ic_class_091012

course PHY 201

10:02 PMSunday, November 08, 2009

I'm still way behind on the assignments (or so it appears). However, I thought it best to spend my time working on the QA exercises 9 - 14, which I went through thoroughly on paper. This helped immensely in getting my head where it ought to be with respect to what's been going on the past couple weeks, so hopefully I am mostly prepared for the test and will be able to power through the remaining few assignments and get caught up with the rest of the class as soon as possible. Thanks for your patience with me. I love physics but it just takes a lot of time soaking up the material before it starts to make sense (and I haven't had nearly as much time to do that as I [or you] would like!)."

I know what you're capable of. I expect I'll be seeing a very good result by the time you're done.

"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 arithmetic--no need for calculator-punching) or with a few scratches on paper, see that you understand the process and the units of the given solutions, and if necessary jot down enough notes to be sure you'll retain what you understand and use it when needed, then you don't need to waste time submitting them.

* 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.

* If you find later that you need to work through a 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 response.

* Don't delude yourself into thinking you understand 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?

****

At rest, PE is at its greatest and KE is at its minimum. As the ball reaches the mid-point where the ramps converge, KE is at its greatest and PE is at its minimum.

&&&&

`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 + `dPE is zero since at any point, these forces ""even out"" -- when `dKE is high, `dPE is low, and vice versa. Thus the work done on the system by non-conservative forces is zero (since we are neglecting friction?).

That's right if we're neglecting friction.

&&&&

`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?

****

Friction works in the opposite direction of motion, and thus should be negative.

&&&&

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 events transpire as thus:

1. Ball is released (KE = 0, PE = max)

- KE begins to increase, PE begins to decrease

2. Ball reaches end of ramp (PE = 0, KE_x = max, KE_y = increasing)

- KE reaches max in the horizontal direction

- KE begins to increase in the vertical direction as gravity accelerates the mass toward the center of the earth

- PE reaches 0

3. Ball falls to floor (and comes to rest) (PE = new max (less than original since its altitude has changed), KE = 0 since ball is at rest)

&&&&

`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?

****

Interval 1: friction of the ramp

Interval 2: friction of air resistance

&&&&

`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 prescence of nonconservative forces during interval 1 would slow the change of KE, acting in the opposite direction of motion. In turn, this acts in favor of PE.

`dW_ON_NC = friction, air resistance, working opposite the direction of motion (negative)

`dKE = change in kinetic energy, moves from 0 to maximum (positive)

`dPE = change in potential energy, moves from max to 0 (negative)

Since some energy is lost to the nonconservative forces, the outcome should be negative.

&&&&

`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?

****

`dW_ON_NC = `dKE + `dPE

1000 ergs = `dKE + 100 000 ergs

`dKE = 1000 ergs - 100 000 ergs

= - 99 000 ergs

Very good. I should have said | `dPE | is 100 000 ergs and | `dW_NC_ON | = 1000 ergs, which would have allowed you to get an answer consistent with the behavior of the system.

The appended solution works this problem for `dPE = -100 000 ergs and `dW_NC_ON = -1000 ergs; the results is `dKE = +99 000 ergs, consistent with the fact that the ball speeds up.

&&&&

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')?

****

1. Release of rubber band

- PE begins to decrease from maximum

- KE begins to increase from minimum

2. Block leaves rubber band

- PE reaches minimum and begins to increase

- KE reaches maximum and begins to decrease

3. Block comes to rest

- PE reaches maximum

- KE reaches minimum

&&&&

`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?

****

There is no gravitational change since no parts of the system are acting in the vertical direction.

The elastic PE is maximum when the rubber band is stretched and poised for release. The elastic PE begins to change at the point of release, decreasing to 0 as the rubber band reaches elastic equilibrium.

&&&&

`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?

****

The friction of the block against the table is the nonconservative force.

The KE of the system goes from 0 to maximum, and then back to 0.

&&&&

`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?

****

F_frict = .0002 J / cm = .0002 J / 0.01 m = 0.02 N

PE0 = .01 J

`ds = 10cm = 0.01 m

`dW_ON_NC = F_net_NC * `ds

= 0.02 N * 0.01 m

= 0.002 J

this would be negative; better expressed as -.0002 J / cm * 10 cm = -.002 J.

Not sure I'm going down the right path with this one...

So `dKE = `dW_NC_ON - `dPE = -.002 J + .01 J = .008 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 work done by the nonservative force will account for any difference in `dPE and `dKE. I've got the concept I believe, but this problem isn't making much sense to me at the moment.

&&&&

`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?

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?

****

b = sqrt( 10^2 + 6^2)

= 8 cm

10cm - 8 cm = 2cm descent

&&&&

`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?

****

Fnet = .3 N

`ds = .02 m

`dPE = Fnet * `ds

= .3 N * .02 m

= 0.006 J

&&&&

Homework:

Your label for this assignment:

ic_class_091012

Copy and paste this label into the form.

Report your results from today's class using the Submit Work Form. Answer the questions posed above.

The topics we addressed today in class relate to q_a_'s #13 and 14. #13 was previously assigned.

Today q_A_14 is assigned.

No additional qa's will be assigned this week. This should give you time to catch up on the qa's.

However before the next class, if you don't have time to work all the qa's be sure you at least read them, and be sure you've at least got notes on the definitions of force, work, energy, kinetic energy and potential energy, and that you know and understand the details and the units of these calculations.

You're in pretty good shape here. See my notes, then see the appended document.

class 091007

Class 091012

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

more or less optional:

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

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,

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

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

the work-kinetic energy theorem `dW_net_ON = `dKE

becomes

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:

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.

`q001.  Consider the ball rolling from rest down one

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

On this interval

****

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.

&&&&