Chapter 6 Conservation of Energy. MFMcGraw-PHY 1401Ch06b - Energy - Revised: 6/25/102 Conservation of Energy Work by a Constant Force Kinetic Energy Potential

Chapter 6

Conservation of Energy

MFMcGraw-PHY 1401 Ch06b - Energy - Revised: 6/25/10 2

Conservation of Energy

• Work by a Constant Force

• Kinetic Energy

• Potential Energy

• Work by a Variable Force

• Springs and Hooke’s Law

• Conservation of Energy

• Power


The Law of Conservation of Energy

The total energy of the Universe is unchanged by any physical process.

The three kinds of energy are:

kinetic energy, potential energy, and rest energy.

Energy may be converted from one form to another or transferred between bodies.



Work by a Constant Force

Work is an energy transfer by the application of a force.

For work to be done there must be a nonzero displacement.

The unit of work and energy is the joule (J).

1 J = 1 Nm = 1 kg m2/s2.


Only the force in the direction of the displacement that does work.

An FBD for the box at left:

x

y

Fw

N

The work done by the force F is:

xFrFW xxF cos

rx

Frx

Work - Example


The work done by the normal force N is: 0NW

The normal force is perpendicular to the displacement.

The work done by gravity (w) is: 0gW

The force of gravity is perpendicular to the displacement.

Work - Example


The net work done on the box is:

xF

xF

WWWW gNF

cos

00cos

net

Work - Example


In general, the work done by a force F is defined as

cosrFW

where F is the magnitude of the force, r is the magnitude of the object’s displacement, and is the angle between F and r (drawn tail-to-tail).

Work Done


Example: A ball is tossed straight up. What is the work done by the force of gravity on the ball as it rises?

FBD for rising ball:

x

y

w

r

ymg

ywWg

180cos

Work - Example


A box of mass m is towed up a frictionless incline at constant speed. The applied force F is parallel to the incline.

Question: What is the net work done on the box?

F

w

NFx

y

0cos

0sin

wNF

wFF

y

x

Apply Newton’s 2nd Law:

An FBD for

the box:

Inclined Plane-V Constant


The magnitude of F is: sinmgF

If the box travels along the ramp a distance of x the work by the force F is

sin0cos xmgxFWF

The work by gravity is

sin90cos xmgxwWg

Example continued:



Example continued:

The work by the normal force is:

090cos xNWN

The net work done on the box is:

0

0sinsin

net

xmgxmg

WWWW NgF



Example: What is the net work done on the box in the previous example if the box is not pulled at constant speed?

sin

sin

wmaF

mawFFx

Proceeding as before:

xFxma

xmgxmgma

WWWW NgF

net

net

0sinsin

Inclined Plane-Acceleration

New Term


Kinetic Energy

2

2

1mvK is an object’s translational

kinetic energy.

This is the energy an object has because of its state of motion.

It can be shown that, in general

Net Work = Change in K

.net KW


Example: The extinction of the dinosaurs and the majority of species on Earth in the Cretaceous Period (65 Myr ago) is thought to have been caused by an asteroid striking the Earth near the Yucatan Peninsula. The

resulting ejecta caused widespread global climate change.

If the mass of the asteroid was 1016 kg (diameter in the range of 4-9 miles) and had a speed of 30.0 km/sec, what was the asteroid’s kinetic energy?

J 105.4

m/s 1030kg 102

1

2

1

24

23162

mvK

This is equivalent to ~109 Megatons of TNT.

Kinetic Energy


Gravitational Potential Energy Part 1- Close to Earth’s Surface

Potential energy is an energy of position.

There are potential energies associated with different forces. Forces that have a potential energy associated

with them are called conservative forces.

Not all forces are conservative, i.e. Friction.


Gravitational Potential Energy

Every conservative force has an associated potential function.

F is equal in magnitude and opposite in direction of mg. The object is raised a distance Δx doing work against the gravitational field. The change in U is positive

0

f i

f i

W = F Δx = mgΔxcos(0 )

W = ΔU = U - U = mgΔx(+1)

ΔU = U - U = mgΔx


The change in gravitational potential energy (only near the surface of the Earth) is

ymgU g

where y is the change in the object’s vertical position with respect to some reference point.

