physics430_lecture07 (1)

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Physics 430: Lecture 7Kinetic and Potential Energy

Dale E. Gary

NJITPhysics Department


2/17

September 22, 2009

We are now going to take up the conservation of energy, and itsimplications. You have all seen this before, but now we will use a powerful,more mathematical description.

You will see that the discussion is more complicated that the otherconservation laws for linear and angular momentum. The main reason isthat each type of momentum comes in only one flavor, whereas there are

many forms of energy (kinetic, several kinds of potential, thermal, etc.). Processes transform one type of energy into another, and it is only the total

energy that is conserved, hence the additional complication.

We will be introducing new mathematical tools of vector calculus, such asthe gradient and the curl, which you may be familiar with, or not. I willgive you the needed background as they come up.

Chapter 4Energy


3/17

September 22, 2009

An obvious form of energy is energy of motion, or kinetic energy. We willuse the symbol T, which is perhaps strange to you but is very muchstandard in Classical Mechanics. The kinetic energy of a particle of mass mtraveling at speed vis defined to be:

Consider such a particle moving on some trajectory through space while its

kinetic energy changes, on moving from position r1to r1+ dr. We can takethe time derivative of the kinetic energy, after writing , so that

But the first term on the right is the force . Thus, we can write

the derivative of kinetic energy as

Finally, multiplying both sides by dtand noting that vdt= dr, we have

4.1 Kinetic Energy and Work

2

21 mvT

vv 2v

vvvvvvvv mmdt

dm

dt

dT)()(

21

21

vpF m

vF dtdT

rF ddT

r1r1+drdr

Work-KE Theorem


4/17

September 22, 2009

The equation just derived is only valid for an infinitesimal displacement, but

we can extend this to macroscopic displacements by integrating, to get:

which says that the change in kinetic energy of a particle is equal to thesum of force (in the direction of the displacement) times the incrementaldisplacement.

However, note that this is the displacement along the pathof the particle.Such an integral is called a line integral. In evaluating the integral, it isusually possible to convert it into an ordinary integral over a single variable,as in the following example (which we will look at in a moment).

With the notation of the line integral

where the last is a definition, defining the work done by Fin moving frompoint 1 to point 2. Note that Fis the net force on the particle, but we canalso add up the work done by each force separately and write:

Line Integrals and Work

2

1

r

rrF dT

)21(2

112

WdTTT rF

i

iWTT )21(12


5/17

September 22, 2009

Evaluate the line integral for the work done by the 2-d force F= (y, 2x)going from the origin Oto the pointP= (1, 1)along each of the three paths:

a)OQthen QP

b) OPalongx = y

c) OPalong a circle

Path a):

Path b):

Example 4.1: Three Line Integrals

22021

0

1

0

1

0

dyxdyydx

dddWP

Q

Q

Oaa rFrFrF

P

O Q

5.12332

1

0

21

00

x

xdxxdyydx

ddW

P

P

Obb rFrF


6/17

September 22, 2009

Path c): This is a tricky one. Path c can be expressed as

so

This is a parametric equation, using qas a parameter along the path. Withthis parameter, F= (sin q, 2(1cos q)). With this substitution:

The point here is that the line integral depends on the path, in general (butnot for special kinds of forces, which we will introduce in a moment).

Example 4.1: Three Line Integrals

21.14/2

cos)cos1(2sin2/

0

2

qqqq

ddWc

c rF

)sin,cos1(),( qq yxr

)cos,(sin),( qq dydxdr


7/17

September 22, 2009

We must now introduce the concept of potential energy corresponding tothe forces on an object. As you know, not every force lends itself to acorresponding potential energy. Those that do are called conservativeforces.

There are two conditions that a force must satisfy to be considered a

conservative force. The first condition for a force Fto be conservative is that Fdepends only

on the position rof the object on which it acts. It cannot depend onvelocity, time, or any other parameter.

