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# 第四章 - PowerPoint PPT Presentation

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Presentation Transcript

F

F

1

a

mv2

2

a

A

=

F

r

cos

Δ

r

Δ

Ek =

=

F

r

Δ

.

§1、

1. 恒力的功

F cosa

F

o

s

s

s

1

2

ds

dA

a

dr

F

a

=

F

ds

cos

.

.

F

d

r

F

ds

=

=

.

ò

F

cos

a

ds

ò

A

=

F

dr

=

2. 变力的功

F

Fx

i

Fy

j

Fz

+

=

+

k

dr

=

dx

i

+

d

y

j

+

dz

k

=

Fx

dx

+

Fy

dy

+

Fz

dz

.

dA

F

dr

=

ò

ò

ò

ò

ò

ò

dx

dx

+

+

Fy

Fy

=

=

Fx

Fx

dy

dy

+

+

Fz

Fz

dz

dz

.

A

ò

=

F

dr

F

cos

a

ds

ò

=

.

ò

A

=

F

dr

F (s)

s

2

ò

A

=

F (s) ds

s

1

F

o

s

s

s

1

2

ds

Δ

A

N

=

Δ

t

.

F

A

dA

dr

Δ

lim

N

=

=

=

dt

t

dt

Δ

t

Δ

0

.

F

=

v

3. 功率

mv dv

m ds

Ftds

matds

ò

ò

ò

ò

=

=

=

=

dv

dt

a

1

1

dr

=

m

v

m

v

2

2

2

2

F

0

= Ek2 Ek1 = ΔEk

A = ΔEk

F

cos

a

ds

ò

=

.

ò

A

=

F

dr

T

H

= ò(m - kh )g dh

0

1

m’g

= mgH - kgH2

2

[例2]

∵匀速∴T = m’g

A = ò Tdh

§2、

y

a

dr

b

y

y

.

dA

G

dr

=

a

b

G

(

m

g

j

)

i

d

y

j

)

.

(

d

x

=

+

o

x

mg

dy

=

A

mg

dy

(

m

y

m

y

)

ò

g

g

=

=

b

a

.

A

G

dr

0

ò

=

=

1. 重力的功

.

F

ds

0

=

ò

：沿任意闭合路径积分．

ò

A保

=

DE

p

A保

=

(

mg

y

mg

y

)

b

a

=

(

E

E

)

pb

pa

=

E

Δ

p

.

A

G

dr

0

ò

=

=

∴ 只有保守力才能引入势能的概念。

∴ 重力是保守力

F

o

x

x

F

=

kx

dA

F

dx

kx

dx

=

=

x

ò

b

A

kx

dx

=

1

1

x

A保

(

)

=

kx

2

kx

2

a

a

2

2

b

.

A

1

F弹

dr

0

ò

(

E

E

)

E

=

=

=

Δ

=

2

Epa = kx

p

pb

pa

2

a

2.弹性力的功

∴ 弹性力是保守力

3. 万有引力的功

b

r

b

dr

.

dr

dA

F

dr

=

θ

Mm

m

.

.

r

dr

r

=

G

F

r

3

M

r

a

a

Mm

r dr

=

G

r

r dr

3

r

GMm

GMm

dr

(

)

(

)

A

=

G

Mm

=

ò

b

r

r

r

= r dr cosq

r

2

b

a

a

.

A

F万

dr

0

ò

=

=

= r dr

∴ 万有引力是保守力

GMm

GMm

(

)

(

)

(

)

=

=

E

E

A保

r

r

pb

pa

b

a

A保

=

DE

p

GMm

E

=

r

pa

a

E

=

Δ

p

.

b

F

dr

ò

=

(

E

E

)

.

.

.

b

b

(0)

A保

pb

pa

a

E

E

E

F

F

F

dr

dr

dr

ò

ò

ò

=

=

=

pa

pa

pa

a

a

a

+ Epb

=

E

Δ

p

1. 势能为系统所有。

2.保守力才有势能的概念。

3. 势能是物体位置的单值函数。

A

F

=

F

A

+

+

F

A

F

A

+

+

A

F

+

+

A

F

=

A

+

A

+

A

= Ekb- Eka

(

)

A

=

E

E

pb

pa

A = Ekb- Eka

(

)

A

=

E

E

pb

pa

(

A

+

A

E

E

)

= Ekb- Eka

pb

pa

A

+

A

A

+

A

+

A

(

= Ekb- Eka

)

