PROGRAM OF PROGRAM OF “PHYSICS”“PHYSICS”Lecturer: Dr. DO Xuan Hoi

Room 413E-mail : dxhoi@hcmiu.edu.vn

PHYSICS I PHYSICS I (General Mechanics) (General Mechanics)

02 credits (30 periods)02 credits (30 periods)

Chapter 1 Bases of KinematicsChapter 1 Bases of Kinematics

Motion in One Dimension Motion in One Dimension

Motion in Two DimensionsMotion in Two Dimensions

Chapter 2 The Laws of Motion Chapter 2 The Laws of Motion

Chapter 3 Work and Mechanical EnergyChapter 3 Work and Mechanical Energy

Chapter 4 Linear Momentum and CollisionsChapter 4 Linear Momentum and Collisions

Chapter 5 Rotation of a Rigid Object Chapter 5 Rotation of a Rigid Object About a Fixed Axis About a Fixed Axis

Chapter 6 Static EquilibriumChapter 6 Static Equilibrium

Chapter 7 Universal GravitationChapter 7 Universal Gravitation

References :References :

Halliday D., Resnick R. and Walker, J. Halliday D., Resnick R. and Walker, J. (2005), Fundamentals of Physics, (2005), Fundamentals of Physics, Extended seventh edition. John Willey Extended seventh edition. John Willey and Sons, Inc.and Sons, Inc.

Alonso M. and Finn E.J. (1992). Physics, Alonso M. and Finn E.J. (1992). Physics, Addison-Wesley Publishing CompanyAddison-Wesley Publishing Company

Hecht, E. (2000). Physics. Calculus, Second Hecht, E. (2000). Physics. Calculus, Second Edition. Brooks/Cole.Edition. Brooks/Cole.

Faughn/Serway (2006), Serway’s College Faughn/Serway (2006), Serway’s College Physics, Brooks/Cole.Physics, Brooks/Cole.

Roger Muncaster (1994), A-Level Physics, Roger Muncaster (1994), A-Level Physics, Stanley Thornes.Stanley Thornes.

http://ocw.mit.edu/OcwWeb/Physics/index.htmhttp://www.opensourcephysics.org/index.htmlhttp://hyperphysics.phy-astr.gsu.edu/hbase/HFrame.htmlhttp://www.practicalphysics.org/go/Default.htmlhttp://www.msm.cam.ac.uk/http://www.iop.org/index.html...

PHYSICS IPHYSICS IChapter 3Chapter 3 Work and Mechanical Work and Mechanical

EnergyEnergy Work Done by Force. Power Work Done by Force. Power

Kinetic Energy and the Work. Kinetic Energy and the Work.

Potential Energy of a SystemPotential Energy of a System

Conservative and Non-conservative ForcesConservative and Non-conservative Forces

Conservation of Mechanical Energy Conservation of Mechanical Energy

Changes in Mechanical Energy for Non-conservative Changes in Mechanical Energy for Non-conservative Forces Forces

Relationship Between Conservative Forces and Relationship Between Conservative Forces and Potential EnergyPotential Energy

IntroductionIntroductionForms of energy:Forms of energy:

MechanicalMechanical

focus for nowfocus for now

chemicalchemical

electromagneticelectromagnetic

nuclearnuclear

Energy can be transformed from one form to Energy can be transformed from one form to anotheranother

Essential to the study of physics, chemistry, Essential to the study of physics, chemistry, biology, geology, astronomybiology, geology, astronomy

Can be used in place of Newton’s laws to solve Can be used in place of Newton’s laws to solve certain problems more simplycertain problems more simply

1 Work Done by Force. Power1 Work Done by Force. Power

Provides a link between force and energyProvides a link between force and energy The work, The work, WW, done by a constant force on , done by a constant force on an object is defined as the product of the an object is defined as the product of the component of the force along the direction component of the force along the direction of displacementof displacement and and the magnitude of the the magnitude of the displacementdisplacement

ΔxΔx is the displacement is the displacement

(F cos θ)(F cos θ) is the component of the force in the is the component of the force in the direction of the displacementdirection of the displacement

(F cos θ)(F cos θ) is the component of is the component of the force in the direction of the the force in the direction of the displacement displacement

( cos )W F x

By using the By using the scalar product (dot product) :

( cos ) .W F x F x ??????????????

