Kinetic Energy and Work

Chapter 7

Work and Energy

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

Work and energy are scalars, measured in NWork and energy are scalars, measured in N··m or Joules, J, m or Joules, J,

1J=kg1J=kg∙m∙m22/s/s22

Energy can exist in many forms - mechanical, electrical, Energy can exist in many forms - mechanical, electrical,

nuclear, thermal, chemical….nuclear, thermal, chemical….

Work and EnergyWork and Energy

Energy is a conserved quantity - the total Energy is a conserved quantity - the total amount of energy in the universe is amount of energy in the universe is constant.constant.

Energy can be converted from one type to Energy can be converted from one type to

another but never destroyed.another but never destroyed.

Work and energy concepts can simplify solutions of Work and energy concepts can simplify solutions of

mechanical problems - they can be used in an mechanical problems - they can be used in an

alternative analysisalternative analysis

Work done by a constant forceWork done by a constant force

Work done on an object by a constant force is Work done on an object by a constant force is defined to be the product of the magnitude of the defined to be the product of the magnitude of the displacementdisplacement and the component of the and the component of the forceforce parallel to the displacementparallel to the displacement

Where Where FFIIII is the component of the force is the component of the force FF parallel parallel

to the displacement to the displacement dd

W = FW = FII II ·· d d

WorkWorkIn other words - In other words -

Where Where is the angle between is the angle between FF and and dd

If If is > 90 is > 90oo, work is negative. A decelerating car has , work is negative. A decelerating car has negative work done on it by its engine.negative work done on it by its engine.

The unit of work is called The unit of work is called JouleJoule (J), (J), 1 J = 1 N1 J = 1 N··mm

1J=kg1J=kg∙m∙m22/s/s22

d

F

W = F d cos

Work - on and byWork - on and by

A person pushes block 30 m along the ground A person pushes block 30 m along the ground by exerting force of 25 N on the trolley. How by exerting force of 25 N on the trolley. How much work does the person do on the trolley?much work does the person do on the trolley?

W = F d = 25N x 30m = 750 NmW = F d = 25N x 30m = 750 Nm

Trolley does -750 Nm work on the personTrolley does -750 Nm work on the person

FptFtp

d

Mechanical EnergyMechanical Energy

Mechanical energy (energy associated with

masses) can be thought of as having two

components: kinetic and potential– Kinetic energy is energy of motion

– Potential energy is energy of position

Kinetic EnergyKinetic Energy

In many situations, energy can be considered In many situations, energy can be considered asas “the ability to do work”“the ability to do work”

Energy can exist in different formsEnergy can exist in different forms

Kinetic energy is the energy associated with Kinetic energy is the energy associated with the the motionmotion of an object of an object

A moving object can do work on another A moving object can do work on another objectobject– E.g hammer does work on nail.E.g hammer does work on nail.

Kinetic EnergyKinetic Energy Consider an object with massConsider an object with mass mm moving in a straight moving in a straight line with initial velocityline with initial velocity vvii. . To accelerate it uniformly to To accelerate it uniformly to

a speeda speed vvff a constant net forcea constant net force FF is exerted on it parallel is exerted on it parallel

to motion over a distanceto motion over a distance dd..

Work done on objectWork done on object

W = F d = m a dW = F d = m a d (NII)

SoSo

da

ada

if

iifif

2

2)(222

222

vv

vxxvv

dd

mW if

2

22 vv

Kinetic EnergyKinetic Energy

If we rearrange this we obtainIf we rearrange this we obtain

We define the quantityWe define the quantity ½mv½mv22 to be the translational to be the translational kinetic energy (KE) of the objectkinetic energy (KE) of the object

This is the ‘This is the ‘Work-Energy TheoremWork-Energy Theorem’:’:

““The net The net workwork done on an object is equal done on an object is equal to its to its change in kinetic energychange in kinetic energy””

22

2

1

2

1if mmW vv

W = KE

The Work-Energy TheoremThe Work-Energy Theorem

• NoteNote– The net work done on an object is the work The net work done on an object is the work

done by the net force.done by the net force.– Units of energy and work must be the same (J)Units of energy and work must be the same (J)

W = KE

I. 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.

1D-Analysis

3D-Analysis

VI. Power

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

I. Kinetic energy

Energy associated with the state of motion of an object.