Although this is restrictive, there are plenty of forces that satisfy thiscondition, such as gravity, the spring force, the electric force. You can oftensee this directly, such as for the gravitational force:

4.2 Potential Energy andConservative Forces

rrF )(2r

GmM Depends only onr.


8/17

September 22, 2009

The second condition for a force to be conservative concerns the work doneby the force as the object on which it acts moves between two points r1and r2(or just points 1 and 2, for short)

Reusing our earlier figure, we saw in Example 4.1 that the force describedthere was NOT conservative, because it did different amounts of work forthe three paths a, b, and c.

Forces involving friction, obviously are not conservative, because if youwere sliding a box, say, on a surface with friction along the three pathsshown, the friction would do work , whereLis different

for the three paths. Such forces are non-conservative.

Non-Conservative Forces

2

1)21( rF dW

2

1

LfW fricfric )21(


9/17

September 22, 2009

The force of gravity, on the other hand, has the property that the workdone is independent of the path. You know that if the height of point 1 andpoint 2 differ by an amount h, then you will drop in height by hno matter

what path you take. In fact

independent of path.

The conditions for a force to be conservative, then, are:

Conservative Forces

mghW )21(grav

Conditions for a Force to be Conservative

A force Facting on a particle is conservative if and only if it satisfies

two conditions:

1. F depends only on the particles position r(and not on the velocity

v, or the time t, or any other variable); that is, F= F(r).

2. For any two points 1 and 2, the work W(1 2)done by Fis the

same for all paths between 1 and 2.


10/17

September 22, 2009

The reason that forces meeting these conditions are called conservative isthat, if all of the forces on an object are conservative we can define aquantity called potential energy, denoted U(r), a function only of position,

with the property that the total mechanical energy

is constant, i.e. is conserved.

To define the potential energy, we must first choose a reference point ro, atwhich Uis defined to be zero. (For gravity, we typically choose thereference point to be ground level.) Then U(r), the potential energy, at anyarbitrary point r, is defined to be

In words, U(r)is minus the work done by Fwhen the particle moves fromthe reference point roto the point r.

Potential Energy

)(rUTE

Potential Energy

r

rrrFrrr

o

)()()( o dWU


11/17

September 22, 2009

Statement of the problem: A charge qis placed in a uniform electric field pointing in the xdirection with

strengthEo, so that the force on qis . Show that this force is

conservative and find the corresponding potential energy.

Solution:

The work done by Fin going between any two points 1 and 2 along any path

(which is negative potential energy) is:

This work done is independent of the path, because the electric force dependsonly on position, i.e. the force is conservative. To find the corresponding

potential energy, we must first choose a reference point at which Uis zero. Anatural choice is to choose our origin (the point 1), in which case the potentialenergy is

You may recall that Eoxis the electric potential V, so that qVis the potential

energy.

Example 4.2: Potential Energy of aCharge in a Uniform Electric Field

)()21( 12o2

1o

2

1o

2

1xxqEdxqEdqEdW rxrF

xEF oqEq

xqEWrU o)0()( r


12/17

September 22, 2009

The potential energy can be defined even when more than one force is

acting, so long as all of the forces are conservative. An important exampleis when both gravity Fgravand a spring force Fsprare acting (so long as thespring obeys Hookes Law, F(r) = kr).

The work-kinetic energy theorem says that if we move an object subject tothese two forces along some path, the forces will do work independent ofthe path (depending only on the two end-points of the path) given by

Rearrangement shows that

hence total mechanical energy is conserved. Extended to nsuch forces:

Several Forces

)( sprgravsprgrav UUWWT

Principle of Conservation of Energy for One ParticleIf all of the nforcesFi(i=1n) acting on a particle are conservative, each with its

corresponding potential energy Ui(r), the total mechanical energy defined as

is constant in time.

)()(1 rr nUUTUTE

0)( sprgrav UUT


13/17

September 22, 2009

As we have seen, not all forces are conservative, meaning we cannot define

a corresponding potential energy. As you might guess, in that case wecannot define a conserved mechanical energy.