Work can be Work can be positivepositive or or negativenegativePositivePositive if the force and the displacement if the force and the displacement are are in the same directionin the same directionNegativeNegative if the force and the if the force and the displacement are displacement are in the opposite in the opposite directiondirection

Example 1:Example 1: lifting a cement block… lifting a cement block…Work done Work done by the personby the person: :

is is positivepositive when lifting the box when lifting the box is is negativenegative when lowering the box when lowering the box

Example 2:Example 2: … then moving it horizontally … then moving it horizontallyWork done Work done by gravityby gravity::

is is negativenegative when lifting the box when lifting the box is is positivepositive when lowering the box when lowering the box is is zerozero when moving it horizontallywhen moving it horizontally00: 321 mghmghWWWWworkTotal

lifting lowering moving total

EXAMPLE 1EXAMPLE 1

2 22.0 3.0 3.6d m

= = 2.0 + 3.0 )(5.0 + 2.0 ) W F.d ( i j i j

(a)

A particle moving in the xy plane undergoes a displacement

d = (2.0i + 3.0j) m as a constant force F = (5.0i + 2.0j) N

acts on the particle. (a) Calculate the magnitude of the displacement

and that of the force.(b) Calculate the work done by F.(c) Calculate the angle between F and d.

16 J16

cos ;. 3.6 5.4

W JF d N m

(c)

(b)

2 25.0 2.0 5.4F N

35o

WORK DONE BY A VARYING FORCE

( cos )i iW F x

Split Split totaltotal displacement (x displacement (xf f - x- xii))

into into many small many small displacementsdisplacements xx For each small displacement:For each small displacement:

Thus, total work is:Thus, total work is:

which is which is total area under the F(x) curve!total area under the F(x) curve!

i

ixi

itot xFWW

0lim

ff

i i

xx

x xx x x

F x F dx

f

i

x

xx

W F dx

PROBLEM 1

The interplanetary probe shown in the figure is attracted tothe Sun by a force of magnitude

F = - (1.3 10-23)/x 2

where x is the distance measured outward from the Sun tothe probe. Determine how much work is done by the Sun on the probe as the probe–Sun separation changes from1.5 1011 m to 2.3 1011 m

SOLUTION

PROBLEM 2

A block on a horizontal, frictionless surface is connected to a spring. The spring exerts on the block a force of magnitudeF = - kx , where x is the displacement of the block from its unstretched (x = 0) position and k is a positive constant called the force constant of the spring.Determine the work done by the spring force when the block undergoes an arbitrary displacement from x = xi to x = xf .

SOLUTIONf

i

x

xx

W F dx

( )f

i

x

x

kx dx 212

f

i

x

x

kx

212

i

f

x

x

kx

2 21 12 2i fW kx kx

WORK DONE BY THE GRAVITATIONAL FORCE

Divide the path into small segmentss??????????????

s xi yj??????????????

( ) ( )

0

mg xi mg yj

mg y

mg mgj

mg

i

j

( )W F s mg xi yj??????????????

The total work done by the gravitational force :

2 1( )W mg y y

The work done by the gravitational force does not

depend on the path; it depends only on the vertical

distance

POWERGeneral definition : the time rate of energy

transfer.If an external force is applied to an object and if the work done by this force in the time interval t is W , then the average power expended during this interval is defined as :W

t

P

0limt

W dWt dt

P

The instantaneous power is the limiting value of the average power as t approaches zero :

dW Fds ; ds

F Fvdt

P

J/s W horsepower (hp):746 W

EXAMPLE 2EXAMPLE 2

= 0 ;Y fricF T f Mg

An elevator car has a mass of 1 000 kg and is carrying

passengers having a combined mass of 800 kg. A constant

frictional force of 4 000 N retards its motion upward.

What must be the minimum power delivered by the motor

to lift the elevator car at a constant speed of 3.00 m/s?

3 3 24.00 10 (1.80 10 ) (9.80 / )T N kg m s

fricT f Mg

42.16 10 N

Tv Tv

P 4 42.16 10 3.00 / 6.48 10N m s W

2 2 Kinetic energy and the workKinetic energy : Energy : Energy associated with the associated with the motionmotion of of an objectan object

Scalar quantityScalar quantity with the with the same units as worksame units as work

Work is related to kinetic energyWork is related to kinetic energy

Let Let FF be a be a constantconstant force: force:

2 22 2 0

0

2 22 20

0

( ) , :

2 , .2

1 1: .

