v2 vot 2a.DE VVoF moi aDX01

mCv2_vo4 2 maybe Fm

12m02 tzmvoh lmal.DE

mv2 tzmvoa F.DEN N

Work and Energy

Outline Work, Kinetic Energy

Work-Kinetic Energy theorem

Work

Kinetic Energy

Conservative Forces

Potential Energy

Energy Conservation

Work-Kinetic Energy

Time-independent equation of motion

→ final velocity

→ initial velocity

→ displacement

→ acceleration

→ force

→ mass

If we multiply the time-independent equation of motion with the mass

Motion (geometry)

Mechanics (force)

KE 12mV VV F AI

N v

KEF M 7

KEI Imf

Vf Vi KEF KEI AKE KEF KE D

- The motion side of the equation means that if an object is in motion, it has some energy associated with its motion. This energy is called Kinetic Energy

Kinetic Energy Work done by a force

Final kinetic energy

Initial kinetic energy

If the net force is zero on an object

If there is no net force acting on an object, then it moves with a constant velocity, and so its kinetic energy remains constant during the motion. As derived above, the change in the kinetic energy is equal to the work done by a force; thus, no net force leads to zero work, as well. Work is the energy transferred to or taken from an object.

Mr 01 N kg a MLT2

MV J kgm a ML'T2

F DX J Nm MET 2

V2 Vol _20h DX 912m65 Vo Fx DX

VI Wolf 2aybyatzmcvf vojl fy.byVI Vohz 2azDZ 912m62 Voz fz.bzN

D KE m Kitty th tzmlroxtvojtv.atway Fx DX 1 Fyby fzBZ2 0

24,4

2 E FeetFyjtfzkvohvoxtvoytv.EDF dxiisyjtbzkf.br_fxbxtfybytfz.DZ

Parameter Equation Unit Dimension

Force

Kinetic Energy

Work

The unit of energy is Joule, so the work has also the same unit.If we generalize time-independent equation of motion to three-dimensions:

DKE f m v2 VI F9IN F DF _IFI.ID rl.co so 70

Dr

L E sZ

O It s

07 s

Vik F DF IFT IDFI Caso AKE _IN

FxDXtFyDytFzD2KEm

2

Work is positive

Work is zero

Work is negative

Work done by a force can be calculated by taking the scalar multiplication of the force and the displacement. Work can be positive, negative or zero.

If we return to the beginning of the discussion, we drove that the change in the kinetic energy is equal to the work done by the net force acting on objects.

ΔKE = W Work-Kinetic Energy Theorem

WorkIn sum,

nowFg m mgj

LgaDF oyj

24M Dy 24 m m 1200kgL goodof g 9.81 2

W Fg DF C mgj C byj rugby1200kg 9.81 4 24M

VV 2.8 1055

DKE KEF KE AKE IN

KEf mof2 1zmvf2 rugbyKEI 12m72 if 0 Vf's 2gDYDRE zmof Vf V2C981m15 24in

vf 22m15

ExampleIn a movie stunt, a car of mass 1200kg falls a vertical distance of 24 meter starting from rest. What is the work done by the gravitational force? What is the velocity of the car just before it

hits the ground?

Solution

The work done by the gravitational force:

Using work-kinetic energy theorem

f BDN Jinnn constant in

f 01 0 IEE 0 e ai g Ef ke fk l Xt Ii p i

ng i imgl l ite 91DX 3meter

X F fk O fr Mieng n ng O s N mgf k MkmgF fu F Mang O F Mkmg

Wf F DE IFI.be

MkmgBXQ2 200kg 9.81 2 3m

lNf 620J

ExampleA box of books of mass 100kg is pushed with a constant speed in a straight line over a rough floor with a kinetic friction coefficient μ = 0.2. Find the work done by the force that pushes the box if the box is moved 3 meter across the floor.

Solution

From FBD:

O IT Wf fT D I fk DX Cos Ee n s DXf k Mkmg DX L Il

or2 200kg G81m44 3m

Wf 620 J g

IN IN f t Uvf 620J 620J OD KE IN O

For fknI DXI l 7 XI e INf lt l a NfIxDX

Wf F DX INF f k DX

This is only the individual work of F. However, there is also a friction force acting on the object. Let us calculate the work done by the friction force:

Total work on the object is

Negative work

Let us see the work from graphs:

Area below the

horizontal axis

means negative

work.

Area in the graph

FArea Work F DX

I

IDX X

FArea fFcx dx

I 7DX X

VV E dra ri erf sriFfxitfyjtfzkdrdxitdyjtd.tkF dF fxdxtfydy fz.dz

What if force is not constant?

The area between a function and the horizontal axis can be calculated by integrating the function over its variable.

If a force is varying with position (i.e. a force is not constant), its work can be calculated as

initial position

final position

F KI Xo OS

XfWs Fs de

Xi

Example (Spring Force)A mass m is attached to the end of a spring with a spring constant k. The equilibrium position of the mass is at × =0. If the mass and spring are stretched by L away from the equilibrium and released, calculate the work done by the spring force when

Solution

a) the object passes the equilibrium point,

b) the object squeezes the spring by L.c) What is the velocity of the object, when it squeezes the

spring by L?d) Calculate the work of the spring force when the object

passes the equilibrium point at a second time.

