万有引力、弹性力 与保守力相对应的是非保守力，如摩擦力。
若质点沿任意闭合路径运动一周，保守力 对质点所做的功为零。
A保
Fdr l
Fdr ACB
F dr
BDA
F d r F d r
ACB
ADB
A C
解：（1）选择地球惯性系建立坐标系； （2）确定研究对象： 链条 设经时间 t 秒，链条下落 x （3）分析所受的力；重力和摩擦力 摩擦力:
f LL-- ax o
xa
（4）链条离开桌面过程中摩擦力所做的功
X
（5）下落过程重力做的功： （6）应用动能原理列方程解方程
链条刚刚离开桌面时的速率：
f L- x o x
例 一轻弹簧, 其一端 系在铅直放置的圆环的顶 点P，另一端系一质量为m 的小球, 小球穿过圆环并在 环上运动(μ =0)．开始球 静止于点 A, 弹簧处于自然 状态，其长为环半径R; 当球运动到环的底端点B时，球对环没有压 力．求弹簧的劲度系数．
解 以弹簧、小球和 地球为一系统
只有保守内力做功 系统
P.229 4.15
一质量为 m 的物体，从质量为M的圆弧形槽顶由静止滑下， 圆弧形槽的半径为R，张角为 900 。如果所有摩擦可以忽略。
大学物理 功和能汇总
2 动能定理： A 1 2 mv 0
2A v 4 m s m
[思考] 在 x =0 至 x =1m 过程中， F 的冲量？
10
§4.3 质点系的动能定理
Theorem of Kinetic Energy for a system of Particle
对第 i 质点 求和
O 张力不做功，重力做功： 用动能变化定理解：
l
m
T
A mg dl mg dl cos
mgl cos d mgl sin 0 1 2 mgl sin mv 2
ˆn e
v
mg
ˆt e
比直接解牛顿方程简单，但仍作积分运算。
13
§4.4 *柯尼希定理
i
14
一对力 的功
内力总是成对出现 dr1 两质点间的内力 f ij 和 f ji ,
B1
B2
dr2
f 12
称为一对力 f ij f ji
m1
r21
f 21
m2
A1
A2
一对力做的功之和
dA = f12 dr1 + f21 dr2
f 21 dr2 dr1 f 21 dr21
mi ac dri
m i ac
z
y
mi
= ac mi dri
ri
ac
C 质心 O
12
= ac d mi ri = 0 A i
B
x
=
0
【例】柔软细绳长为l，小球质量为m，求摆下至 角时小球的速度和绳的张力。
大学物理(双语)
v = v x2 + v y2
v and tanθ = v y x
r r r Ar B = C × r
r C
r B
α
Both pairs of values contain the same information.
Direction ( figure )
r A
2
�
?
Coordinate system can be chosen freely.
Components of vectors (50)
A component of a vector is the projection of the vector on an axis. This component may be in 2 or 3 dimensional (3D) coordinate system.
gh = 1 v 2 2
L T 2 L
L 2 T
[L T ] = [L T ]
2 2 2 2
一个量的量纲是该量所描述的物理特性
The dimensions on both sides of an equation must be equal. ----check the validity(正确性)of a calculation
Watt and Angstrom ( A ) are derived units.
Derived unites are defined in terms of the basic units. Some have given names, some do not.