2

2

1mvK

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

II. Work

Energy transferred “to” or “from” an object by means of a force acting on the object.

To +WFrom -W

- Constant force: xx maF

dFWdFKKvvm

dvvmmaF

d

vvadavv

xxiff

fxxf

xxf

)(2

1

1)(

2

1

22

20

2

20

220

220

2

Work done by the force = Energy transfer due to the force.

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

dFFddFW x

cos

-Assumptions:Assumptions: 1) F = constant force 1) F = constant force 2) Object is particle-like (rigid object, all parts 2) Object is particle-like (rigid object, all parts of the object must move together).of the object must move together).

090

90180

90

W

W

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

Net work done by several forces = Sum of works done by Net work done by several forces = Sum of works done by individual forces.individual forces. Calculation:Calculation: 1) W1) Wnetnet= W= W11+W+W22+W+W33+…+…

2) F2) Fnet net W Wnetnet=F=Fnetnet d d

II. Work-Kinetic Energy TheoremII. Work-Kinetic Energy Theorem

WKKK if

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

III. Work done by a constant forceIII. Work done by a constant force

- Gravitational force:

cosmgdW

Rising objectRising object:: W= mgd cos180W= mgd cos180º º = -mgd= -mgd FFgg transfers mgd energy from the object’s kinetic energy.

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

External applied force + Gravitational force: External applied force + Gravitational force:

gaif WWKKK

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

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

IV. Work done by a variable forceIV. Work done by a variable force

- Spring force:- Spring force:

dkF

Hooke’s lawHooke’s law

k = spring constant measures spring’s stiffness. 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.

x

Fx

xi xfΔx

Fj

1) The block displacement must be divided into many segments of infinitesimal width, Δx.

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

- Calculation:

x

Fx

xi xfΔx

Fj

)(2

1

2

1)(

0

222if

x

x

x

x

x

x

x

x

js

xxkxkdxxkdxkxdxF

xxFW

f

i

f

i

f

i

f

i

22

2

1

2

1fis xkxkW

Ws>0 Block ends up closer to the relaxed position (x=0) than it was initially.Ws<0 Block ends up further away from x=0.Ws=0 Block ends up at x=0.

02

1 2 ifs xifxkW

Work done by an applied force + spring force:Work done by an applied force + spring force:

saif WWKKK

Block stationary before and after the displacement: Wa= -Ws The work done by the applied force displacing the block is the negative of the work done by the spring force.

Work done by a general variable force:Work done by a general variable force:

1D-Analysis

f

i

x

x

avgjx

avgjj

avgjj

dxxFWxFW

xionapproximatbetter

xFWW

xFW

)(

0

,0

,

,

lim

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

3D-Analysis

f

i

f

i

f

i

f

i

z

z

z

y

y

y

x

x

x

r

r

zyx

zyxzyx

dzFdyFdxFdWWdzFdyFdxFrdFdW

kdzjdyidxrd

zFFyFFxFFkFjFiFF

ˆˆˆ

)(),(),(;ˆˆˆ

Work-Kinetic Energy Theorem - Variable force

KKKmvmvdvvmdvmvW

vdx

dv

dt

dx

dx

dv

dt

dv

mvdvdxvdx

dvmdx

dt

dvmdxma

dxmadxxFW

ifif

v

v

v

v

x

x

x

x

f

i

f

i

f

i

f

i

22

2

1

2

1

)(

V. PowerV. Power

Time rate at which the applied force does work.Time rate at which the applied force does work.

