Chapter 7 – Kinetic energy, potential energy, workI. Kinetic energy.

II. Work.

III. Work - Kinetic energy theorem.

IV. Work done by a constant force: Gravitational force

V. Work done by a variable force.

- Spring force.- General: 1D, 3D, Work-Kinetic Energy Theorem

VI. Power

VII. Potential energy Energy of configuration

VIII. Work and potential energy

IX. Conservative / Non-conservative forces

X. Determining potential energy values: gravitational potential energy,elastic potential energy

I. Kinetic energyEnergy associated with the state of motion of an object. )1.7(

21 2mvK

Units: 1 Joule = 1J = 1 kgm2/s2 = N m

II. WorkEnergy transferred “to” or “from” an object by means of a force acting onthe object. To +W

From -W

- Constant force: xx maF

dvvadavv xx 2

220

220

2

Work done by the force = Energytransfer due to the force.

Energy: scalar quantity associated with a state (or condition) of one or more objects.

)(211)(

21 2

022

02 vvmdma

dvvmmaF xxx

dFWdFKKvvm xxif )(21 2

02

- To calculate the work done on an object by a force during a displacement,we use only the force component along the object’s displacement. Theforce component perpendicular to the displacement does zero work.

)3.7(cos dFdFdFW x

- Assumptions: 1) F=cte, 2) Object particle-like.

090

90180

90

W

W

Units: 1 Joule = 1J = 1 kgm2/s2

A force does +W when it has a vector component in the same directionas the displacement, and –W when it has a vector component in theopposite direction. W=0 when it has no such vector component.

Net work done by several forces = Sum of works done by individual forces.

2) Fnet Wnet=Fnet d

Calculation: 1) Wnet= W1+W2+W3+…

Fdcos φ

II. Work-Kinetic Energy Theorem

)4.7(WKKK if

Change in the kinetic energy of the particle = Net work done on the particle

III. Work done by a constant force

- Gravitational force: )5.7(cosmgddFW

Rising object: W= mgd cos180º = -mgd Fg transfersmgd energy from the object’s kinetic energy.

Falling object: W= mgd cos 0º = +mgd Fg transfers mgd energy to the object’s kinetic energy.

IV. Work done by a variable force

- External applied force + Gravitational force:

)6.7(gaif WWKKK

Object stationary before and after the lift: Wa+Wg=0

The applied force transfers the same amount ofenergy to the object as the gravitational force transfers from the object.

- Spring force: )7.7(dkF

Hooke’s law

k = spring constant measures spring’sstiffness.

Units: N/m

kxFD x 1

Work done by a spring force:

Hooke’s law

- Assumptions: • Spring is massless mspring << mblock• Ideal spring obeys Hooke’s law exactly.• Contact between the block and floor is frictionless.• Block is particle-like.

2) F(x) ≈ cte within each short ∆x segment.

- Calculation:x

Fx

xi xf∆x

Fj1) The block displacement must be divided into many segments of infinitesimal width, ∆x.

Work done by an applied force + spring force:

dxkxdxFWxxFW f

i

f

i

xx

xxsjs )(0

22

21

21

fis xkxkW Ws=0 If Block ends up at xf=xi.

021 2 ifs xifxkW

saif WWKKK

Block stationary before and after the displacement: ∆K=0Wa= -Ws

The work done by the applied force displacing the block is the negativeof the work done by the spring force.

)(21

21 222

ifxx

x

xS xxkxkdxxkW f

i

f

i

Work done by a general variable force:

1D-Analysis

)10.7()(

0,

,0

,

,

lim

f

i

x

xavgj

x

avgjj

avgjj

dxxFWxFW

xxmoreionapproximatbetterxFWW

xFW

Geometrically: Work is the area between the curve F(x) and the x-axis.