speed force m/s kgm/s2 Newton
Scientific notation Mass of the Earth:
大学物理第4章 功和能
f d r 0 （此式也可作为
L
(1 ) ( 1 ) f d r ( 2 ) f d r L1 L2
(2)
L1
L2
保守力的定义） 20
二. 几种保守力 1.万有引力
(2) ×
d r er d r
W 12 对 ( 1 )
(2)
13
本质区别：动能和物体的运动状态相联系，任 一运动状态对应一定的动能，是状态量；而功 是与物体在力作用下的具体运动过程相联系， 它一般是路径的函数，因而功是过程量。 密切关系：过程便意味着状态变化。合外力对 质点做功，质点的动能便发生变化。做功是使 质点动能改变的手段，动能的变化又是用功来 量度的，故二者具有相同的单位。 动能是质 点因运动而具有的做功本领。
——质点的动能定理
“合外力对质点所做的功等于质点动能的增量”
12
2. 分析说明： ①动能定理本质上是牛顿第二定律的推论，它 从一个侧面反映了质点在力学过程（空间积累 过程）中所服从的规律。 ②由动能定理知，力对物体做功，能改变物体 的动能，也只有力对物体做功，物体的动能才 能改变， 功是机械运动能量变化的量度。 ③功和动能的概念不可混淆
14
3. 质点系的动能定理 质点：m1 、 m2
F F 内力： f 1 、f 2 外力： 1 、 2 初速度： 1 a 、 2 a 末速度： 1b 、 2 b
b
v1b
dr b · b·
1 2
v2b
F2
2 2
F1
dr1 m1
m ·
f1 f2
·
m1:
m2:
2 2 2 ( F 1 f 1 ) d r1 1 m 1 1 b 1 m 1 1 a ( 1 ) a 1 2 2 b 2 2 2 ( F 2 f 2 ) d r2 1 m 2 2 b 1 m 2 2 a ( 2 ) a 2 2 2
大学物理(哈里德版)03
保守力的功 W (Epf Epi ) EP
保守力所做的功等于系统势能增量的负值
势能差有绝对意义，势能只有相对意义
选定某一位形作为参考位形，规定此参考位 形的势能为零——势能零点
保守力 势能(Ep )
势能的符号表示
势能零点
重力 重力势能
Ep mgy
y=0
弹性力 弹性势能 引力 引力势能
Ep
1 2
• 作业：P94-95 21;31
动能的单位：焦耳（J）
二、功-动能定理
力对质点做功：
W
xf
F x dx
xf
ma dx
x f dv m dx
xi
xi
xi dt
v f
mv dv
vi
vf
m v dv
vi
1 2
m
v2f
1 2
m
vi2
Ek Ekf Eki W 功-动能定理
注意 功与动能的联系与区别
1、功是用对物体施力的方法传给物体（或由物体传 出）能量。W >0 时传给物体能量；W <0 时物体传 出能量。
W Emec
没有摩擦时，外力对系统做功等于系统机械能 的增量。
二、有摩擦的功能关系
由匀加速运动公式
v 2 v02 2ad
由牛顿第二定律
F Fk ma
Fd
1 2
m v2
1 2
m v02
Fk d
Fd Ek Fkd
W Ek EEthth
W Ek Eth
考虑势能（如物块沿斜面运动），有
b
（2）如果运动的轨迹是如图中
o
x
的adb， 那么重力所作的功又是多少？ z
物理英文版
Time -averaged net forceWhen a time-varying net force isdifficult to measure, we can use a time-averaged net force as the substitute provided that it would give the same impulse to the particle in sametime interval.When a particle experiences a impact in a veryshort time, the non-impulse forces such as gravitational force and friction force are negligible compared to impulsive force.1f i t t F Fdtt =Δ∑∫J G J G ()F t ∑J G J p F t=Δ=Δ∑JG J G J GTime -averaged net force (continued)For a given amount of momentum change, we can delay the timeinterval to decrease the impulsive force.A baseball player catching a ball can soften the impact by pulling hishand back.Example:Example: Conical Pendulum.A small object of mass m is suspended from a string of length L . The object revolves in a horizontal circle of radium r with constant speed v . Determine the impulse exerted (1) by gravity, (2) by string tension on the object, during the time in which the object has passed half of the circle.Solution: From impulse-momentum theorem212122t net net t J F dt mv mv mv ==−=∫J G J G G G G The impulse exerted by gravity on the object 211222t mg t T r r J mgdt mg mgmg v v ππ====∫J G G G G GThe impulse exerted by string intension on the objectT net mgJ J J =−J G J G J G ()222222224T rmg r g J mv m v vvππ⎛⎞=+=+⎜⎟⎝⎠Impulse-momentum theorem for a system of particlesConsider a system of N interacting particles with masses m 1, m 2, …, m N .For i-th particle : the net external force the internal force exerted by j-th particle For the system of particles :According to Newton’s third law, the internal forces cancel in pairs.The total external force acting on the system:The total momentum of the system:i F J G ijf J G ()i ij iF f dt d p +=JG J G J G()i ij iij iiF f dt d p ≠+=∑∑∑JG J G J Gij ij if ≠=∑∑J G iiF ∑J Gtotiip p =∑JG J G §2Impulse -momentum theorem fora system of particlesImpulse-momentum theorem for a system of particles (continued)zerototd p J GWhich case of collision satisfies the conservation of momentum? Or conservation of component momentum?Example②Example: A wooden block of mass M 1is suspended from a cord of length L attached to a cart of mass M 2which can roll freely on a frictionless horizontal track. A bullet of mass m is fired into the block from left. After the impact of the bullet, the block swings up with the maximum angle of θ. What is the initial speed v of the bullet?Solution:(1)Stage 1:For the systemconsisting of m and M 1, the momentum isconserved in horizontal during a small intervaltime of impact.11()mv M m v =+Example: is suspended vertically with its lowest end§3 Center of MassDescribe the motion of a system of particlesby every motion for individual particles by overall motion in terms of center of mass Center of massFor the system of discretely distributed particles represented by components For the extended object with uniformly distribution of mass CM i i i i i i i i m r m r r m M==∑∑∑G G G CM CM CM i i i i i i ii i m x m y m z x y z M M M ===∑∑∑10CM 10CM lim 1, lim 11, i i N i i N i m N i N i m CM x m x dm x x dm M dm m y y dm z z dm M M →∞=Δ→→∞=Δ→Δ===Δ==∑∫∫∫∑∫∫CM 