2 2 2

net

net

W Fs ma s but

v vv v a s or a s

v vThus W m mv mv

1=

22KE mvThis quantity is called kinetic

energy:

Kinetic Energy Theorem (or (or The work–kinetic energy theorem)

“ The net work done on a particle by external forces equals the change in kinetic energy of the particle”

net fiW KE KE KE

1=

22KE mv

2 20

1 1.

2 2netW mv mv

Speed will increase if work is positiveSpeed will increase if work is positive

Speed will decrease if work is negativeSpeed will decrease if work is negative

PROBLEM 3

Derived the work–kinetic energy theorem when the force varies

SOLUTION

ff

i i

x x

x xx x

W F dx ma dx

ff

i i

x v

x v

dvW mv dx mvdv

dx

dva

dt

2 21 12 2fiW mv mv

dv dx dvv

dx dt dx

TestTestTwo marbles, one twice as heavy as the other, are dropped to the ground from the roof of a building. Just before hitting the ground, the heavier marble has

1. as much kinetic energy as the lighter one.2. twice as much kinetic energy as the lighter

one.3. half as much kinetic energy as the lighter one.4. four times as much kinetic energy as the lighter

one.5. impossible to determine.

EXAMPLE 3EXAMPLE 3

W Fd (12 )(3.0 ) 36 N m J

2f

Wv

m

(a)

A 6.0-kg block initially at rest is pulled to the right along a horizontal, frictionless surface by a constant horizontal force of 12 N.(a) Find the speed of the block after it has moved 3.0 m.

2(36 )6.0

Jkg

3.5 /m s

210

2net fi fW KE KE KE mv

EXAMPLE 3EXAMPLE 3

fric kW F d (8.82 )(3.0 ) 26.5 N m J

(b)

A 6.0-kg block initially at rest is pulled to the right along a horizontal, frictionless surface by a constant horizontal force of 12 N.(b) Find the final speed of the block if the surface has a coefficient of kinetic friction of 0.15.

2(36 26.5 )6.0f

J Jv

kg

1.8 /m s

210 36 26.5

2 f netmv W J J

k k kf n mg 2 (0.15)(6.0 )(9.80 / ) 8.82 kg m s N

PROBLEM 4

A block of mass 1.6 kg is attached to a horizontal spring thathas a force constant of 1.0 103 N/m, as shown in the figure. The spring is compressed 2.0 cm and is then released from rest. Use the result of PROBLEM 2.(a) Calculate the speed of the block as it passes through the equilibrium position x = 0 if the surface is frictionlessSOLUTION (a)

2 21 12 2i fW kx kx

3 -21(1.0 10 / )(2.0 10 ) 0 0.20

2N m m J

2f

Wv

m

2(0.20 )1.6

Jkg

0.50 /m s210 ;

2 fW mv

PROBLEM 4

A block of mass 1.6 kg is attached to a horizontal spring thathas a force constant of 1.0 103 N/m, as shown in the figure. The spring is compressed 2.0 cm and is then released from rest. Use the result of PROBLEM 2.(b) Calculate the speed of the block as it passes through the equilibrium position if a constant frictional force of 4.0 N retards its motion from the moment it is released.SOLUTION (b)

2 21 1;

2 2net i fW kx kx 0.20 0.08 0.12netW J J J

2 netf

Wv

m 2(0.12 )

1.6J

kg 0.39 /m s21

0 ;2net fW mv

fric kW F d2 (4.0 )(2.0 10 ) 0.080 N m J

3. Potential Energy of a System3. Potential Energy of a System

► Potential energyPotential energy is associated with the is associated with the positionposition of the object within some systemof the object within some system Potential energy is a property of the system, Potential energy is a property of the system,

not the objectnot the object A system is a collection of objects or A system is a collection of objects or

particles interacting via forces or processes particles interacting via forces or processes that are internal to the systemthat are internal to the system

► Units of Potential Energy are the same as Units of Potential Energy are the same as those of Work and Kinetic Energythose of Work and Kinetic Energy

3.1 Gravitational Potential 3.1 Gravitational Potential EnergyEnergy► Gravitational Potential EnergyGravitational Potential Energy is the energy associated is the energy associated

with the with the relative positionrelative position of an object in space near the of an object in space near the Earth’s surfaceEarth’s surface Objects interact with the earth through the Objects interact with the earth through the

gravitational forcegravitational force Actually the potential energy of the earth-object Actually the potential energy of the earth-object

systemsystem► Consider block of mass m at initial height Consider block of mass m at initial height yyii

► Work done by the gravitational forceWork done by the gravitational force

, ,gravity grav i grav fW U U

cosgrav i fi fW F s mg y y mgy mgy

Potential energy:Potential energy: grav gravPE U mgy

Note:Note:

Important: work is related to the difference in PE’s!