Work-Kinetic Energy Theorem - Variable force

f

i

f

i

x

x

x

x

dxmadxxFW )(

3D-Analysis

)(),(),(;ˆˆˆ zFFyFFxFFkFjFiFF zyxzyx

kdzjdyidxrd ˆˆˆ

f

i

f

i

f

i

f

i

z

zz

y

yy

x

xx

r

rzyx dzFdyFdxFdWWdzFdyFdxFrdFdW

dxdtdvmdxma

KKKmvmvdvvmdvmvW ifif

v

v

v

v

f

i

f

i

22

21

21

v

dxdv

dtdx

dxdv

dtdv

mvdvdxvdxdvm

V. Power

Time rate at which the applied force does work.

- Average power: amount of work done in an amount of time ∆t by a force.

- Instantaneous power: instantaneous time rate of doing work.

)12.7(t

WPavg

)13.7(dt

dWP

Units: 1 Watt= 1 W = 1J/s

1 kilowatt-hour = 1 kW·h = 3.60 x 106 J = 3.6 MJ

)14.7(coscoscos

vFFvdtdxF

dtdxF

dtdWP

x

Fφ

54. In the figure (a) below a 2N force is applied to a 4kg block at a downwardangle θ as the block moves rightward through 1m across a frictionless floor. Findan expression for the speed vf at the end of that distance if the block’s initialvelocity is: (a) 0 and (b) 1m/s to the right. (c) The situation in (b) is similar in thatthe block is initially moving at 1m/s to the right, but now the 2N force is directeddownward to the left. Find an expression for the speed of the block at the end ofthe 1m distance.

N

mg

Fx

Fy

)(5.0

)cos(20

2 vvmKW

dFdFW

f

N

mg

Fx

Fy

smv

JvkgN

JmvKsmvc

f

f

f

/cos1

2)4(5.0cos)2(

25.0/1)(2

20

smv

vkgN

mvKva

f

f

f

/cos

)4(5.0cos)2(

5.00)(2

20

smv

JvkgN

smkgmvKsmvb

f

f

f

/cos1

2)4(5.0cos)2(

)/1()4(5.05.0/1)(2

220

18. In the figure below a horizontal force Fa of magnitude 20N is applied to a 3kgpsychology book, as the book slides a distance of d=0.5m up a frictionless ramp.(a) During the displacement, what is the net work done on the book by Fa, thegravitational force on the book and the normal force on the book? (b) If the bookhas zero kinetic energy at the start of the displacement, what is the speed at theend of the displacement?

workdoFFOnlyWdN

axgx ,0

xy

mg

N

Fgy

Fgx

JmNNWmgFgFaF

dFWorWWWa

net

xxnet

netnetxFgxFa

31.15.0)7.1432.17(30sin30cos20

)(

smvmvJW

KKWKb

ff

f

/93.05.031.1

0)(2

0

55. A 2kg lunchbox is sent sliding over a frictionless surface, in the positivedirection of an x axis along the surface. Beginning at t=0, a steady wind pusheson the lunchbox in the negative direction of x, Fig. below. Estimate the kineticenergy of the lunchbox at (a) t=1s, (b) t=5s. (c) How much work does the forcefrom the wind do on the lunch box from t=1s to t=5s?

)10(101 ttx

paraboladownwardconcaveMotion

JsmkgK

smvsta

f

f

64.0)/8.0)(2(5.0

/8.01)(2

JK

vstb

f

f

0

05)(

JW

sKsKKWc ff

64.064.00

)1()5()(

2/2.01021021

smdtdva

tdtdxv

)1.0)(4.0(

4.0)/2.0)(2(2

2

ttNxFW

NsmkgmacteF

74. (a) Find the work done on the particle by the force represented inthe graph below as the particle moves from x=1 to x=3m. (b) The curveis given by F=a/x2, with a=9Nm2. Calculate the work using integration

JNmsquaresW

curveunderAreaWa

75.5)1)(5.0)(5.11(

)(

Jx

dxx

Wb 6)131(9199)(

3

1

3

12

73. An elevator has a mass of 4500kg and can carry a maximum load of1800kg. If the cab is moving upward at full load at 3.8m/s, what power isrequired of the force moving the cab to maintain that speed?