1r r dm M=∫GGExample: Find the center of mass of a uniform solidhemisphere of radius R and mass M .Solution: From symmetry it is apparent that the center ofmass lies on the z axis. x CM =0, y CM =0.The three-dimensional integral can be treated as an one-dimensional integral. Subdivide the hemisphere into a pile of thin disk .CM 11z zdm z dV M Mρ==∫∫2dV r dzπ=323M R ρπ⎛⎞=⎜⎟⎝⎠022CM 323322003cos sin (sin )233cos sin sin (sin )22313248z R R R d R R d R d R R πππθθθθθθθθθ=−===×=∫∫∫Find r , z in terms of θ.sin cos sin r R z R dz R d θθθθ===−Supplement problem: Find the center of mass of a uniform semicircular plate of radius R and mass M.Applications of center of massFor a system of discrete particlesA cannon shell in a parabolictrajectory explodes in flight,splitting into two fragments. Thefragments follow new paths, butcenter of mass continues on the original parabolic trajectory.For a rigid bodyWe can describe a rigid body as a combination of translational motion of the center of mass and rotational motion about an axis throughthe center of mass.Fig. (a) The motion of the diver is pure translation.Fig. (b) The motion of the diver is translation plus rotation.Example: A rocket is fired into the air. At the moment it reaches its highest point,a horizontal distance d from its starting point, an explosion separates it intotwo parts of equal mass. Part I is stopped in midair by explosion and fallsvertically to Earth. Where does par II land?Solution: After the rocket is fired, the path of the center of mass of the system continues to follow the parabolic trajectory of a projectile acted on only by a constant gravitational force. The center of mass will thus arrive at a point 2d from the starting point. Since the masses of I and II are equal,the center of mass must be midway between them. Therefore, II lands a distance3d from the starting point.Example:Solution II: Treat the two blocks as a whole.External force: exerted by wall on m 2.2CM2122()()d x k l x m m dt −=+2CM 212m x x m m =+2CM12CM 1222()()d x m m k l x m m m dt+−=+2CM 2CM 22120d x m k x l dt m m m ⎛⎞+−=⎜⎟+⎝⎠2CM 122cos m k x l A t m m m φ⎛⎞−=+⎜⎟⎜⎟+⎝⎠。
MOOC大学物理热力学第一定律(双语)
（2）当系统由状态b沿曲线ba返回状态a时，外界对系统作 功 60J，则系统吸收热量Q＝ ？
Pቤተ መጻሕፍቲ ባይዱ
c a O
b d
V
21.3 Applying the First Law of Thermodynamics 1. Isochoric Processes(等体过程) (1)Process equation (Restriction)过程方程 P (2) PV diagram cons tan t T
Three degrees of translational freedom 确定质心位置 Two degrees of rotational freedom 确定转轴位置 f=t+r＝3+2=5
Kt 1 1 1 2 2 2 m v Cx m v Cy m v Cz 2 2 2
1 2 1 2 K r I y I z 2 2
thermodynamics quasistatic process molar specific heat isothermal process isobaric process isochoric process adiabatic process monatomic molecules diatomic molecules polyatomic molecules degrees of freedom principle of equipartition of energy internal energy
2
分子任何一个自由度的平均能量都相等，均为
1 k T 2
Monatomic gas Diatomic gas Polyatomic gas
3 3 U N kT nRT 2 2 5 5 U N kT nRT 2 2
《物理双语教学课件》Chapter 4 Work and Energy 功和能
Chapter 4 Work and EnergyThe concept of energy is one of the most important in the world of science. In everyday usage, the term energy has to do with the cost of fuel for transportation and heating, electricity for lights and appliances, and the foods we consume. However, these ideas do not really define energy. They tell us only that fuels are needed to do a job and that those fuels provide us with something we call energy.Energy is present in the Universe in a variety of forms, including mechanical energy, chemical energy, electromagnetic energy, heat energy, and nuclear energy. Although energy can be transformed from one form to another, the total amount of energy in the Universe remains the same. If an isolated system loses energy in some form, then by the principle of conservation of energy, the system must gain an equal amount of energy in other form. The transformation of energy from one form into another is an essential part of the study of physics, chemistry, biology, geology, and astronomy.In this chapter we are concerned only with mechanical energy. We introduce the concept of kinetic energy, which is defined as the energy associated with motion, and the concept of potential energy, the energy associated with position. We shall see that the ideas of work and energy can be used in place ofNewton’s law to solve certain problems.4.1 Work and Power1. Work W done by a constant force is defined as the product of the component of the force along the direction of displacement and the magnitude of the displacement.