(

)

=

E

+

E

E

+

E

pb

ka

pa

kb

= DE

A

+

A

= DE

(

)

(

)

A

+

A

=

E

+

E

E

+

E

pb

ka

pa

kb

A

+

A

=

0

E

+

E

E

+

E

C

=

=

pb

pa

ka

kb

§3、

（孤立系统）

A = DEk

A

+

A

+

A

=

D

A

=

E

p

A

+

A

=

DE + DE

k

p

A、功是能量交换、变化的一种量度，

B、无功则能量不生不灭，但能互相转换 。

A

= DU +

DE + DE

k

p

A

+

A

=

DE + DE

k

p

1

1

1

1

2

2

2

2

2

o

m

m

x0

x0

x

x

q

2

2

Ek0+ kx - mgx0sinq

= Ek+ kx - mgxsinq

0

1

2

2

Ek 0 = Ek + kx – kxx0 + kx

Ep = kx

2

0

f0 = - kx0 = - mg sinq

∵只有保守内力作功 ∴机械能守恒

1

1

1

2

2

2

o

m

x0

x

q

x

2

2

Ek 0 = Ek + kx – kxx0 + kx

0

Ek 0 = Ek + k (x–x0 )

2

x–x0 :离开平衡位置位移

Ek 0 = Ek + Ep弹

GMm

m

=

E

pa

2

R

a

GMm

R

E

=

b

pb

R

R

M

GMm

GMm

1

mv

2

0

=

+

+

2

R

R

2

GM

v

=

R

[例1]

f = k/r4，试求两个粒子相距为 r 时的

.

E

f

dr

ò

=

p

r

1

dr

kr -3

kr -4

ò

=

=

3

r

k

=

3r3

r

[例2]

2

2

=

2

s

+

h

l

d

s

d

l

s

=

l

d

l

d

H

d

t

d

t

=

v =

d

t

d

t

s

V

s

d

s

=

mgH = mv2+ MV2

l

h

l

d

t

a

M

M

H = －

1

1

h

h

v

m

2

2

sina2

sina1

H

m

[例3]

∵m = 0 ∴ DE = 0

= cosa2·V

……

v

k

V

∵ SF = 0

M

m

∴ DP = 0

12

12

12

MV2

mv2

kx2

= +

[例4]

∵只有弹力作功 ∴ DE = 0

mv + MV = 0

……

v10

v20

v1

v2

F1

F2

∵ SF = 0 ∴ DP = 0

m1

m2

m1

m2

m1v1 + m2v2 = m1 v10 + m2 v20

、碰撞

(极为短暂时间的相互作用)

§4

m1v1 + m2v2 = m1 v10 + m2 v20

( m1+ m2)v = m1v10 + m2v20

DEk = 0

[例1]

-m(m+M )gs = ks2 - (m+M )V2

∴Dp = 0

M

1

1

v

2

2

m

A

+

A

= DE

∵ m和M碰撞

∵有摩擦

∴用功能原理

mv = (m+M )V

p129-4-10质量为m1 和m2 的物体同倔强系数为k 的弹簧连结，安放在光滑的水平面上，弹簧开始处在自由长度。现有一质量为m3 的子弹以速度v 沿弹簧长度方向水平射入m1 物体内，求弹簧最大压缩量。

m2

m1

v

m3

∴Dp1 = 0

∴Dp2 = 0

[例2]

∵ m1 和m3 碰撞

m3v = (m1+ m3)v1

∵有弹簧

∴DE2 = 0

m2

m1

v

1

1

1

2

2

2

(m1+m3)v1 = (m1+m2+m3)v2+ kx

m3

2

2

2

1

x = m3v [ — ]

2

1

1

k (m1+m2+m3)

k (m1+m3)

m3v = (m1+ m3)v1

(m1+ m3)v1 = (m1+ m2 + m3)v2

F

B

A

[ 3]

F ≠ 0

M = 0 ,

DL = 0

L = r mv

LA = LB , EkA < EkB

[ E ]

.

F

ds

0

=

.

(0)

A保

E

F

dr

ò

=

pa

a

A

+

A

=

0

E

+

E

E

+

E

C

=

=

pb

pa

ka

kb

= DE

=

E

Δ

p

(

)

(

)

A

+

A

=

E

+

E

E

+

E

ò

pb

ka

pa

kb