ic_class_091012

During this interval KE increases, so even if `dW_on_NC is zero, we still have `dPE + `dKE = 0 so that `dKE = - `dPE.

`dKE is greater than if `dW_NC_ON is negative.

Note that the mass of the rubber band is small compared to that of the block, so its KE is relatively insignificant; the main KE change is in the block

On interval 2 the block began with significant KE, and lost it. What do we therefore conclude about `dW_ON_NC, and what is the source of that work?

Class 091012

Submitting the qa's for this course is recommended, but is

more or less optional:

• If you aren't making A's on your tests you probably
  need to do at least most of the q_a_'s.
• If you can read through the entire qa, do the
  problems pretty much in your head (with ballpark estimates on the
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  solutions, and if necessary jot down enough notes to be sure you'll retain
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• If there is any problem you aren't completely sure
  of, then you should submit including a self-critique, for my feedback. 
  You don't need to answer every question; only the ones you aren't sure of. 

• If you don't answer a question I will assume you
  understand it thoroughly.  If you don't submit a q_a_ I will make the
  same assumption.
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  previously assigned q_a_ (perhaps because you didn't really understand what
  you thought you did), it will always be accepted and I will always post my
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  the given solution when all you're doing is recognizing most of the words. 
  If you can't work a problem without looking at the solution, you don't
  understand it.  If you're skipping problems or qa's, you should
  periodically 'quiz' yourself by going back and working a few of the problems
  without looking at the solutions.

Energy Conservation in terms of `dW_ON_NC, `dPE and `dKE

We have previously seen the following:

Using F_net = m a and vf^2  = v0^2 + 2 a `ds on

an interval between two events, we obtain

F_net * `ds = 1/2 m vf^2 - 1/2 m v0^2.

Defining `dW_net as F_net * `ds, and KE = 1/2 m v^2,

the work done on the interval by the net force thus becomes

• `dW_net = `d(KE).

F_net is the work done ON the mass m by the net force. 

The distinction between the force and work ON the system vs. force and work done

BY the system is very important.  We express the situation here explicitly

by writing F_net_ON and `dW_net_ON.  Our statement is thus expressed as

follows:

The force F_net_ON does work on a system of mass m as it

moves through displacement `ds in the direction of F_net, the work being equal

to F_net_ON * `ds.  We designate this as `dW_net_ON.  Then, for this

interval,

• `dW_net_ON = `d(KE),

the work done ON the system by the net force is equal to

the change in its kinetic energy.  This is the work-kinetic energy

theorem.

We have also defined potential energy as being

equal and opposite to the work done ON the system by

conservative forces. 

It's pretty straightforward to break the work done by the

net force into the work done by conservative forces (which will be associated

with the change in PE) and the work done by nonconservative forces.  This

will give us the more general work-energy theorem:

Any net force is a combination of

conservative and nonconservative forces (one or both of which can

be zero).  So F_net_ON can be expressed as

• F_net_ON = F_net_ON_cons + F_net_ON_noncons,

The work done by the net force is therefore equal to

the work done by conservative forces plus the work done by nonconservative

forces.  In symbols we have

`dW_net_ON = F_net_ON * `ds

= (F_net_ON_cons + F_net_ON_noncons) * `ds

= F_net_ON_cons * `ds + F_net_ON_noncons * `ds

= `dW_net_ON_cons + `dW_net_ON_noncons,

where we make the obvious definitions

`dW_net_ON_cons = F_net_ON_cons * `ds and

`dW_net_ON_noncons = F_net_ON_noncons * `ds.

Since

• `dW_net_ON = `dW_net_ON_cons + `dW_net_ON_noncons,

the work-kinetic energy theorem `dW_net_ON = `dKE

becomes

• `dW_net_ON_cons + `dW_net_ON_noncons = `dKE.

We call this the work-energy theorem.

Now note that `dW_net_ON_cons is the work done

by conservative forces on the system, which our definition of

`dPE says to be equal and opposite the change in PE. 

So `dW_Net_ON = - `dPE and our work-energy theorem `dW_net_ON_cons

+ `dW_net_ON_noncons = `dKE becomes

-`dPE + `dW_net_ON_noncons = `dKE.

Let's abbreviate `dW_net_ON_noncons a bit as `dW_NC_ON,

keeping in mind that this expression represents the total work done by all

nonconservative forces acting on our system.  Our restatement is

-`dPE + `dW_ON_NC = `dKE.

We add `dPE to both sides to get a more intuitive form

of the statement:

• `dW_ON_NC = `dKE + `dPE.