ϕcos FS S F W =⋅=Where the force makes an angle of ϕ with displacement .SThe SI unit of work is the joule (J), named for James Prescott Joule, an English scientist of the 1800s. It is derived directly from the units for mass and velocity:1joule=1J=1(kg) (m/s 2) (m)=1 kg m 2/s 22. Work done by a variable force(1). The increment of work dW done on the particle by F during the displacement d r isr d F dW ⋅=, where Force F is function of its position.(2). The work W done by F while the particle moves from an initial position a to a final position b is then⎰⎰⋅==b a b a r d F dW W(3). We use the components ofF and r d to express the forceand displacement, then we havedzF dy F dx F k dz j dy i dx k F j F i F r d F W z y b a x z y b a x b a ++=++⋅++=⋅=⎰⎰⎰)()( 3. Work done by multiple forces : If there are several forces act on a particle, we can replace F in above equation with the net force ∑F , where +++=∑321F F F F , where j F are theindividual forces. Then+++=⋅+++=⋅=⎰⎰321321)(W W W r d F F F r d F W b aba 4. Power(1). The rate at which work is done by a force is said to be the power due to the force . If an amount of work W is done in a time intervalt ∆ by a force, then the average power due to the force is t WP ∆=.(2). The instantaneous power P is the instantaneous rate of doing work, which can be written asv F dtr d F dt dW P ⋅=⋅==. (3). The SI unit of power is the joule per second . This unit is used so often that it has a special name, the watt (W), after James Watt, who greatly improved the rate at which steam engines could do work.4.2 Kinetic Energy and Work-Kinetic Energy TheoremEnergy is a scalar quantity that is associated with a state of oneor more object. The term state here has its common meaning: it is the condition of an object.1. Kinetic energy K is associated with the state of motion of an object. The faster the object moves, the greater is its kinetic energy. For an object of mass m and whose speed v is well below the speed of light , we define kinetic energy as 221mv K = The SI unit of kinetic energy is the same as work —joule.A convenient unit of energy for dealing with atoms or with subatomic particles is the electron-volt (eV).1 electron-volt = 1 eV =1.60 x 10 –19 J.2. Work-kinetic energy theorem : If a force changes the speed of an object, it also changes the kinetic energy of the object. If the kinetic energy is the only type of energy of the object being changed by the force, then the change in kinetic energy is equal to the work W done by the force:W K K K i f =-=∆ Here i K is the initial kinetic energy (=2021mv ) and f K is the kinetic energy (221mv ) after the work is done.3. We can prove work-kinetic energy theorem as follow:002022222221212121)(21k f v v b a z y x z z y y x x b a b a b a K K mv mv v m mvdv W dv v v v d dv v dv v dv v v d v v d v m dt v dtv d m s d f W -=-====++=++=⋅⋅=⋅⋅=⋅=⎰⎰⎰⎰(Numerator and denominator of a fraction)4.3 Work done by weight and by a spring force1. Work done by weight :[]a b a b b a b a m gh m gh h h m g dh m g dW W m gdhm gdl m gdl l d g m dW --=--=-==-=--==⋅=⎰⎰)()cos(cos απαWe can find that the work done by weight on a particle between two points does not depend on the path taken by the particle . Or no matter what path we choose to move the particle, the work done by its weight is the same . In other word if we move a particle around a closed path, the work done by weight on the particle is zero .2. Work done on aparticle-like object by aparticular type of variableforce, namely, springforce —the force exertedby a spring.⎥⎦⎤⎢⎣⎡--=-=-==-=⋅=-=⎰⎰222212121'a b x x x x b a kx kx x k kxdx dW W kxdxs d F dW law s Hook kx F b a b a 3. Conservative force and Non-conservative force : If the workdone by the force is independent of the path the particle moves,the force is a conservative force ; otherwise a non-conservative force . Weight and spring-force are conservative forces; friction is a non-conservative force.4.4 Potential energyPotential energy U is energy that can be associated with the configuration (or arrangement) of a system of objects that exert a force on one another. If the configuration of the system changes, then the potential energy of the system also changes.1. We know that work done by a conservative force has nothing to do with the path the particle taken. So we can introduce a quantity which is the function of the state of the system to indicate this kind of nature for a conservative force. We call it potential .2. Gravitational Potential Energy(1). The work done by weight can be expressed as: []U mgh mgh W a b ∆-=--==, where U ∆ is the change in the gravitational potential energy. Since the work done by weight has definite magnitude from an initial position to a final position, so only a changeU ∆ in gravitational potential energy is physically important .