You are free to choose to location of this reference point where ever it is convenient.



The table is 1.0 m tall and the mass of the box is 1.0 kg.

Ques: What is the change in gravitational potential energy of the box if it is placed on the table?

J 8.9m 0m 0.1m/s 8.9kg 0.1 2

ifg yymgymgU

First: Choose the reference level at the floor. U = 0 here.

GPE Problem

U = 0

U = 0


Example continued:

J 8.9m 0.1 m0.0m/s 8.9kg 1 2

ifg yymgymgU

Now take the reference level (U = 0) to be on top of the table so that yi = 1.0 m and yf = 0.0 m.

The results do not depend on the location of U = 0.

GPE Problem


Mechanical energy is UKE

The total mechanical energy of a system is conserved whenever nonconservative forces do no work.

That is Ei = Ef or K = U.

Total Mechanical Energy

Then if K increases U decreases and vice versa


A cart starts from position 4 with v = 0.0 m/s to the left. Find the speed of the cart at positions 1, 2, and 3. Ignore friction.

4 3

4 4 3 3

2 24 3 3

3 4 3

E = E

U + K = U + K

1 1mgy + m(0) = mgy + mv

2 2

v = 2g y - y = 14.0 m/s

Mechanical Energy Problem


4 2

4 4 2 2

2 24 2 2

2 4 2

E = E

U + K = U + K

1 1mgy + m(0) = mgy + mv

2 2

v = 2g y - y = 9.1 m/s

4 1

4 4 1 1

2 24 1 1

1 4 1

E = E

U + K = U + K

1 1mgy + m(0) = mgy + mv

2 2

v = 2g y - y = 19.8 m/s

Or use E3=E2

Or use E3=E1

E2=E1

Mechanical Energy Problem


A roller coaster car is about to roll down a track. Ignore friction and air resistance.

40 m 20

m y=0

m = 988 kg

(a) At what speed does the car reach the top of the loop?

m/s 8.192

2

10 2

fif

ffi

ffii

fi

yygv

mvmgymgy

KUKU

EE

Roller Coaster Problem


(b) What is the force exerted on the car by the track at the top of the loop?

N w

y

x

N 109.2 42

2

2

mgr

vmN

r

vmwN

r

vmmawNF ry

Example continued:

FBD for the car:Apply Newton’s Second Law:



(c) From what minimum height above the bottom of the track can the car be released so that it does not lose contact with the track at the top of the loop?

Example continued:

2min2

10 mvmgymgy

KUKU

EE

fi

ffii

fi

Using conservation of mechanical energy:

g

vyy fi 2

2minSolve for the starting height



Example continued:

v = vmin when N = 0. This means that

grv

r

vmmgw

r

vmwN

min

2min

2

What is vmin?

m 0.255.22

22

2min r

g

grr

g

vyy fi

The initial height must be



What do you do when there are nonconservative forces? For example, if friction is present

fricWEEE if

The work done by friction.

Nonconservative Forces

In the presence of non-conservative forces Total Mechanical Energy is not conserved.


Gravitational Potential Energy Part 2 - Away from Earth’s Surface

The general expression for gravitational potential energy is:

0 where

21

rUr

MGMrU

It is convenient in this case to assign the zero of PE to a point an infinite distance away.



0 where

21

rUr

MGMrU

The potential energy can be viewed as the work required to bring a second mass from infinity to its present position r. Since we can a conservative force that has a potential we don’t have to do the force x distance calculation.

Work = ΔU = U r - U r =

where U r = = 0

Work = ΔU = U r


Example:

What is the gravitational potential energy of a body of mass m on the surface of the Earth?

e

ee R

mGM

r

MGMRrU 21


The negative sign is useful in describing bound states, i.e. A moon bound to an planet, as having a negative energy. A positve energy state would represent a free state, i.e. an unbound moon that had somehow bee freed from its planet..