Reference Levels for Gravitational Potential Reference Levels for Gravitational Potential EnergyEnergy

► A location where the A location where the gravitational potential energy gravitational potential energy is zerois zero must be chosen for each problem must be chosen for each problem The choice is arbitrary since the change in the The choice is arbitrary since the change in the

potential energy is the important quantitypotential energy is the important quantity Choose a convenient location for the zero Choose a convenient location for the zero

reference heightreference height►often the Earth’s surfaceoften the Earth’s surface►may be some other point suggested by the may be some other point suggested by the

problemproblem

A location where the A location where the gravitational potential gravitational potential energy is zero must be energy is zero must be chosen for each problemchosen for each problem

The choice is arbitrary The choice is arbitrary since since the changethe change in the in the potential energy gives the potential energy gives the work donework done

2 22grav i fW mgy mgy

Reference Levels for Gravitational Potential Reference Levels for Gravitational Potential EnergyEnergy

1 2 3grav grav gravW W W

1 11grav i fW mgy mgy

3 33grav i fW mgy mgy

EXAMPLE 4EXAMPLE 4

i iU mgy 2 (7 )(9.80 / )(0.5 m) 34.3 kg m s J (a)

A bowling ball of a mass of approximately 7 kg held by a careless bowler slips from the from a height of 0.5 m and drops on the bowler’s toe which is about 0.03 m above the floor. (a) Choosing floor level as the y = 0 point of your coordinate system, estimate the total work done on the ball by the force of gravity as the ball falls.

34.3 2.06 32.24J J J g i fW W W

ffU mgy 2 (7 )(9.80 / )(0.03 m) 2.06 kg m s J

EXAMPLE 4EXAMPLE 4

i iU mgy 2 (7 )(9.80 / )( 1 m) 68.6 kg m s J (b)

A bowling ball of a mass of approximately 7 kg held by a careless bowler slips from the from a height of 0.5 m and drops on the bowler’s toe which is about 0.03 m above the floor. (b) Repeat the calculation, using the top of the bowler’s head (which we estimate to be 1.50 m above the floor) as the origin of coordinates.

68.6 ( 100.8 ) 32.24J J J g i fW W W

ffU mgy 2 (7 )(9.80 / )( 1.47 m) 100.8 kg m s J

PROBLEM 5

A small block with a mass of 0.120 kg is attached to a cord passing through a hole in a frictionless, horizontal surface.The block is originally revolving at a distance of 0.40 m from the hole with a speed of 0.70 m/s. The cord is then pulled from below, shortening the radius of the circle in which the block revolves to 0.10 m. At this new distance, the speed of the block is observed to be 2.80 m/s.(a) What is the tension in the cord in the original situation when the block has speed 0.70 m/s?

SOLUTION21

11

R

vF T m

r

2(0.70 / )(0.120 )

0.40m s

kgm

(a)

0.15N

PROBLEM 5

A small block with a mass of 0.120 kg is attached to a cord passing through a hole in a frictionless, horizontal surface.The block is originally revolving at a distance of 0.40 m from the hole with a speed of 0.70 m/s. The cord is then pulled from below, shortening the radius of the circle in which the block revolves to 0.10 m. At this new distance, the speed of the block is observed to be 2.80 m/s. (b) What is the tension in the cord in the final situation when the block has speed v = 2.80 m/s?

22

22

R

vF T m

r

2(2.80 / )(0.120 )

0.10m s

kgm

(b)

9.40N

PROBLEM 5

A small block with a mass of 0.120 kg is attached to a cord passing through a hole in a frictionless, horizontal surface.The block is originally revolving at a distance of 0.40 m from the hole with a speed of 0.70 m/s. The cord is then pulled from below, shortening the radius of the circle in which the block revolves to 0.10 m. At this new distance, the speed of the block is observed to be 2.80 m/s.(c) How much work was done by the person who pulled on the cord?

