Chap. 14 Kinetics of particles: work and energy

14-2

APPLICATIONS

A roller coaster makes use of gravitational forces to assist the cars in reaching high speeds in the “valleys” of the track.

How can we design the track (e.g., the height, h, and the radius of curvature, ρ) to control the forces experienced by the passengers?

14-3

14.1 Work of a Force

質點P的位移：dr = r’ - r

外力F所做的功：

單位： J (N-m) or (ft-lb)

θcosFds

rdFdU

=

⋅=

14-4

Work of a Variable Force

∫=

∫ ⋅=→2

1

2

1

cos21

S

S

r

r

dsF

rdFU

θ

運動位移由S1到S2的曲線 下之面積

14-5

Work of a Constant Force (Fc)

運動軌跡為直線

)(

)(cos

cos

12

212

1

矩形面積

SSF

dsFU

C

S

SC

−=

= ∫→θ

θ

14-6

Work of Weight

kdzjdyidxrdjWW

++=

−=

位移：

重力：

)(

)(

)()(

21

12

21

2

1

2

1

往下做運動時做正功yWU

or

yyW

Wdy

kdzjdyidxjWrdFU

y

y

r

r

Δ−=

−−=

−=

++⋅−=⋅=

→

→

∫

∫∫

14-7

Work of a Spring Force 彈簧力Fs

2

1

2

2

21

21

21

2

11

ksks

ksdsdsFU SSS

S S

−=

∫=∫=→

彈簧力與位移同

向伸長

壓縮

Fs = ks

14-8

質點附著在彈簧上

質點受力與位移反向

)21

21( 21

2

221 ksksU −−=→Fs = -ks

14-9

14.2 Principle of Work and Energy

vdvadsdsdvv

dtds

dsdv

dsds

dtdva

=

=

⋅=

⋅=

有做功僅有切線方向的 tFΣ⋅=⋅Σ rdamrdF

14-10

)(

21T

21

21

21

21

)(

cos

2211

2

21

2221

21

22

2

1

2

1

僅能求切線方向的力

設動能

動能定律

TUT

mv

mvmvU

mvmv

vdvdsamvdvrdF

dsamdsFdsFrdF

t

v

v

r

r

tt

=Σ+⇒

=

←−=Σ⇒

−=

==⋅Σ⇒

=Σ=Σ=⋅Σ

→

→

∫∫ Qθ

14-11

Work of Friction Caused by Sliding

使物體的溫度上升

為內能

其中的差值

應小於的位移實際上

)S'-(S

SS'

21

21 22

N

N

mvNsPsmv

k

k

k

μ

μ

μ =−+

14-12

p. 184, 14-5 The 1.5-kg block slides along a smooth plane and strikes a nonlinear spring with a speed of v = 4 m/s. The spring is termed “nonlinear” because it has a resistance of FS = ks2, where k = 900 N/m2. Determine the speed of the block after it has compressed the spring s = 0.2 m.

14-13

14-14

p. 187, 14-20 Packages having a mass of 7.5 kg are transferred horizontally from one conveyor to the next using a ramp for which μk = 0.15. The top conveyor is moving at 1.8 m/s and the packages are spaced 0.9 m apart. Determine the required speed of the bottom conveyor so no sliding occurs when the packages come horizontally in contact with it. What is the spacing s between the packages on the bottom conveyor?

14-15

14-16

p. 191, 14-36 The 50-kg stone has a speed of vA = 8 m/s when it reaches point A. Determine the normal force it exerts on the incline when it reaches point B. Neglect friction and the stone’s size.

14-17

14-18

14-19

14.4 Power and Efficiency

Def.

單位：W (watt) = J/s = N-m/s1 hp = 550 ft-lb/s= 746 W

效率

vFdt

rdFdtdUP

⋅=

⋅=

= )(功對時間的變化率

1)(

14-20

p. 197, 14-52 The material hoist and the load have a total mass of 800 kg and the counterweight C has a mass of 150 kg. If the upward speed of the

hoist increases uniformly from 0.5 m/s to 1.5 m/s in 1.5 s. Determine the average power generated by the motor M during this time. The motor operates with an efficiency of . ε

= 0.8

14-21

14-22

14-23

14.5 Conservative Forces and Potential Energy Conservative Forces (保守力)Def. 施於一質點的外力所做的功與路徑無關，只是和位

置有關，則此外力稱為一保守力。Ex. 重力、彈簧力

摩擦力：non-conservative force 非保守力Potential Energy (位能)Def. 一保守力由一已知位置至基準線所做的功。

)21

21( 21

22 ksksU

ywU

s

w

−−=

Δ−=

14-24

Gravitational Potential Energy (重力位能)

WyVg =

14-25

Elastic Potential Energy (彈力位能)

)(

)0(21 2

使彈簧回復原狀

彈簧力作用於質點的功

>= ksVe

14-26

Potential Function (位能函數)

2

21 ksWs

VVV eg

+−=

+=

14-27

Potential Function (位能函數)

),,(

),,(),,(

)21

21()(

)21()

