This presentation is the property of its rightful owner.
Sponsored Links
1 / 39

效实高考在线 PowerPoint PPT Presentation


  • 142 Views
  • Uploaded on
  • Presentation posted in: General

物理 第三讲. 效实高考在线. 第三讲 牛顿运动定律. 本章知识不仅是力学的基础,对整个高中阶段的物理学习都有重要的作用。. 复习要点. 一、理解基本概念,准确作出判断. 二、善做两类分析,搞清运动因果. 三、掌握解题技巧,灵活解决问题. 牛顿第一定律. 惯性. 牛顿第二定律 F 合 =ma. 牛顿运动定律. 牛顿第三定律 F=-Fˊ. a. 力 运动. 应 用. 超重与失重. 连结体问题. 本章的知识结构如下所示:. S=v 0 t+at 2 /2 V t =v 0 +at V=(v 0 +v t )/2.

Download Presentation

效实高考在线

An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Presentation Transcript


4140240


4140240


4140240


4140240

F=ma

F=-F

a

S=v0t+at2/2

Vt=v0+at

V=(v0+vt)/2


4140240

1;

2;

3;

4


4140240

1

2

3


4140240

1

2

3


4140240

  • 1

  • A. B.

  • C. D.


4140240

F=ma

1

2

3

4


4140240

(F=-F)

1

2

3

4


4140240

1

2

3


4140240

a

Ty

T

1

Tx

mg

2

[2] 1a.

[] 2mgTT:

Ty--mg= 0

Tx=ma

Tcos= mg Tsin= ma

T: tan= a/g


4140240

T

ma

mg

3

:

1T3:

tan= a/g


4140240

a

4

N

ma

mg

2a=gtantan, a4TN5aa=gtan

5


4140240

a

6

[3]6ma.


4140240

Fy

F

ma

mg

7

[]17F1

F=mg/cosF ma/sin

,?


A 0 f 0

: a=0F=0

mg


4140240

Fy

F

ma

mg

7

7F

Ftan

tan=ma/mg=a/g


4140240

8a5FagtanF

F5

F4

F3

F2

F1

ma3

ma4

ma5

ma2

mg

8


4140240


4140240

9

[4]9ABCAB

A

B

C


4140240

T

F1

F1

B

A

A

mg

C

T

mg

[] A

F=F1-mg-T

A

A

F= F1mgmg

A

aA=2g

mg

aB=2gaC=0


4140240

a

m

N

f

mgsin

mgcos

mg

4 10ma


4140240

N

f

mgsin

mgcos

mg

N-mgcos=0

mgsin-f=ma

f=N


4140240

[]

a

a=g(sin-cos)1


4140240

ff

a=g(sincos)


4140240

12

a=g3


4140240

a=g(sincos)1

a=g(sincos)2

a=g3


4140240


4140240

A

F

B

[5] MAlmBFBAT


4140240

T

N

A

F

B

mg

F

mg

a>a0

a<a0

ABFa


4140240

A

B

mg

a=a0

a0

Tsin=mg Tcos=ma0

a0 =gcot T=mg/sin

T


A a 0

F

a=

M+m

T1

mg

F

a>a0

a>a0

T1sin=mg T1cos=ma

T1=m g2+F2/(m+M)2


A a 01

X

F

a=

a

T

mF

M+m

N

F= mgsin+ cos

M+m

A

F

B

mg

a<a0

a<a0:

XNcos+Tsin=mg

YTcos- Nsin=ma

Y


4140240


4140240


4140240


4140240

F

F

a

a

mg

mg

60

2.5 m/s280 kg______kg40 kg_____ m/s2g10 m/s2

5


4140240


  • Login