(2). However, to simplify a calculation or a discussion, we cansay that a certain gravitational potential U is associated with any given configuration of the system, with the particle at a given height h. To do so, we rewrite the above equation as:)(i i h h mg U U -=-Then we take i U to be the gravitational potential energy of thesystem when it is in a reference configuration, with the particle at a reference pointi h . Usually, we set 0=i U and 0=i h , then we have mgh U =. So the gravitational potential energy associated with a particle-Earth system depends on the height h of the particle relative to the reference position of h i =0, not the horizontal position.3. Elastic Potential Energy(1). Similar to a particle-Earth system, the work done by the spring force can be rewritten as U kx kx W a b ∆-=⎥⎦⎤⎢⎣⎡--=)21()21(22.(2). To associated a potential energy U with any given configuration of the system, with the block at position x, we set the reference point for the block as x i =0, which is always at the equilibrium position of the block. And we set the corresponding elastic potential energy of the system as U i =0. Thus we have 221)(kx x U =. Attention: (1). Potential energy belongs to the whole system.(2). The magnitude of potential energy depends onthe choice of the reference point.4.5 Work-Energy Theorem and Conservation of Mechanical Energy1. Work-kinetic energy theorem for one particle: We have W K K K i f =-=∆2. Work-Energy Theorem: Suppose we have particles of N in the system we discussed and we use work-kinetic energy theorem for each particle, then we have total amount of N equations like those:jijf j if if K K W K K W K K W -=-=-=222111We can divide the forces exerted on every particle into external forces and internal forces. And the internal forces can also be classified as conservative internal forces and non-conservative forces. Summing the two side of above equations, we will have: i fN j ji jf N j j K K K K W -=-=∑∑==11)(Or to be exactly,K W W W noncon in con in ext ∆=++--We also know that the work done by the conservative internalforce can be written as the minus difference of potential energy. So we get the Work-Energy Theorem :E U K W K W W con in noncon in ext ∆=∆+∆=-∆=+--The work done by external forces and non-conservative internal forces in a given system is exactly equal to the difference of its Mechanical Energy .3. Conservation of Mechanical Energy : When only conservative forces act within a system, the kinetic energy and potential energy can change. However, their sum, the mechanical energy E of the system, does not change . 0=∆+∆=∆U K E4.6 Reading a Potential energy curveConsider a particle that is part of a system in which a conservative force acts. Suppose that the particle is constrained to move along an x axis while the conservative force does work on it.1. Finding the Force Analytically : For one-dimensional motion, the work W done by a conservative force that acts on a particle as the particle moves, and the potential energy have the relation as followdx x F dW x U )()(-=-=∆So can get the force from the potential energytion ensionalmo one dx x dU x F dim )()(--=We can, for example, check this result by putting221)(kx x U =, which is the elastic potentialenergy function for a springforce. Above equation yieldskx x F -=)( as expected.2. The Potential EnergyCurve : The following figureis a plot of a potential energyfunction U(x) for a system inwhich a particle is inone-dimensional motionwhile a conservative forceF(x) does work on it. We caneasily find F(x) by taking theslope of the U(x) curve atvarious points. Fig. (b) is a plot of F(x) found in this way.3. Tuning Point : Since there is only conservative force acting on the particle, the system will remain conservation of its mechanical energy. So we have)()(x U E x K -=. Since kinetic energy)(x K is not less than zero. As the particle moves from 2xto 1x (Fig. a), when the particle reaches 1x , its kineticenergy is zero, meanwhile the force on the particle is positive. It means the particle does not remain atx but instead begins to1move back to the right. Hencex is a tuning point, a place1where K=0 and the particle changes direction of its motion.4.Equilibrium Points:Neutral equilibriumUnstable equilibriumStable equilibrium。
大学物理英文版
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