mg

Fa

kNsmkgmgF

FFFgmF

kgkgkgm

a

ganeta

total

74.61)/8.9)(6300(

00

630018004500

2

kWPsmkNvFP

61.234)/8.3)(74.61(

A single force acts on a body that moves along an x-axis. The figure below showsthe velocity component versus time for the body. For each of the intervals AB, BC,CD, and DE, give the sign (plus or minus) of the work done by the force, or statethat the work is zero.

v

tA

B C

D

20

20 2

1 vvmKKKW ff

0 WvvBC BC

0 WvvCD CD

00,0 WvvDE DE

0 WvvAB AB

E

50. A 250g block is dropped onto a relaxed vertical spring that has a springconstant of k=2.5N/cm. The block becomes attached to the spring andcompresses the spring 12 cm before momentarily stopping. While the spring isbeing compressed, what work is done on the block by (a) the gravitational force onit and (b) the spring force? (c) What is the speed of the block just before it hits thespring? (Friction negligible) (d) If the speed at impact is doubled, what is themaximum compression of the spring?

JmsmkgmgddFWa gFg 29.0)12.0)(/8.9)(25.0()( 2

mgJmmNkdWb s 8.1)12.0)(/250(5.0

21)( 22

smvvkgJJ

WWmvKKKv

mvmvKWc

i

i

sFgiiff

ifnet

/47.3)25.0(5.08.129.0

5.000

5.05.0)(

2

2

22

mdmvKkdmgdW

vncompressiospringMaximumsmvIfd

inet

fi

23.0''5.0'5.0'

0?/95.6')(22

d mg

Fs

62. In the figure below, a cord runs around two massless, frictionless pulleys; acanister with mass m=20kg hangs from one pulley; and you exert a force F on thefree end of the cord. (a) What must be the magnitude of F if you are to lift thecanister at a constant speed? (b) To lift the canister by 2cm, how far must you pullthe free end of the cord? During that lift, what is the work done on the canister by(c) your force (via the cord) and (d) the gravitational force on the canister?

NmgFFTcordHand

NTmgTFctevPulleya net

982

0:

98020:1)(

mg

TT T

P1

P2

JmNdFWc F 92.3)04.0)(98()(

(b) To rise “m” 0.02m, two segments of the cord mustbe shorten by that amount. Thus, the amount of thestring pulled down at the left end is: 0.04m

JsmkgmmgdWd Fg 92.3)/8.9)(20)(02.0()( 2

WF+WFg=0 There is no change in kinetic energy.

I. Potential energyEnergy associated with the arrangement of a system of objects that exeforces on one another.

Units: J Examples:

- Gravitational potential energy: associated with the state of separationbetween objects which can attract one another via the gravitational for

- Elastic potential energy: associated with the stateofcompression/extension of an elastic object.II. Work and potential energy

If tomato rises gravitational force transfers energy“from” tomato’s kinetic energy “to” the gravitationalpotential energy of the tomato-Earth system.

If tomato falls down gravitational force transfersenergy “from” the gravitational potential energy “to”the tomato’s kinetic energy.

WU Also valid for elastic potential energy

Spring force does –W on block energytransfer from kinetic energy of the block topotential elastic energy of the spring.

Spring force does +W on blockenergy transfer from potential energyof the spring to kinetic energy of theblock.

General:

- System of two or more objects.

- A force acts between a particle in the system and the rest of the system.

fs

fs

Spring compression

Spring extension

- When the configuration change is reversed force reverses the energytransfer, doing W2.

III. Conservative / Nonconservative forces

- If W1=W2 always conservative force.

Examples: Gravitational force and spring force associated potentialenergies.

- If W1≠W2 nonconservative force.

Examples: Drag force, frictional force KE transferred into thermalenergy. Non-reversible process.

- When system configuration changes force does work on the object (W1) transferring energy between KE of the object and some other form of energy of the system.