SOLUTION (c)2 22 1

1 12 2

W mv mv

2 21(0.120 ) (2.80 / ) (0.70 / )

2kg m s m s

0.44 J

PROBLEM 6

You throw a 20-N rock vertically into the air from ground level. You observe that when it is 15.0 m above the ground, it is traveling at 25.0 m/s upward. Use the work-energy theorem to find (a) the rock's speed just as it left the ground and (b) its maximum height. SOLUTION

(a) 2 22 1

1 1;

2 2W mgh mv mv

30.3 /m s

21 22v gh v

2 21 2 9.80 / 15.0 (2.80 / )v m s m m s

(b)At the maximum height : v ’2 = 02 2

2 1

1 1' ' ' ;

2 2W mgh mv mv 2

1

1' 0

2mgh mv

2(30.3 / )2 9.8 /

m sm s

21'

2v

hg

46.8m

PROBLEM 7

A boy skateboards down a curved playground ramp, he

moves through a quarter-circle with radius R = 3.00 m. The

total mass of this boy and his skateboard is 25.0 kg. He starts

from rest and there is no friction.(a) Find his speed at the bottom of the ramp(b) Find the normal force that acts on him at the

bottom of the curve.

SOLUTION

(a) 2 22 1

1 1;

2 2mv mv W mgh mgR

7.67 /m s

2 2v gR

22 2 9.80 / 3.00v m s m

22

10 ;

2mv mgR

PROBLEM 7

A boy skateboards down a curved playground ramp, he

moves through a quarter-circle with radius R = 3.00 m. The

total mass of this boy and his skateboard is 25.0 kg. He starts

from rest and there is no friction.(a) Find his speed at the bottom of the ramp(b) Find the normal force that acts on him at the

bottom of the curve.

SOLUTION

(b)22

R

va

R

735N

2mg

23 25.0 9.80 /kg m s

Rn mg ma

22

gRg

R

At the bottom :

3n mg

PROBLEM 8

We want to load a l2-kg crate into a truck by sliding it up a

ramp 2.5 m long, inclined at 300. A worker, giving no thought

to friction, calculates that he can get the crate up the ramp by

giving it an initial speed of 5.0 m/s at the bottom and letting it

go. But friction is not negligible; the crate slides 1.6 m up the

ramp, stops, and slides back down.(a) Assuming that the friction force acting on the crate

isconstant, find its magnitude

SOLUTION

Kinetic Energy Theorem

net fiW KE KE KE

2 22 1

1 1( ) ;

2 2 FRICmv mv W mgh f d 21

10 ( ) ;

2 FRICmv mgh f d 21

12

FRIC

mv mghf

d

(a)

PROBLEM 8

We want to load a l2-kg crate into a truck by sliding it up a

ramp 2.5 m long, inclined at 300. A worker, giving no thought

to friction, calculates that he can get the crate up the ramp by

giving it an initial speed of 5.0 m/s at the bottom and letting it

go. But friction is not negligible; the crate slides 1.6 m up the

ramp, stops, and slides back down.(a) Assuming that the friction force acting on the crate

isconstant, find its magnitude

SOLUTION

21

12

FRIC

mv mghf

d

2 2 0112 (5.0 / ) 12 9.8 / (1.6 )sin30

22.5

kg m s kg m s m

m

35N

PROBLEM 8

We want to load a l2-kg crate into a truck by sliding it up a

ramp 2.5 m long, inclined at 300. A worker, giving no thought

to friction, calculates that he can get the crate up the ramp by

giving it an initial speed of 5.0 m/s at the bottom and letting it

go. But friction is not negligible; the crate slides 1.6 m up the

ramp, stops, and slides back down.(b) How fast is the crate moving when it reaches the

bottom of the ramp?

SOLUTION

2 21

1 1' 0 ( 2 ) ;

2 2 FRICmv mv W f d (b)

2 21

1 1' 2

2 2 FRICmv mv f d 2112 (5.0 / ) 2 3.5 1.6

2kg m s N m 38 J

2 38'

12J

vkg

2.5 /m s

3.2 3.2 Elastic Potential EnergyIn a displacement from xl to x2 the spring does an amount of work given by :

We define the elastic potential energy

(work done by a spring)

2 21 2

1 12 2

W kx kx

212elU kx

so that :

1 2el elW U U

4.1 Conservative Forces► A force is A force is conservative conservative if the work it does on an if the work it does on an

object moving between two points is object moving between two points is independent independent of the pathof the path the objects take between the points the objects take between the points The work depends only upon The work depends only upon the initial and final the initial and final

positionspositions of the object of the object Any conservative force can have a Any conservative force can have a potential potential

energyenergy function associated with it function associated with it

Note: a force is conservative if the work it does on an object moving through any closed path is zero.