This says that the total work done on

the system by nonconservative forces is equal to the sum

of the changes in KE and PE.

Our focus for today will be to understand the three

quantities, `dW_ON_NC, `dKE and `dPE in terms of systems we have observed in

class.

Two-incline system

Students responding to the 09107 Class Notes clearly

understood the following:

When the ball rolls down one ramp, we have seen

that it loses gravitational PE and gains KE (it goes lower and

speeds up). 

When it travels up the other ramp it gains

gravitational PE and loses KE.

However its PE when it reaches its maximum

displacement on the second ramp is less than it was when

it was released on the first ramp.

With every roll down one ramp and up the other, the

gravitational PE decreases, as does the maximum KE of the ball when it

reaches the point between the two ramps.

We now ask where that PE goes.

• The simple answer is that it is dissipated, mostly by
  rolling friction between the ball and the ramp.
• Let's look into some of the details:

`q001.  Consider the ball rolling from rest down one

ramp then rolling to rest as it travels up the other.

On this interval

• Does its gravitational PE increase or decrease, and
  why?
• Does its KE increase or decrease, and why?

****

The ball ends up lower on the second ramp than it was when

released on the first.  So from release to rest its gravitational PE

decreases.  Thus the change in gravitational PE was negative.

The ball was released from rest and ended up at rest, so

its initial and final velocities were zero.  Its initial and final KE were

therefore both zero, so the change in KE was zero.

&&&&

`q002.  The work-energy theorem says that  `dW_ON_NC

= `dKE + `dPE. 

The right-hand side of this expression is `dKE + `dPE. 

Based on your answers to the preceding question, is `dKE + `dPE positive,

negative or zero?

Do you therefore conclude that `dW_ON_NC is positive,

negative, or zero?

****

`dKE was zero and `dPE was negative.  Therefore `dKE

+ `dPE is negative (zero plus a negative is negative).

Since `dW_ON_NC

= `dKE + `dPE, it follows that `dW_ON_NC is negative.

That is, the work done on the system by nonconservative

forces is negative.

&&&&

`q003.  Assume that `dW_ON_NC is the result of

rolling friction, so that `dW_ON_NC = F_frict_ON * `ds, where F_frict_ON is the

force of rolling friction ON the system and `ds the displacement of the ball

along the ramp.  (there is a slight problem with this wording due to the

fact that the ramps are not parallel, but we're going to 'gloss over' this for

now)

From the sign of `dW_ON_NC, do you conclude that

F_frict_ON is in the direction of `ds or opposite the direction of `ds?

Based on your understanding of frictional forces, do you

think that F_frict_ON acts in the direction of the ball's motion, or opposite

this direction?

Are you answers to these questions consistent?

****

Since `dW_ON_NC = F_frict_ON * `ds is negative, F_frict_ON

and `ds must be of opposite signs.

When an object moves across a surface, the frictional

force exerted by the surface on the object resists its motion.  Thus

friction acts in the direction opposite motion.  This is completely

consistent with the fact that `dW_ON_NC is negative so that F_frict_ON and `ds

must be of opposite signs.

&&&&

Ball down ramp and to floor

The ball rolls from point A at the top of the ramp to

point B at the bottom of the ramp then fall to point C, at which it makes its

very first contact with the floor.

`q004.  What three events occur at these three

points?

What happens during the interval between the first and

second events (we will call this 'interval 1')?

What happens during the interval between the second and

third events (we will call this 'interval 2')?

****

The first event is the release of the ball.

The second event is the ball's arrival at and departure

from the end of the ramp.

The third event is the ball's very first contact with the

floor.

On interval 1 the ball rolls from rest down the ramp.

On interval 2 the ball falls freely to the floor with a

small initial vertical velocity and a larger initial horizontal velocity.

&&&&

`q005.  The only conservative force doing work during

the entire interval from the first to the third event is the gravitational

force.  (An excellent question that came up in one of the groups is whether

the normal force is conservative.  For the steel ramp it turns out that it

is, but the normal force acts in the direction perpendicular to incline, so

there is no significant displacement in the direction of this force, and it has

no significant effect on this energy situation.  In some situations the

normal force could be associated with a significant amount of energy, but

analysis of these situations are beyond the scope of the course, except for the

general statement that the slight flexing of the ramp does create some thermal

energy (heat) at the expense of the ball's KE.)

It is obvious that the KE of the ball increases during

both intervals.

What nonconservative forces (if any) act during interval

1?

What nonconservative forces (if any) act during interval

2?