Spring force is not a constant force.

IN FT DI L K x dx I k x

CalWs I kLL

D KE VV

K Ei OK Ef mya

Im't I kip

Uf Is Ln M

pme kn gn M

Uf IN L

L L L

wi f FT de fl K x dx I k xO O 0

lbVvs Iz k L g

a) From x=L to x=0

Let us also calculate the speed of the object at x=0

Angular velocity of simple harmonic motion.

b) From x=0 to x=-L

Negative work

L O

INI W W f Kei OKEf I m vLIN Vy Con e w b

IN I ki Ike D

D KE IN O s kEf zmv2 OD

e ga

X O LO

wd fEs de s

Lpath

c) When the object squeezes the spring by L, its total kinetic energy will be equal to the total work done by the spring force

The total work of the spring force is zero. The work- kinetic energy theorem is

Thus the total kinetic energy of the object will be zero, too. The object mill be stopped by the spring force.

d) The path over which the spring force is workingThe spring force switches its direction at the equilibrium point (x=0).

We need to divide this integral at the points where either the spring force or the displacement switches direction.

w'd Fs.detoffs dxet.ffs.dk

yuca yy

W tb D

Vy'd Es de fi kxdx tzkx2

L

Vy'd L KL

AKE Wy'd 21mV L KLpIS L wk

n M

(see part (c))

When the object passes the equilibrium point at a second time, the object will have the same kinetic energy and so the same speed. However, velocity switches its direction.

ea

L O L L O L

INI 21kL2 wz I let

F fb

actF INL Wz

Hab bTbe

a Wba Way Wba

IN _Wab Wba D

Conservative ForcesWe have seen that when an object moves from a point to another as a force acts on it, the work by the force is also made. As we saw in the previous example, the work done by some forces may not depend on the choice of the path when the objects moves between the same points over different paths.

Thus, the work of the spring force from x =L to x=0 does not depend on the choice of the path

Path 1 Path 2

If the work of force between a and b does not depend on the path (1 or 2) , this force

is called "Conservative Force"

Path 1

Path 2

The work of a conservative force over a closed path (starting and ending at the same point) is zero.

TEN'sa

LN in a IN IN

Feet FgIFexH tryI mying mgby myCy yeNext Wg myLya ft

Work of a conservative force is the same all over the paths unless one changes the initial and final points.

This stored energy is called "potential energy" and it is nothing but the work you have done against the gravitational force.

Spring force, gravitational force, electromagnetic force, nuclear forces are some of examples for the conservative forces.

Potential Energy When you carry a brick from y1 to y2, you work against the gravitational force. Assume you are carrying the brick with a constant speed, thus

If you hold the brick at y2 for a while, the kinetic energy will be zero. Then, when you drop the brick, it will start moving down, and you observe some kinetic energy. This kinetic energy should be stored in the brick as another type of energy and it should turn to be the kinetic energy when the brick falls.

Wext Ug ing

Ug myDJ s

Went Ws I k x

Us I k x Ws

Gravitational Potential Energy.

Let us see another picture

You see a spring in its equilibrium in the beginning, and we squeeze it with a ball and hold it for a while. There is no motion, there is no kinetic energy. Finally we release the ball, and we observe that the ball will be thrown with a velocity. When you squeeze the spring, you work against the spring force, and your work;

→ distance from equilibrium

Spring potential energy

Thus we see that, when we work against a force, our work is stored as the potential energy in the object. What if we work against another force, say the friction force? Is our work stored as the potential energy in the object?

FJ

fk e Fk

U W f F dir

U JF dir ft sF dirdirF Fx it FgjtFzI Exit If It k

JvFx 22 i Fy I fz2J 22

D KE IN 7

DU IN 7

If we push a box over a rough surface, then release it, we are not going to see our work stored in the object, since it is going to stop eventually. Thus the work against the friction force is NOT stored as the potential energy.

The potential energy is stored by the work against the conservative forces, only.

Each conservative force is associated with a potential energy. The minus sign is very important here.

Conservative force

Work-Kinetic Energy theorem

Work-Potential Energy

F KE t U s

Ef KEf t Uf S

Ei KEI t Ui e

DE Ef Ei KEf tUf LKEi Vi

KEF KEe t Uf ViB KE t DUIN t L W

B E O

Ef Ei s DE O s

P AI pD t Dun sB t s

Total Mechanical Energy

Final Mechanical Energy

Initial Mechanical Energy

Thus, if all forces acting on objects are conservative, their total mechanical energy remains constant. This is called energy conservation.

Total energy is conserved.

Power

We discussed the work and energy so far without considering its rate. You may do the same work eventually, but you may be fast or slow while doing this work. Power is defined as the rate of your work with respect to time. We can also define power as the energy lost or gained per unit time.

