Work and Kinetic EnergyWork done by a constant forceWork is a scalar quantity.No motion (s=0) → no work (W=0)

cosFssFW

Units: [ W ] = newton·meter = N·m = J = joule (SI) [ W ] = dyne·cantimeter = dyn·cm = erg (CGS) 1 J = 1 N · 1 m = 103 g 100 (cm/s2) 100 cm = 107 erg

Particular cases: (i) φ = 900 → cos φ = 0 → W = 0 (no work) (ii) φ = 1800 → cos φ = -1 → W = - Fs ≤ 0 (negative work)

forcenetaisFFj

j

James Joule (1818 – 1889)sF

F

s

Kinetic Energy and Work-Energy Theorem

constaand

constFIf

Work and Energy with Varying Force

2

1

x

x

xdxFW

(1D-motion)

Fx =const → W = Fx (x2 – x1)Particular cases:

Fx = -k·x (Hooke’s law) → 2

1

21

22 2

1

2

1x

x

kxkxkxdxW

Work-Energy Theorem for 1D-Motion under Varying Forces

X

xF

m 2

1

2

1

x

x

xj

x

x

xjxj dxmadxFxFW

xxxxx

x dvvvdvdt

dxdvdxa )(

2

1

21

22 2

1

2

1x

x

v

v

xxxx mvmvdvmvW

KKKW 12is kinetic energy2

2

1mvK

Example 6.7: Air-track glider attached to springData: m=0.1 kg, v0=1.5m/s, k=20 N/m, μk = 0.47Spring was unstretched.Find: maximum displacement dSolution:

,2

1,

2

1

0

220

d

springfrictionspring kdkxdxWmvWW

k

kmvmgmgdmvmgdkd kk

k

20

220

2 )(

2

1

2

1

mgddfW kkfriction

Work-Energy Theorem for 3D-Motion along a Curve

1221

22

,,

21

22

,,

),,(

),,(

||

||

2

1

2

1

2

1

2

1

)(

cos

cos

2

1

222

111

2

1

2

1

2

1

KKvmvmmvmv

dvmvdzFdyFdxF

ldFdlFdlFdWW

ldFdlFdlFdW

zyxjjj

zyxj

v

v

jjz

zyx

zyx

yx

P

p

P

p

P

p

j

j

Line integral

12 KKWMotionnalTranslatioforTheoremEnergyWork

Powerv

Average powert

WPP av

Instantaneous power dt

dW

t

WP

lim

Units: [ P ] = [ W ] / [ T ] , 1 Watt = 1 W = 1 J / s1 horsepower = 1 hp = 550 ft·lb/s = 746 W = 0.746 kW

Related energy unit 1 kilowatt-hour = 1 kWh = (1000 J/s) 3600 s = 3.6·106 J = 3.6 MJ

vFPdt

sdF

dt

dWP

0t

Power is a scalar quantity.

Power is the time rate at which work is done,or the rate at which the energy is changing.These rates are the same due to work-energytheorem.

James Watt(1736-1819), the developerof steam engine.

Exam Example 13: Stopping Distance (problems 6.31, 7.27)

x

v

0a

Data: v0 = 50 mph, m = 1000 kg, μk = 0.5 Find: (a) kinetic friction force fkx ;(b)work done by friction W for stopping a car;(c)stopping distance d ;(d)stopping time T;(e) friction power P at x=0 and at x=d/2;(f) stopping distance d’ if v0’ = 2v0 .

NF

gm

kf

Solution:(a)Vertical equilibrium → FN = mg → friction force fkx = - μk FN = - μk mg .(b) Work-energy theorem → W = Kf – K0 = - (1/2)mv0

2 .

(c) W = fkxd = - μkmgd and (b) yield μkmgd = (1/2)mv02 → d = v0

2 / (2μkg) . Another solution: second Newton’s law max= fkx= - μkmg → ax = - μkg and from kinematic Eq. (4) vx

2=v02+2axx for vx=0 and x=d we find

the same answer d = v02 / (2μkg) .

(d) Kinematic Eq. (1) vx = v0 + axt yields T = - v0 /ax = v0 / μkg . (e) P = fkx vx → P(x=0) = -μk mgv0 and, since vx

2(x=d/2) = v02-μkgd = v0

2 /2 , P(x=d/2) = P(x=0)/21/2 = -μkmgv0 /21/2 .(f) According to (c), d depends quadratically on v0 → d’ = (2v0)2/(2μkg) = 4d

Exam Example 14: Swing (example 6.8)Find the work done by each force if(a) F supports quasi-equilibrium or(b) F = const ,as well as the final kinetic energy K.

Solution:

(a) Σ Fx = 0 → F = T sinθ , Σ Fy = 0 → T cosθ = w = mg , hence, F = w tanθ ; K = 0 since v=0 .

WT =0 always since ldT

Rddsdl

0000

)cos1(sincostancos wRdwRRdwdsFldFWF

0

0 0

)cos1(sin

)sin(

wRdwR

RdwldwWgrav

0 0

sincos)( FRRdFldFWb F

Data: m, R, θ

)cossin( wwFRWWWWK TgravF