****

Rolling friction is the only non-negligible

nonconservative force acting along the ball's line of motion during the first

interval.  The normal force between the ramp and the ball is mostly

conservative (the steel ramp bends slightly but returns to its original shape as

the ball passes), but on, say, an earth ramp the normal force would be mostly

nonconservative.

Air friction, which is pretty much negligible for a steel

ball at the speeds encountered in this experiment, is the only significant

nonconservative force encountered during the fall.  (The floor will exert

nonconservative forces, but only after the balls very first contact.)

Note that gravitational

&&&&

`q006.  How does the presence of nonconservative

forces during interval 1 affect the KE change during that interval (e.g., is the

KE change different than it would be if the nonconservative forces were not

present, and if so would the KE change be greater or less in the absence of

these forces)?

How does the presence of nonconservative forces during

interval 1 affect the PE change during that interval?

You may answer according to your intuition, but also

attempt to identify `dW_NC_ON, `dPE and `dKE.  Specify whether each is

positive or negative, and reconcile your conclusions with the equation `dW_NC_ON

= `dPE + `dKE.

****

The force of rolling friction, which is nonconservative,

causes the ball to pick up less speed than it would in the absence of friction. 

So `dKE is lower than it would be were there no friction.

The change in altitude on the the first interval is from

the top of the ramp to the bottom, and as long as the ball gets from the top to

the bottom, the change in altitude isn't affected by the nonconservative forces. 

The change in gravitational PE depends only on the change in altitude, so it

isn't affected either.

`dPE is negative, `dW_NC_ON is negative (the frictional

force does negative work on the ball, as we saw earlier), and `dKE is positive.

`dKE + `dPE is the sum of a positive and a negative

quantity, and must equal the negative quantity `dW_NC_ON.  So the negative

`dPE must be greater in magnitude than the positive `dKE.

&&&&

`q007.  Suppose that for interval 1, `dW_NC_ON is

-1000 ergs and `dPE is -100 000 ergs.  What then would be `dKE on this

interval?  (you don't need to know what an erg is to answer this question,

but you're probably wondering; an erg is a unit of energy, it take 10 000 000

ergs to make a Joule, and a Cheerio contains something like a thousand Joules of

chemical energy, maybe 15% of which you could use to actually perform work).

What would you estimate to be the PE change between the

end of the ramp and the floor?  What therefore do you think would be the KE

change between event 1 and event 3?

****

Since `dKE = -`dPE + `dW_NC_ON = -(-100 000 ergs) - 1000

ergs = +99 000 ergs.

The ball's position decreases by about a centimeter on the

ramp, and by about 100 cm between ramp and floor.  So its 1000 erg `dPE on

the ramp is only about 1% of its `dPE from the end of the ramp to the floor. 

We conclude that `dPE from ramp's end to floor is about 100 * (-100 000 ergs) =

-10 000 000 ergs.

So using the same reasoning as before the ball will gain

about 10 000 000 ergs - 1000 ergs = 9 999 000 ergs, which is not significantly

different than 10 000 000 ergs.

&&&&

Rubber band block across table

Let event 1 be the release of the rubber band.

Let event 2 occur at the instant the rubber band loses

contact with the block.

Let event 3 occur at the instant the block comes to rest.

`q008.  What happens during the interval between the

first and second events (we will call this 'interval 1')?

What happens during the interval between the second and

third events (we will call this 'interval 2')?

****

&&&&

`q009.  Is there any change in gravitational PE

between event 1 and event 3?

Is there any change in elastic PE on interval 1?

Is there any change in elastic PE on interval 2?

****

On interval 1 the rubber band loses PE as it snaps back.

On interval 2 the rubber band presumably doesn't change

its position or length (a small amount of vibration might occur, but will have

insignificant energy) so there is no elastic PE, and the block remains at the

same altitude so there is no change in gravitational PE. 

&&&&

`q010.  Answer for each interval:

Is there any nonconservative force acting on the system,

and if so what?

Does the KE of the system increase or decrease?

****

Friction is the only significant nonconservative force

acting in the direction of motion.

The normal force is a mix of conservative and

nonconservative forces, but it acts in the direction perpendicular to motion so

it has no direct effect on the energy of the system.  (the normal force is

related to the frictional force so is indirectly related to the work done on the

system).

Both of these forces act during both intervals.

On the first interval the KE of the system, which is

initially at rest, increases.

On the second interval the system comes to rest, so its KE

decreases.

&&&&

`q011.  Assume that the rubber band forces are

completely conservative (not so, but assume the ideal case).

Assume that frictional forces dissipate .0002 Joule of

energy for every centimeter the block slides.

