7.4 Work Done by a 7.4 Work Done by a Varying ForceVarying Force

Work Done by a Varying Work Done by a Varying ForceForce

Assume that during a very small displacement, x, F is constant

For that displacement, W ~ F x

For all of the intervals,f

i

x

xx

W F x

Work Done by a Varying Work Done by a Varying Force, contForce, cont

Sum approaches a definite value:

Therefore:

(7.7)(7.7)

The work done is equal The work done is equal to the area under the to the area under the curve!!curve!!

lim0

ff

ii

xx

x x xxx

F x F dx

f

i

x

xxW F dx

Example 7.7 Example 7.7 Total Work Done from a Total Work Done from a Graph Graph (Example 7.4 Text Book)(Example 7.4 Text Book)

The net worknet work done by this force is the area the area under the curveunder the curve

WW = Area under the = Area under the Curve Curve

WW = = AAR R + A+ ATT

W =W = (B)(h)(B)(h) ++ (B)(h)/2 =(B)(h)/2 = (4m)(5N)(4m)(5N) ++ (2m)(5N)/2(2m)(5N)/2

W =W = 20J + 5J =20J + 5J = 25 J25 J

Work Done By Work Done By Multiple ForcesMultiple Forces

If more than one force acts on a system and the system can be the system can be modeled as a particlemodeled as a particle, the total total workwork done ONON the system is the work done by the net forcework done by the net force

(7.8)(7.8) f

i

x

net xxW W F dx

Work Done by Work Done by Multiple Forces, cont.Multiple Forces, cont.

If the system cannot be modeled as cannot be modeled as a particlea particle, then the total worktotal work is equal to the algebraic sumthe algebraic sum of the work done by the individual forces

net by individual forcesW W

Hooke’s LawHooke’s Law

The force exerted The force exerted BYBY the spring is the spring is

FFss = = ––kkx x (7.9)(7.9) xx is the position of the block with respect to the

equilibrium position (xx = 0 = 0) kk is called the spring constantspring constant or force constantforce constant

and measures the stiffness of the spring (Units: N/m)(Units: N/m) This is called Hooke’s LawHooke’s Law

Hooke’s Law, cont.Hooke’s Law, cont. When xx is positive positive

(spring is stretched), FFss is negativenegative

When xx is 00 (at the equilibrium position), FFss is 00

When x is negativenegative (spring is compressed), FFss is positivepositive

Hooke’s Law, finalHooke’s Law, final The force exerted by the spring (FFss )

is always directed oppositeopposite to the displacementdisplacement from equilibrium

FFss is called the restoring forcerestoring force If the block is released it will

oscillateoscillate back and forth between ––xx and xx

Work Done by a Work Done by a SpringSpring

Identify the blockblock as the systemsystem

The workwork as the block moves from:

xxii = = –– xxmaxmax toto xxff = 0 = 0

max

0 2max

1

2

f

i

x

s xx xW F dx kx dx kx

(7.10)(7.10)

Work Done by a Work Done by a Spring, contSpring, cont

The workwork as the block moves from:

xxii = 0 = 0 toto xxff = = xxmaxmax

(7.10) (a)(7.10) (a)2max2

1

0

max

)( kxdxkxWx

s

Work Done by a Work Done by a Spring, finalSpring, final

Therefore: Net WorkNet Work done by the

spring forcespring force as the block moves from –x–xmaxmax to xxmaxmax is ZERO!!!!ZERO!!!!

For any arbitrary displacement: xxii to xxff :

2212

212

212

21)( fiif

x

xs kxkxkxkxdxkxWf

i

(7.11)(7.11)

Spring with an Spring with an Applied ForceApplied Force

Suppose an external agent, FFappapp, stretches the spring

The applied forceapplied force is equalequal and oppositeopposite to the spring force:spring force:

FFappapp == ––FFss == ––((––kxkx))

FFappapp = = kxkx

Spring with an Spring with an Applied Force, finalApplied Force, final

Work done by FFappapp

when xxii = 0 = 0 toto xxff = = xxmaxmax is:

WWFappFapp = ½ = ½kxkx22maxmax

For any arbitrary displacement: xxii to xxff :

2212

21

if

x

x

x

x appFapp kxkxdxkxdxFWf

i

f

i

(7.12)(7.12)

Active Figure 7.10Active Figure 7.10

Kinetic EnergyKinetic Energy is the energy of a particle due to its motion KK = ½ = ½ mvmv2 2 (7.15) (7.15)

KK is the kinetic energy mm is the mass of the particle vv is the speed of the particle

Units of KK:: Joules (J)Joules (J)