4. Conservative and Non-conservative 4. Conservative and Non-conservative ForcesForces

Examples of Conservative Forces:Examples of Conservative Forces:

► Examples of conservative forces include:Examples of conservative forces include: GravityGravity Spring forceSpring force Electromagnetic forcesElectromagnetic forces

► Since work is independent of the path:Since work is independent of the path: : only initial and final points: only initial and final pointsi fW U U

grav i fW mgy mgy

2 21 2

1 12 2elW kx kx

4.2 Conservative forces and potential energy

grav i fW U U U

grav i fW mgy mgy

2 21 2

1 12 2elW kx kx

el i fW U U U

The work done by a conservative force equalsthe negative of the change in the potential energy associated with that force

i fW U U U PE

In general :

i fW PE PE PE U In general :

f

i

x

fi xx

U U U W F dx f

i

x

f x ix

U F dx U

;xdU F dx x

dUF

dx

Example : Potential energy :Potential energy : gravU mgy

y

dUF

dy ( )d mgy

dy mg

Elastic potential energy :Elastic potential energy : 212elU kx

elx

dUF

dx 21

( )2

dkx

dy kx

NOTE :NOTE :

x

dUF

dx

Gradient operator :

dnabla i

dx

In 3-D space : ; ;x y z

dU dU dUF F F

dx dy dz

( )dU d

F i i Udx dx

F gradU U

d d dnabla i j k

dx dy dz

F gradU U

4.3 Nonconservative 4.3 Nonconservative ForcesForces

► A force is A force is nonconservativenonconservative if the work it does on an if the work it does on an object object depends on the pathdepends on the path taken by the object taken by the object between its final and starting points.between its final and starting points.

► Examples of nonconservative forces:Examples of nonconservative forces: kinetic friction, air drag, propulsive forceskinetic friction, air drag, propulsive forces

► The friction force transforms kinetic energy of the The friction force transforms kinetic energy of the object into a type of energy associated with object into a type of energy associated with temperaturetemperature the objects are warmer than they were before the objects are warmer than they were before

the movementthe movement Internal EnergyInternal Energy is the term used for the energy is the term used for the energy

associated with an object’s temperatureassociated with an object’s temperature

Friction Depends on the PathFriction Depends on the Path

► The blue path is The blue path is shorter than the red shorter than the red pathpath

► The work required is The work required is less on the blue path less on the blue path than on the red paththan on the red path

► Friction depends on Friction depends on the path and so is a the path and so is a nonconservative forcenonconservative force

5. Conservation of Mechanical Energy5. Conservation of Mechanical Energy

► ConservationConservation in general in general To say a physical quantity is To say a physical quantity is conservedconserved is is

to say that the numerical value of the to say that the numerical value of the quantity quantity remains constantremains constant

► In In Conservation of EnergyConservation of Energy, the , the total total mechanical energy remains constantmechanical energy remains constant In any In any isolated systemisolated system of objects that of objects that

interact only through interact only through conservative forcesconservative forces, , the total mechanical energy of the system the total mechanical energy of the system remains constant.remains constant.

Conservation of Energy :Conservation of Energy :

► Total mechanical energyTotal mechanical energy is the sum of the is the sum of the kinetic and potential energies in the systemkinetic and potential energies in the system

;i i i ff fE K U E K U

i f

i i ff

E E

K U K U

Problem-Solving Strategy Problem-Solving Strategy with Conservation of Energywith Conservation of Energy

► Define the system► Select the location of zero gravitational

potential energy Do not change this location while solving

the problem► Determine whether or not nonconservative

forces are present► If only conservative forces are present, apply

conservation of energy and solve for the unknown

EXAMPLE 5EXAMPLE 5

2.29 /Bv m s

A pendulum of length 2.00 m and mass 0.500 kg is released from rest when the cord makes an angle of 30.0°with the vertical.Find the speed of the sphere and the tension in the cord when the sphere is at its lowest point

EXAMPLE 5EXAMPLE 5A pendulum of length 2.00 m and mass 0.500 kg is released from rest when the cord makes an angle of 30.0°with the vertical.Find the speed of the sphere and the tension in the cord when the sphere is at its lowest point

6.21BT N

PROBLEM 9

A glider with mass m = 0.200 kg sits on a frictionlesshorizontal air track, connected to a spring with force constant k = 5.00 N/m. You pull on the glider, stretching the spring 0.100 m, and then release it with no initial velocity. The glider begins to move back toward its equilibrium position (x = 0). What is its x-velocity when x = 0.080 m?