A planet of mass m has an elliptical orbit around the Sun. The elliptical nature of the orbit means that the distance between the planet and Sun varies as the planet follows its orbital path. Take the planet to move counterclockwise from its initial location.

QUES: How does the speed of a planet vary as it orbits the Sun once?

The mechanical energy of the planet-sun system is:

constant2

1 sun2 r

GmMmvE

rB A

Planetary Motion


constant2

1 sun2 r

GmMmvE

Starting from point “B” (where the planet moves the slowest), as the planet moves in its orbit r begins to decrease. As it decreases the planet moves faster.

At point “A” the planet reaches its fastest speed. As the planet moves past point A in its orbit, r begins to increase and the planet

moves slower.

rB A

Planetary Motion

At point “B” the planet is the farthest from the Sun. At point “A” the planet is at its closest approach to the sun.


Work by a Variable Force

Work can be calculated by finding the area underneath a plot of the applied force in the direction of the displacement versus the displacement.


x (m)

Fx (N)F3

F2

F1

x3x2x1

The work done by F1 is 0111 xFW

Example: What is the work done by the variable force shown below?

The net work is then W1+W2+W3.

The work done by F2 is 1222 xxFW

The work done by F3 is 2333 xxFW


By hanging masses on a spring we find that stretch applied force. This is Hooke’s law.

For an ideal spring: Fx = kx

Fx is the magnitude of the force exerted by the free end of the spring, x is the measured stretch of the spring, and k is the spring constant (a constant of proportionality; its units are N/m).

A larger value of k implies a stiffer spring.

Spring Force


(a) A force of 5.0 N applied to the end of a spring cause the spring to stretch 3.5 cm from its relaxed length.

Ques: How far does a force of 7.0 N cause the same spring to stretch?

For springs Fx. This allows us to write .2

1

2

1

x

x

F

F

Solving for x2: cm. 9.4cm 5.3N 5.0

N 0.71

1

22

x

F

Fx

Spring Force


Example continued:

N/cm. 43.1cm 3.5

N 0.5

1

1 x

Fk

(b) What is the spring constant of this spring?

N/cm. 43.1cm 4.9

N 0.7

2

2 x

Fk

Or

Spring Force


An ideal spring has k = 20.0 N/m. What is the amount of work done (by

an external agent) to stretch the spring 0.40 m from its relaxed length?

J 6.1m 4.0N/m 0.202

1

2

1

2

1

curveunder Area

22111

kxxkx

W

Spring Force


The work done in stretching or compressing a spring transfers energy to the spring.

Below is the equation of the spring potential energy.

2

2

1kxU s

Elastic Potential Energy

The spring is considered the system


A box of mass 0.25 kg slides along a horizontal, frictionless surface with a speed of 3.0 m/s. The box encounters a spring with k = 200 N/m.

Ques: How far is the spring compressed when the box is brought to rest?

m 11.0

02

1

2

10 22

vk

mx

kxmv

KUKU

EE

ffii

fi

Elastic Potential Energy


Power

cosFvP

Average Power

The unit of power is the watt. 1 watt = 1 J/s = 1 W.

1 Horsepower (hp) = 550 ft-lbs/s = 745.7 W

Instantaneous Power

Power is the rate of energy transfer.

av

av avg

Energy Work F Δx ΔxP = = = = F

Time Time Δt Δt

P = F v


A race car with a mass of 500.0 kg completes a quarter-mile (402 m) race in a time of 4.2 s starting from rest. The car’s final speed is 125 m/s. (Neglect friction and air resistance.)

Ques: What is the engine’s average power output?

watts103.921

5

2

av

t

mv

t

K

t

KU

t

EP

f

Power - Car Example


Summary

• Conservation of Energy

• Calculation of Work Done by a Constant or Variable Force

• Kinetic Energy

• Potential Energy (gravitational, elastic)

• Power

Documents

Chapter 6 Conservation of Energy. MFMcGraw-PHY 1401Ch06b - Energy - Revised: 6/25/102 Conservation of Energy Work by a Constant Force Kinetic Energy Potential