Power

Total work

Duration

DW dinp linDt s 0 Dt dt

power kgY Watt

p MI1 3

dull F dx s p dd tf Fr dxf dfsdt s

p tf F

At the limit Δt —> 0

Unit:

Dimension:

velocity

Ifad4

INCal Wcbl We4

IN ca Area F.bzl30Nzl5

0mz.sj

IN Area F DX 3 on som 155

IN AreaF 13 0 15042 7.55

Solved Examples

Example 1A particle is subject to a force Fx that varies with the position of the object as shown in the

figure. Find the work done by the force

a) from x =0 to x= 5 mb) from x=5m to x=10m

c) from x=10m to x=15md) From x=0m to x=15m

Solution

The force is not constant, so

We can simply use the area beneath the graph.

a)

b)

c)

Vy'd Wca web wee

7.551 1551 1755305

M 40kg f ON ANf e s

y e If5 om fv

X F f mo my

y N my O N mg f MN Mmg

d)

Example 2A 40kg initially at rest box is pushed 5.0 m along a rough, horizontal floor with a constant horizontally applied force of 130N. If the friction coefficient between the floor and the box is 0.30

a) find the work done by the applied forceb) The energy lost due to the frictionc) The change in the kinetic energy

d) The final speed of the box.

SolutionFBD

Wp F DI

Wp 130N 5 Om 650J

I Mmg E DI 50m15Wf I DI L Mmg Dx

0.30 40kg 9.81 2 5 Om590J

Energy lost half 590J

DK We Mft WfDK 650 5 c 59057 605

D KE KEF KEE 2miff IME E DD KE KEF Im offVf

2 D KE 2 60JM 40kg

173 m s

a) The work done by the applied force

(F is a constant force)

b) The friction is also constant. If the direction of motion is

assign to be positive

The work of the friction force will be equal to the energy lost due to the friction as

c) Using Work-Kinetic Energy theorem

d)

I fBDf g rg i g X

rf e If1200 1200a

f Tug

X F f mysin20 moi

y N myCos 20 O s N mg cos 20

f MN

Example 3A crate of mass 10.0 kg is pulled up a rough incline with an initial velocity of 1. 50 m/s. The pulling force is 100N parallel to the incline, which makes an angle of 20 degrees with the horizontal. The friction coefficient is 0.4, and the crate is pulled 5 meter along the incline.a) How much work is done by the 100 N force?b) How much energy is lost due to the friction?c) How much work is done by the gravity?d) what is the speed of the crate when it moves 5.0 m?

Solution

IN f F DI F DX 100N 5 ON 500J

Wf F DI III IDI't cos CioImg cos 20 DX

CO 4 10.0kg 9.81 4 os2o1C5 Om

184 5J

9 DI Wy Ff DI

J noo Fgl IDI't Cosmo

my DX Cos 110uFg my Wy lo0kg119 81 2 5 On CosGeo

Wy 167.8 J

D KE We IN f t half t Wg500J t L 184 5J tC 167.8J

147 7 J

a)

b)

Energy lost due to the friction is 184 J.

c)

d)

DRE zm of of re 1.50 m s

rf 2bmkE tvf 32 FI

Vf 31.7 2 5 7 m s

Volx Vo Croly D

04 0 ay gVx Croix Vy Croly g tX Xotwolxt y y Cody t Igt

Example 4Two children are playing a game in which they try to hit a small box on the floor with a marble fired with a spring that is mounted on a table. The target box is 2.20 m away from the leg of

the table. The first child compresses the spring by 1.10 cm, but the marble falls 27.0 cm short from the box. How much should the second child compress the spring to hit the box?

Solution

The marble moves in a projectile motion.

Horizontal Vertical

a 0

E Ef Ei I m v t I KE t ngyOEf I m ro't I k x engy

r

I k x't ngy I moo'tngykm XX

Xt s

Xz s

VI 9

Vz 7

xi sxi s

XI 2 20 m O 27 m 1 S3 m

XI 2 20 M

The initial velocity of the marble can be calculated by the energy conservation.

compression of the first child (1.10 cm)

compression of the second child

velocity in the first child's game

velocity in the second child's game

horizontal distance in the first child's game

horizontal distance in the second child's game

t tz t y go I g t

I g ta Jh LytaE 21g

h s

xia v t IET a FXz t FT Xa Y

I h aight

Xz Iv 25 cm

Flight times t1 and t2 will be the same for both children, since they shoot the marble from the same height.

height of the table

ex IT X 3 2

F deDX

F Lyx't 133 3 2

fixe X x 6 3

EE 0 a

f DI Ddx

duDI X x

2 6 x p X12

Xz OXz 3

Example 5

Consider a potential energy in one-dimension given by

b) Determine the equilibrium points and discuss.

a) Find the conservative force associated with the potential

energy given above,

Solutiona) For the conservative forces, in one-dimension

b) Equilibrium

Extrema are the equilibrium points

du O sdaXe

du Ddx

Xe

du LO eDX Xe

du 3 2 2 6 2DX X XL 2

IV 3 2 2 6 6DX Xz x DDIdx x 3 2 2 6 15

x 3

We need to check if these equilibrium points are stable or unstable. From the sketch we can say x1 and x3 correspond to stable equilibrium while x2 is an unstable equilibrium.

Stable equilibrium

natural equilibrium

unstable equilibrium

stable

unstable

stable