1 J = N1 J = N••m = (m = (kgkg••m/sm/s22)m= kg)m= kg••mm22/s/s2 2 = kg(m/s) = kg(m/s)22

A change change in kinetic energy kinetic energy is one possible result of doing work to transfer energytransfer energy into a system

7.5 Kinetic Energy And the 7.5 Kinetic Energy And the Work-Kinetic Energy Work-Kinetic Energy TheoremTheorem

Kinetic Energy, contKinetic Energy, cont Calculating the work:

Knowing that:F = ma = mdv/dt F = ma = mdv/dt

=m(dv/dt)(dx/dx) =m(dv/dt)(dx/dx) Fdx = m(dv/dx)(dx/dt)dx = mvdvFdx = m(dv/dx)(dx/dt)dx = mvdv

(7.14)(7.14)2 21 1

2 2

f f

i i

f

i

x x

x x

v

v

f i

W F dx ma dx

W mvdv

W mv mv

Work-Kinetic Energy Work-Kinetic Energy TheoremTheorem

The Work-Kinetic Energy PrincipleWork-Kinetic Energy Principle states

WW = = KKff – – KKii = = K K (7.16)(7.16) In the case in which work is done on a

system and the only change in the system is in its speed, the work donework done by the net force equals the change in kineticchange in kinetic energy of the system.

We can also define the kinetic energy KK = ½ = ½ mvmv2 2 (7.15) (7.15)

Work-Kinetic Energy Theorem, Work-Kinetic Energy Theorem, contcont

Summary:Summary: Net work done by a constant force in accelerating an object of mass mm from vv11 to vv22 is:

WWnetnet = ½ = ½mmvv222 2 –– ½ ½mmvv11

2 2 KK

““Net work on an object = Change in Net work on an object = Change in Kinetic Energy”Kinetic Energy”

It’s been shown for a one-It’s been shown for a one-dimension constant force.dimension constant force. However, this is valid in general!!!However, this is valid in general!!!

Work-Kinetic Energy Work-Kinetic Energy REMARKS!!REMARKS!!

WWnetnet ≡ work done by the net (total) force. WWnetnet is a scalar.scalar.

WWnetnet can be positivepositive or negativenegative since KK can be both ++ or ––

KK ½ ½mmvv22 is always positivepositive. MassMass and vv22 are both positive. (Question 10 Homework)(Question 10 Homework)

UnitsUnits are JoulesJoules for both work & kinetic energy. The work-kinetic theorem:The work-kinetic theorem: relates work to a

change in speedspeed of an object, not to a change in its velocityvelocity.

Example 7.8Example 7.8 Question Question #14#14

(a).(a). K Kii ½ ½m m vv2 2 ≥ 0 ≥ 0

KK depends on:depends on: vv2 2 ≥ 0 ≥ 0 && mm > 0 > 0If vv 22vv

KKff = = ½½m m ((22vv))22 = 4(½ = 4(½mmvv2 2 ) = ) = 44KKii

Then: Doubling Doubling the speed makes an object’s kinetic energy four timesfour times larger

(b).(b). If If WW = 0 = 0 vv must be the samesame at the final point as it was at the initial point

Example 7.9Example 7.9 Work-Kinetic Work-Kinetic Energy Theorem Energy Theorem (Example 7.7 (Example 7.7 Text Book)Text Book)

m = 6.0kgm = 6.0kg first at rest is pulled to the right with a force F = 12NF = 12N (frictionless).

Find v after mm moves 3.0m Solution:Solution:

The normalnormal and gravitationalgravitational forces do no workdo no work since they are perpendicular to the direction of the displacement

WW = = F F x x == (12)(3)J = 36J(12)(3)J = 36J WW = = KK = ½ = ½ mvmvff

22 –– 0 0 36J = ½(6.0kg)36J = ½(6.0kg)vvff

22 = (3kg) = (3kg)vvff22

VVf f =(36J/3kg)=(36J/3kg)½½ = = 3.5m/s3.5m/s

Example 7.10Example 7.10 Work to Stop a Work to Stop a CarCar

WWnenett = Fdcos180 = Fdcos180°= °= ––FdFd = = ––FdFd

WWnetnet = = KK = = ½½mmvv222 2 –– ½ ½mmvv11

2 2 = = ––Fd Fd

-Fd = 0 -Fd = 0 –– ½ ½m m vv112 2 dd vv11

22

If the car’s initial speed doubledspeed doubled, the stopping distance is 4 times greater4 times greater. Then: d = 80 md = 80 m

A moving hammer strikes a nail and comes to rest. The hammer exerts a force FF on the nail, the nail exerts a force –F–F on the hammer (Newton's 3rd Law) mm

Work done on the nail is Work done on the nail is positive:positive:

Wn = KKnn = = FFd = ½md = ½mnnvvnn22 – 0 > 0 – 0 > 0

Work done on the hammer is Work done on the hammer is negative:negative:

WWhh = = KKhh = = ––FFd = 0 – ½md = 0 – ½mhhvvhh22 < 0 < 0

Example 7.11Example 7.11 Moving Moving Hammer can do Work on NailHammer can do Work on Nail

Example 7.12Example 7.12 Work on a Car to Work on a Car to Increase Kinetic EnergyIncrease Kinetic Energy

Find WWnetnet to accelerate the 1000 kg car.