- Average power: - Average power: amount of work doneamount of work donein an amount of time in an amount of time ΔΔt by a force.t by a force.

-Instantaneous power: Instantaneous power: -instantaneous time rate of doing work.instantaneous time rate of doing work.

t

WPavg

dt

dWP

Units:Units: 1 watt= 1 W = 1J/s 1 watt= 1 W = 1J/s 1 kilowatt-hour = 1000W1 kilowatt-hour = 1000W··h = 1000J/s x 3600s = 3.6 x 10h = 1000J/s x 3600s = 3.6 x 1066 J J = 3.6 MJ= 3.6 MJ

vFFvdt

dxF

dt

dxF

dt

dWP

coscos

cos

x

F

φ

In the figure (a) below a In the figure (a) below a 2N 2N force is applied to a force is applied to a 4kg4kg block at a block at a downward angle downward angle θθ as the block moves right-ward through as the block moves right-ward through 1m1m across across a frictionless floor. Find an expression for the speed va frictionless floor. Find an expression for the speed v ff at the end of at the end of

that distance if the block’s initial velocity is: that distance if the block’s initial velocity is: (a) 0 and (b) 1m/s to the right. (c) The situation in Fig.(b) is similar in that(a) 0 and (b) 1m/s to the right. (c) The situation in Fig.(b) is similar in that the block is initially moving at 1m/s to the right, but now the 2N force the block is initially moving at 1m/s to the right, but now the 2N force is directed downward to the left. Find an expression for the speed of is directed downward to the left. Find an expression for the speed of the block at the end of the 1m distance.the block at the end of the 1m distance.

N

mg

Fx

Fy

smv

vkgmN

mvKva

f

f

f

/cos

)4(5.0cos)1)(2(

5.00)(2

20

smv

JvkgN

JmvKsmvb

f

f

f

/cos1

2)4(5.0cos)2(

25.0/1)(

2

20

N

mg

Fx

Fy

dFdFW )cos(

)(5.0 20

2 vvmKW f

(a)

In the figure (a) below a In the figure (a) below a 2N 2N force is applied to a force is applied to a 4kg4kg block at a block at a downward angle downward angle θθ as the block moves right-ward through as the block moves right-ward through 1m1m across across a frictionless floor. Find an expression for the speed va frictionless floor. Find an expression for the speed v ff at the end of at the end of

that distance if the block’s initial velocity is: that distance if the block’s initial velocity is: (a) 0 and (b) 1m/s to the right. (c) The situation in Fig.(b) is similar in that(a) 0 and (b) 1m/s to the right. (c) The situation in Fig.(b) is similar in that the block is initially moving at 1m/s to the right, but now the 2N force the block is initially moving at 1m/s to the right, but now the 2N force is directed downward to the left. Find an expression for the speed of is directed downward to the left. Find an expression for the speed of the block at the end of the 1m distance.the block at the end of the 1m distance.

N

mg

Fx

Fy

smv

vkgmN

mvKva

f

f

f

/cos

)4(5.0cos)1)(2(

5.00)(2

20

smv

JvkgN

JmvKsmvb

f

f

f

/cos1

2)4(5.0cos)2(

25.0/1)(

2

20

N

mg

Fx

Fy

dFdFW )cos(

)(5.0 20

2 vvmKW f

(a) (b)

In the figure (a) below a 2N force is applied to a 4kg block at a downward angle θ as the block moves right-ward through 1m across a frictionless floor. Find an expression for the speed vf at the end of that distance if the block’s initial velocity is: (a) 0 and (b) 1m/s to the right. (c) The situation in (b) is similar in that the block is initially moving at 1m/s to the right, but now the 2N force is directed downward to the left. Find an expression for the speed of the block at the end of the 1m distance.