21(

21

2212

222

2112121

zyxdV

dzzdyydxxVzyxVdU

ksksssW

ksWsksWsVVU

−=

+++−=

−−−=

+−−+−=−=→

14-28

)(

),,(

),,(),,(

dzzVdy

yVdx

xV

zyxdVdzFdyFdxF

dzFdyFdxF

dzdydxFFFrdFdU

zyx

zyx

zyx

∂∂

+∂∂

+∂∂

−=

−=++⇒

++=

⋅=⋅=又

Potential Function (位能函數)

14-29

Potential Function (位能函數)

)(

)()()()(

)(

,,

保守力需符合此條件

VF

kz

jy

ix

del

Vkz

jy

ix

kzVj

yVi

xVF

zVF

yVF

xVF zyx

−∇=⇒

∂∂

+∂∂

+∂∂

=∇

∂∂

+∂∂

+∂∂

−=

∂∂

−∂∂

−∂∂

−=⇒

∂∂

−=∂∂

−=∂∂

−=⇒

jW

Wyjy

F

jWWFWy

−=

∂∂

−=

===

)(

Vw

14.6 Conservation of Energy (能量守恆)

14-31

14.6 Conservation of Energy

2211

21

222111

0)()(

VTVTU

VTUVT

noncons

noncons

+=+=Σ

+=Σ++

→

→

則

若

14-32

14.6 Conservation of Energy

由 p.174, eq 14-7

321

2333

2222

111

222111

221211

221211

2211

)2(210)3(

2)(

21)2(

0)1(

2014..

)()()(

)()(

EEE

Whghg

WTVE

WhhWghg

WVTE

WhWhTVE

figex

VTUVTTUVVT

TUUTTUT

noncons

noncons

nonconscons

==⇒

=+=+=

=+=+=

=+=+=

−

+=Σ++⇒=Σ+−+⇒

=Σ+Σ+⇒=Σ+

→

→

→→

→

14-33

p. 211, 14-80 The 1.5-kg block A slides in the smooth horizontal slot. When s = 0 the block is given an initial velocity of 18 m/s to the right. Determine the maximum horizontal displacement s of the block. Each of the two springs has a stiffness of k = 2500 N/m and an unstretched length of 0.15 m.

14-34

14-35

p.215, 14-100 The 2-kg collar is released from rest at A and travels along the smooth vertical guide. Determine the speed of the collar when it reaches position B. Also, find the normal force exerted on the collar at this position. The spring has an unstretched length of 200 mm.

14-36

14-37

Chap. 14�Kinetics of particles: work and energy投影片編號 214.1 Work of a ForceWork of a Variable ForceWork of a Constant Force (Fc)Work of Weight Work of a Spring Force 彈簧力Fs 投影片編號 814.2 �Principle of Work and Energy 投影片編號 10Work of Friction Caused by Sliding p. 184, 14-5�The 1.5-kg block slides along a smooth plane and strikes a nonlinear spring with a speed of v = 4 m/s. The spring is termed “nonlinear” because it has a resistance of FS = ks2, where k = 900 N/m2. Determine the speed of the block after it has compressed the spring s = 0.2 m.投影片編號 13p. 187, 14-20�Packages having a mass of 7.5 kg are transferred horizontally from one conveyor to the next using a ramp for which k = 0.15. The top conveyor is moving at 1.8 m/s and the packages are spaced 0.9 m apart. Determine the required speed of the bottom conveyor so no sliding occurs when the packages come horizontally in contact with it. What is the spacing s between the packages on the bottom conveyor?投影片編號 15p. 191, 14-36�The 50-kg stone has a speed of vA = 8 m/s when it reaches point A. Determine the normal force it exerts on the incline when it reaches point B. Neglect friction and the stone’s size. 投影片編號 17投影片編號 1814.4 Power and Efficiencyp. 197, 14-52�The material hoist and the load have a total mass of 800 kg and the counterweight C has a mass of 150 kg. If the upward speed of the hoist increases uniformly from 0.5 m/s to 1.5 m/s in 1.5 s. Determine the average power generated by the motor M during this time. The motor operates with an efficiency of . = 0.8投影片編號 21投影片編號 2214.5 Conservative Forces and Potential Energy Gravitational Potential Energy (重力位能)Elastic Potential Energy �(彈力位能) Potential Function (位能函數) Potential Function (位能函數)投影片編號 28Potential Function (位能函數)14.6 Conservation of Energy �(能量守恆) 14.6 Conservation of Energy14.6 Conservation of Energyp. 211, 14-80�The 1.5-kg block A slides in the smooth horizontal slot. When s = 0 the block is given an initial velocity of 18 m/s to the right. Determine the maximum horizontal displacement s of the block. Each of the two springs has a stiffness of k = 2500 N/m and an unstretched length of 0.15 m. 投影片編號 34p.215, 14-100�The 2-kg collar is released from rest at A and travels along the smooth vertical guide. Determine the speed of the collar when it reaches position B. Also, find the normal force exerted on the collar at this position. The spring has an unstretched length of 200 mm. 投影片編號 36投影片編號 37