WWnetnet = = KK = = KK2 2 – K– K1 1 = ½= ½m m vv222 2 –– ½ ½m m vv11

22

WWnetnet == ½(10½(1033kg)(30m/s)kg)(30m/s)2 2 –– ½(10 ½(1033kg)(20m/s)kg)(20m/s)2 2 WWnetnet == 450,000J – 200,000J = 450,000J – 200,000J = 2.50x102.50x1055JJ

Example 7.13Example 7.13 Work and Work and Kinetic Energy on a BaseballKinetic Energy on a Baseball

A 145-g145-g baseball is thrown so that acquires a speed of 25m/s25m/s. ( Remember: vv11 = 0 = 0)Find: (a). Its KK. (b). WWnetnet on the ball by the pitcher.

(a). KK ½ ½mmvv22 = ½(0.145kg)(25m/s)2 KK 45.0 J45.0 J

(b). WWnetnet = KK = = K2 –– K1 = 45.0J – 0J = 45.0J – 0J

WWnetnet = = 45.0 J45.0 J

A nonisolated systemnonisolated system is one that interacts with or is influenced by its environment An isolated systemisolated system would not interact with

its environment The Work-Kinetic EnergyWork-Kinetic Energy Theorem can

be applied to nonisolated systemsnonisolated systems

7.6 Non-isolated 7.6 Non-isolated System System

Internal EnergyInternal Energy The energy associated

with an object’s temperature is called its internal energyinternal energy, , EEintint

In this example, the the surfacesurface is the systemthe system

The frictionThe friction does work and increasesincreases the internal energyinternal energy of the surface

Active Figure 7.16Active Figure 7.16

Potential EnergyPotential Energy Potential energyPotential energy is energy related to

the configurationconfiguration of a system in which the components of the system interact interact by forcesby forces

Examples include:Examples include: elastic potential energy – stored in a elastic potential energy – stored in a

springspring gravitational potential energygravitational potential energy electrical potential energyelectrical potential energy

Ways to Transfer Energy Ways to Transfer Energy Into or Out of A SystemInto or Out of A System

WorkWork – transfers by applying a force and causing a displacement of the point of application of the force

Mechanical WavesMechanical Waves – allow a disturbance to propagate through a medium

HeatHeat – is driven by a temperature difference between two regions in space

More Ways to Transfer More Ways to Transfer Energy Into or Out of A Energy Into or Out of A SystemSystem

Matter TransferMatter Transfer – matter physically crosses the boundary of the system, carrying energy with it

Electrical TransmissionElectrical Transmission – transfer is by electric current

Electromagnetic RadiationElectromagnetic Radiation – energy is transferred by electromagnetic waves

Examples of Ways to Examples of Ways to Transfer EnergyTransfer Energy

a) Worka) Work

b) b) Mechanical Mechanical WavesWaves

c) Heatc) Heat

d) Matter d) Matter transfertransfer

e) Electrical e) Electrical TransmissionTransmission

f) f) Electromagnetic Electromagnetic radiationradiation

Conservation of EnergyConservation of Energy Energy is conservedEnergy is conserved

This means that energy cannot be created or cannot be created or destroyeddestroyed

If the total amount of energy in a system changes, it can only be due to the fact that energy has crossed the boundary of the system by some method of energy transfer

Mathematically: Mathematically: systemsystem(7.17)(7.17) EEsystemsystem is the total energy of the system TT is the energy transferred across the system boundary Established symbols: TTworkwork = = WW and TTheatheat = = QQ

The Work-Kinetic Energy theoremWork-Kinetic Energy theorem is a special case of Conservation of EnergyConservation of Energy

Examples to Read!!!Examples to Read!!! Example 7.7Example 7.7 (page 195) Example 7.9Example 7.9 (page 201) Example 7.12Example 7.12 (page 204)

Homework to be solved in Homework to be solved in Class!!!Class!!! Problems: 11, 26Problems: 11, 26

Material for the FinalMaterial for the Final