N

mg

Fx

Fy

smv

JvkgmN

JmvKsmvb

f

f

f

/cos1

2)4(5.0cos)1)(2(

25.0/1)(2

20

N

mg

Fx

Fy

smv

JvkgN

JmvKsmvc

f

f

f

/cos1

2)4(5.0cos)2(

25.0/1)(

2

20

dFdFW )cos(

)(5.0 20

2 vvmKW f

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

2

2

/2.010

210

21

10

1

smdt

dva

tdt

dxv

ttx

parabola

downwardconcaveMotion

)1.0)(4.0(

4.0)/2.0)(2(2

2

ttNxFW

NsmkgmacteF

JsmkgK

smvsta

f

f

64.0)/8.0)(2(5.0

/8.01)(2

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

2

2

/2.010

210

21

10

1

smdt

dva

tdt

dxv

ttx

parabola

downwardconcaveMotion

)1.0)(4.0(

4.0)/2.0)(2(2

2

ttNxFW

NsmkgmacteF

JK

vstb

f

f

0

05)(

JsKsKKWc ff 64.064.00)1()5()(

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

WdN

axgx ,

0

xy

mg

N

Fgy

Fgx

JmNNW

mgFFF

dFWorWWWa

net

gxaxnet

netnetxFgxFa

31.15.0)7.1432.17(

30sin30cos20

)(

N12. In the figure below a horizontal force Fa of magnitude 20N is applied to a 3kg book, as the book slides a distance of d=0.5m up a frictionless ramp. (a) During the displacement, what is the net force done on the book by Fa, the gravitational force on the book and the normal force on the book? (b) If the book has zero kinetic energy at the start of the displacement, what is the speed at the end of the displacement?workdoFFOnly

WdN

axgx ,

0

xy

mg

N

Fgy

Fgx

smvmvJW

KKWKb

ff

f

/93.05.031.1

0)(2

0

(a) Estimate the work done represented by the graph below in displacing the particle from x=1 to x=3m. (b) The curve is givenby F=a/x2, with a=9Nm2. Calculate the work using integration

J

Nmsquares

curveunderAreaWa

75.5

)1)(5.0)(5.11(

)(

Jx

dxx

Wb 6)13

1(9

19

9)(

3

1

3

12

An elevator has a mass of 4500kg and can carry a maximum load of 1800kg. If the cab is moving upward at full load at 3.8m/s, what power is required of the force moving the cab to maintain that speed?

mg

Fa

kNsmkgmgF

FamgmF

kgkgkgm

a

neta

total

74.61)/8.9)(6300(

0

630018004500

2

kWsmkNvFP 61.234)/8.3)(74.61(

A single force acts on a body that moves along an x-axis. The figure below shows the 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 state that the work is zero.

v

tA

B C

D

E

02

1 20

20

WvvAB

vvmKKKW

AB

ff

0 WvvBC BC 0 WvvCD CD

00,0 WvvDE DE

15E. In the figure below, a cord runs around two massless, frictionless pulleys; a canister with mass m=20kg hangs from one pulley; and you exert a force F on the free end of the cord. (a) What must be the magnitude of F if you are to lift the canister at a constant speed? (b) To lift the canister by 2cm, how far must you pull the 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?

Nmg

FFTcordHand

NTmgTFctevPulleya net

982

0:

98020:1)(

mg

TT T

P1

P2

In the figure below, a cord runs around two massless, frictionless pulleys; a canister with mass m=20kg hangs from one pulley; and you exert a force F on the free end of the cord. (a) What must be the magnitude of F if you are to lift the canister at a constant speed? (b) To lift the canister by 2cm, how far must you pull the 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?mg

TT T

P1

P2

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

JmNdFWc F 92.3)04.0)(98()(

15E. In the figure below, a cord runs around two massless, frictionless pulleys; a canister with mass m=20kg hangs from one pulley; and you exert a force F on the free end of the cord. (a) What must be the magnitude of F if you are to lift the canister at a constant speed? (b) To lift the canister by 2cm, how far must you pull the 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?mg

TT T

P1

P2

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

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