Suppose that the rubber band initially stores .01 Joule of

potential energy, and that the block slides 10 cm while the rubber band is in

contact with it. 

How much KE will the block have at the end of interval 1?

****

The rubber band loses its PE, so `dPE = -.01 Joule.

Frictional forces dissipate .0002 Joules of energy for

every centimeter, so if the block slides 10 cm frictional forces will dissipate

.002 Joules of energy.  So `dW_NC_ON = -.002 Joules.

Solving `dW_NC_ON = `dPE + `dKE for the desired quantity `dKE

we get

`dKE = `dW_NC_ON - `dPE

(this says that the KE change is equal to the work done on

the system by nonconservative forces, which is in this case negative, plus the

potential energy loss).

Substituting our quantities we get

`dKE = -.002 J - (-.01 J) = .098 J.

In terms of just plain common sense, the rubber band gives

up .01 J of energy, which would all go into kinetic energy except that friction

'uses up' .002 J, leaving .0098 J.

&&&&

`q012.  Continuing the preceding question, what is

the change in KE during interval 2, what is the change in PE during this

interval, and what therefore is the work done by the nonconservative force?

****

The ball ends interval 1 with .0098 J of KE.

It ends up at rest.

So on the second interval, `dKE = -.0098 J.

Since `dPE on this interval is zero, we have

`dW_NC_ON = `dPE + `dKE = 0 + (-.0098 J) = -.0098 J.

&&&&

`q013.  Continuing the preceding two questions,

assuming that the frictional force is the only nonconservative force acting

during interval 2, how far will the block therefore slide during this interval?

****

Every centimeter, the nonconservative force (friction)

dissipates .0002 J.  Thus

`dW_NC_ON = -.0002 J / cm * `ds.

So

`ds = `dW_NC_ON / (-.0002 J / cm) = -.0098 J / (-.0002

J/cm) = 49 cm.

&&&&

Pendulum-projectile

`q014.  If a pendulum of length 10 cm is at its

equilibrium position, then it is 10 cm vertically below the fixed point at which

it is being held.

If it is then pulled back 6 cm in the horizontal

direction, then since the string is still 10 cm long, the mass is still 10 cm

from that fixed point.

How far, in the vertical direction, is the pulled-back

pendulum below that fixed point?  (Hint:  sketch the figure, identify

the necessary right triangle and use the Pythagorean Theorem)

How far does it therefore descend in the vertical

direction, between release and return to its equilbrium position?

****

A sketch will reveal a right triangle with horizontal and

vertical legs, whose hypotenuse is 10 cm, with the horizontal leg of length 6 cm

representing the pullback.

The vertical leg represents the distance of the washer

below the fixed point.

The Pythagorean Theorem says that

a^2 + b^2 = c^2,

where and b are the legs and c the hypotenuse. 

Letting the horizontal leg be represented by a, we solve for b:

b^2 = c^2 - a^2 so that

b = +-sqrt(c^2 - a^2).

Substituting we get

b = +- sqrt((10 cm)^2 - (6 cm)^2) = +- sqrt(100 cm^2 - 36

cm^2) = +-sqrt(64 cm^2) = +- 8 cm.

So the vertical leg is 8 units long, putting the washer 8

cm below the fixed point, and the washer is therefore 2 cm above its original

position (which was 10 cm below the fixed point).

&&&&

`q015.  If the pendulum has a weight of .3 Newton,

then how much work is done on it by the gravitational force, as it swings back

to equilibrium?

By how much does its gravitational PE therefore change?

What therefore will be its KE at the equilibrium position?

****

Its gravitational PE changes by an amount equal and

opposite to the work done by gravity as it swings back to equilibrium.

The gravitational force acts in the downward direction. 

In the vertical direction the pendulum moves 2 cm downward.

The gravitational force, being in the same direction as

the vertical displacement, therefore does positive work on the pendulum. 

This work is F_grav * `ds = .3 N * (2 cm) = .3 N * .02 m = .006 N * m = .006

Joules.

We conclude that the PE of the pendulum therefore

decreases by .006 Joules.  That is, `dPE = -.006 J.

Assuming that nonconservative forces (e.g., air

resistance) are negligible, our energy conservation equation

`dW_NC_ON = `dPE + `dKE becomes

`dPE + `dKE = 0 so that

`dKE = -`dPE = - (-.006 J) = .006 J.

This is how we formally represent the commonsense idea

that the thing loses .006 J of PE, and nothing gets in the way so it gains .006

J of KE.

&&&&

Good work. See my comments.

Compare your results to those in the given solutions, which I've appended. See also the solutions to those last couple of problems and let me know if you have questions.