Nevertheless, if there are some conservative forces acting, for which apotential energy can be defined, then we can divide the forces into aconservative part Fcons, and a nonconservative part Fnc, such that

which allows us to write

What this says is that mechanical energy (T + U) is no longer conserved,

but any changes in mechanical energy are precisely equal to the work doneby the nonconservative forces.

In many problems, the only nonconservative force is friction, which acts inthe direction opposite the motion so that the work is negative.

Nonconservative Forces

nc

nccons

WUWWT

ncWUT )(

rf d


14/17

September 22, 2009

Example 4.3: Block Sliding Downan Incline

We did this problem using forces in lecture 2. Lets now apply these ideasof energy to arrive at the same result.

As before, we have to identify the forces, and set a

coordinate system, but this time we write down the

potential and kinetic energies in the problem.

The kinetic energy, as always, is T= mv2.

The gravitational potential energy is U = mgy, where we can set y = 0(andhence U= 0) at the ground level.

The friction force does negative work Wfric= fd, but recall that f= mNwhere

. Putting all of this together, becomes

where dis the distance along the incline, and yis the change in height. If the block starts out with zero initial velocity at the top of the incline, and

we ask what is the speed vat the bottom, then y= h= d sin q, so

or

qcosmgN fric)( WUT

q

mg

Nf

h

qm cos2

212

21 mgdymgmvmv if

qmq cossin221 mgdmgdmv )cos(sin2 qmq gdv


15/17

September 22, 2009

Comparison with Example 1.1

When we did this problem using forces, we obtained equation of motion

from which, after integration, we got the expression

Comparing with our just derived expression

they may seem quite different. What is happening is that using forces wecan get the velocity versus time, whereas with energy we are only gettingthe speed at the end points. Energy considerations are very powerful ifyou just want to know the result at a particular point, in which case youcan ignore the details of the motion in getting there. If you instead needto know the path taken, or the details along the path, you have to use thetools of Newtons Laws.

However, we will find in a few weeks that these energy considerations docontain all of the information of Newtons Laws, and we will build thetools necessary in Lagrangian mechanics to get the equation of motionstarting from energy. This allows us to attack much more complicatedproblems. For this reason, it is important to get good at energyproblems. Here is an example you probably have seen before.

)cos(sin qmqgx

tgx )cos(sin qmq

)cos(sin2 qmq gdv)cos(sin

2

)cos(sin 221

qmq

qmq

g

dt

tgd


16/17

September 22, 2009

Statement of the problem: (a) The force exerted by a one-dimensional spring, fixed at one end, is F= kx,

wherexis the displacement of the other end from its equilibrium position.

Assuming that this force is conservative (which it is) show that the correspondingpotential energy is U= kx2, if we choose U= 0at its equilibrium position.

Solution to (a):

We start with the definition of potential energy:

But we choose U= 0atx=x1, which amounts to choosing x1 = 0, so that

Problem 4.9

)()21()( 212

221

2

1

2

1xxkxdxkFdxWxU

2

21)( kxxU


17/17

September 22, 2009

Statement of the problem:

(b) Suppose this spring is hung vertically from the ceiling with a mass msuspendedfrom the other end, and constrained to move in the vertical direction only. Find theextension xoof the new equilibrium position with the suspended mass. Show thatthe total potential energy (spring plus gravity) has the same form ky2if we usethe coordinate yequal to the displacement measured from the new equilibriumposition atx=xo(and redefine our reference point so that U= 0aty= 0).

Solution to (b): The new equilibrium position is reached when the force of the stretched

spring kxoequals the force of gravity on the mass mg. Thus

To define the potential energy at the new equilibrium position, we

have to examine the work done in displacing the mass a distance y:

Problem 4.9, contd

gk

mxo

2

212

21

o0

o

0 sprgrav

)((

)()0()(

kykyykxmgyydyxkmg

ydFFyWyU

y

y

xoy=0

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physics430_lecture07 (1)