- Thermal energy: Energy associated with the random movement of atomsand molecules. This is not a potential energy.

- Conservative force: The net work it does on a particle moving aroundevery closed path, from an initial point and then back to that point iszero.

Conservative forceWab,1= Wab,2

Wab,1+ Wba,2=0 Wab,1= -Wba,2

Wab,2= - Wba,2

IV. Determining potential energy values

f

i

xx UdxxFW )( Force F is conservative

Gravitational potential energy:Change in the gravitationalpotential energy of theparticle-Earth system.

f

i

f

i

yy if

yy ymgyymgymgdymgU )()(

- The net work it does on a particle moving between two points does not depend on the particle’s path.

Proof:

Wab,2= Wab,1

mgyyUyU ii )(0,0

The gravitational potential energy associated with particle-Earthsystem depends only on particle’s vertical position “y” relative to thereference position y=0, not on the horizontal position.

Reference configuration

Elastic potential energy:

Change in the elastic potential energy of the spring-block system.

222

21

21

2)( i

xx f

xx kxkxxkdxkxU f

i

f

i

2

21)(0,0 kxxUxU ii

Reference configuration when the spring is at its relaxed length and thblock is at xi=0.

Remember! Potential energy is always associated with asystem.V. Conservation of mechanical energyMechanical energy of a system: Sum of its potential (U) and kinetic (K)energies.

Emec= U + K

Assumptions: - Only conservative forces cause energy transfer withinthe system.

UWKW

11221212 0)()(0 UKUKUUKKUK

- In an isolated system where only conservative forces cause energychanges, the kinetic energy and potential energy can change, buttheir sum, the mechanical energy of the system cannot change.

∆Emec= ∆K + ∆U = 0

- When the mechanical energy of a system is conserved, we canrelate the sum of kinetic energy and potential energy at one instantto that at another instant without considering the intermediatemotion and without finding the work done by the forces involved.

- The system is isolated from its environment No external force from an object outside the system causes energy changes inside the system.

Emec= constant

1122

0

UKUK

UKEmec

Potential energy curves

Finding the force analytically:

x

y

)1()()()()( motionDdx

xdUxFxxFWxU

- The force is the negative of the slope of the curve U(x) versus x.

- The particle’s kinetic energy is: K(x) = Emec – U(x)

Turning point: a point x atwhich the particlereverses its motion (K=0).

Equilibrium points: where the slope of the U(x) curve is zero F(x)=0∆U = -F(x) dx ∆U/dx = -F(x)

K always ≥0 (K=0.5mv2 ≥0 )

Examples:

x= x1 Emec= 5J=5J+K K=0

x<x1 Emec= 5J= >5J+KK<0 impossible

Example: x ≥ x5 Emec,1= 4J=4J+K K=0 and also F=0x5 neutral equilibrium

Emec,1

Emec,2

Emec,3

∆U(x)/dx = -F(x) Slope

Equilibrium points

x4 Emec,3=1J=1J+K K=0, F=0, it cannot move to x>x4 or x<x4, since then K<0

Stable equilibrium

x2>x>x1, x5>x>x4 Emec,2= 3J= 3J+K K=0 Turning points

x3 K=0, F=0 particle stationary Unstable equilibrium

Review: Potential energy

W = -∆U

- The zero is arbitrary Only potential energy differences havephysical meaning.

- The force (1D) is given by: F = -dU/dx

- The potential energy is a scalar function of the position.

P1. The force between two atoms in a diatomic molecule can berepresented by the following potential energy function:

612

0 2)(xa

xaUxU where U0 and a are constants.

i) Calculate the force Fx

713

07613120

5

2

11

20

121212

6212)()(

xa

xa

aUxaxaU

xa

xa

xa

xaU

dxxdUxF

ii) Minimum value of U(x).

00

7130

min

21)(

0120)()()(

UUaUax

xa

xa

aUxF

dxxdUifxU

U0 is approx. the energy necessary to dissociate the twoatoms.