SOLUTION

PROBLEM 9

A glider with mass m = 0.200 kg sits on a frictionlesshorizontal air track, connected to a spring with force constant k = 5.00 N/m. You pull on the glider, stretching the spring 0.100 m, and then release it with no initial velocity. The glider begins to move back toward its equilibrium position (x = 0). What is its x-velocity when x = 0.080 m? SOLUTION

6. Changes in Mechanical Energy for 6. Changes in Mechanical Energy for

Non-conservative ForcesNon-conservative Forces

► When When nonconservative forcesnonconservative forces are present, are present, the total mechanical energy of the system isthe total mechanical energy of the system is

notnot constantconstant► The The work done by all nonconservative forceswork done by all nonconservative forces

acting on parts of a system equals acting on parts of a system equals the the change in the mechanical energychange in the mechanical energy of the of the systemsystem

ncW Energy

( ) ( )nc ffi iW K U K U

EXAMPLE 6EXAMPLE 6A block having a mass of 0.80 kg is given an initial velocity1.2 m/s to the right and collides with a spring of negligiblemass and force constant 50 N/m.(a) Assuming the surface to be frictionless, calculatethe maximum compression of the spring after the collision.

A C

A A C C

E E

K U K U

(a)

2 21 10 0

2 2A mmv kx

m A

mx v

k 0.80

(1.20.80 / )50 /

kgm s

N m

0.15mx m

EXAMPLE 6EXAMPLE 6A block having a mass of 0.80 kg is given an initial velocity1.2 m/s to the right and collides with a spring of negligiblemass and force constant 50 N/m.(b) Suppose a constant force of kinetic friction acts between the block and the surface, with k = 0.50. If the speed of the block at the moment it collides with the spring is 1.2 m/s, what is the maximum compression in the spring?

9.2Bx cm

K K Kf n mg (b)

2 21 1(0 ) ( 0)

2 2fi B A K BE E E kx mv x

20.50(0.80 )(9.80 / ) 3.92kg m s N

3.92K B BE f x x

225 3.92 0.576 0B Bx x

PROBLEM 9

A 2.00-kg package is released on a 53.10 incline, 4.00 m from a long spring with force constant 120 N/m that is attached at the bottom of the incline. The coefficient of frictionbetween the package and the incline is k = 0.20. The mass of the spring is negligible. (a) What is the speed of the package just before it reaches the spring?

SOLUTION

cosfriction k kW f L Lmg

cosk k kf n mg

1 10 ; sin ;K U mgL 22 2

1; 0

2K mv U

2 2 1 1( ) ;frictionW K U K U

2 (sin cos )kv gL

21cos sin

2kLmg mv mgL

7.30 /m s

(a)

1

2

L

Ugrav = 0

PROBLEM 9

A 2.00-kg package is released on a 53.10 incline, 4.00 m from a long spring with force constant 120 N/m that is attached at the bottom of the incline. The coefficient of frictionbetween the package and the incline is k = 0.20. The mass of the spring is negligible. (b) What is the maximum compression of the spring?

SOLUTION1

2

3d

' ( ) ( ) cosfriction k kW f L d L d mg

1 10 ; ( )sin ;K U mg L d 2

3 3

10 ;

2K U kd

3 3 1 1' ( ) ;frictionW K U K U

24.504 4.00 0 ;d d

21( ) cos 0 ( )sin 0

2k L d mg kd mg L d

1.06d m

(b)

U grav= 0

L

PROBLEM 9

(c) The package rebounds back up the incline. How close does it get to its initial position?

SOLUTION1

2

3d

'' ( ) ( )friction k kW f L d f d L y

1 10 ; ( )sin ;K U mg L d

4 40 ; ( )sinK U mg d L y

4 4 1 1'' ( ) ;frictionW K U K U

(2 2 ) cos ( )sin ( )sink L d y mg mg L d y mg L d

1.32y m

(c)

U grav= 0

L

y4

(2 2 ) cosk L d y mg

sinmgy 2

( ) ;tan

k

k

y L d