Query 12

#$&*

course PHY 121

6/29 10:30

012. `query 12*********************************************

Question: `qQuery set 3 #'s 13-14 If an object of mass m1 rests on a frictionless

tabletop and a mass m2 hangs over a good pulley by a string attached to the first

object, then what forces act on the two-mass system and what is the net force on the

system? What would be the acceleration of the system? How much would gravitational PE

change if the hanging mass descended a distance `dy?

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Your solution:

Gravity acts on both masses. It pulls down on mass m2, but the fact that mass m1 is

sitting on the tabletop causes that force to be pulling against the table (parallel).

The net force on the system, therefore, is F = m * a or Fnet = m2 * 9.8 m/s^2.

Acceleration of the system will equal a=F/m or (m2 * 9.8 m/s^2) / m2, which should equal

9.8 m/s^2. If the hanging mass descended by a distance of `dy, then the PE change would

be `dw = F * `ds or 9.8 m/s^2 * `dy.

confidence rating #$&*:

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Given Solution:

`a** The net force on the system is the force of gravity on the suspended weight: Fnet

= m2 * 9.8 m/s/s, directed downward.

Gravity also acts on m1 which is balanced by the upward force of table on this mass, so

the forces on m1 make no contribution to Fnet.

Acceleration=net force/total mass = 9.8 m/s^2 * m2 / (m1+m2), again in the downward

direction.

The change in gravitational PE is equal and opposite to the work done by gravity. This

is the definition of change in gravitational PE.

If the mass m2 descends distance `dy then the gravitational force m * g and the

displacement `dy are both downward, so that gravity does positive work m g `dy on the

mass. The change in gravitational PE is therefore - m g `dy.

COMMON MISCONCEPTIONS AND INSTRUCTOR COMMENTS:

Misconception: The tension force contributes to the net force on the 2-mass system.

Student's solution:

The forces acting on the system are the forces which keep the mass on the table, the

tension in the string joining the two masses, and the weight of the suspended mass.

The net force should be the suspended mass * accel due to gravity + Tension.

INSTRUCTOR COMMENT:

String tension shouldn't be counted among the forces contributing to the net force on

the system.

The string tension is internal to the two-mass system. It doesn't act on the system but

within the system.

Net force is therefore suspended mass * accel due to gravity only

'The forces which keep the mass on the table' is too vague and probably not appropriate

in any case. Gravity pulls down, slightly bending the table, which response with an

elastic force that exactly balances the gravitational force. **

STUDENT COMMENT

I don't understand why m1 doesn't affect the net force. Surely it has to, if mass1 was

90kg, or 90g, then are they saying that the force would be the same regardless?

INSTRUCTOR RESPONSE

m1 has no effect on the net force in the given situation.

Whatever the mass on the tabletop, it experiences a gravitational force pulling it down,

and the tabletop exerts an equal and opposite force pushing it up. So the mass of that

object contributes nothing to the net force on the system.

The mass m1 does, however, get accelerated, so m1 does have a lot to do with how quickly

the system accelerates. The greater the mass m1, the less accelerating effect the net

force will have on the system.

Also if friction is present, the mass m1 is pulled against the tabletop by gravity,

resulting in frictional force. The greater the mass m1, the greater would be the

frictional force.

All these ideas are addressed in upcoming questions and exercises.

STUDENT COMMENT

I understand the first few parts of this problem, but I am still a little unsure about

the gravitational PE.

I knew what information that was required to solve the problem, but I just thought the

solution would be more that (-m2 * 9.8m/s^2 * ‘dy).

INSTRUCTOR RESPONSE

Only m2 is changing its altitude, so only m2 experiences a change in gravitational PE.

Equivalently, only m2 experiences a gravitational force in its direction of motion, so

work is done by gravity on only m2.

STUDENT COMMENT

I forgot that PE = m * g * 'dy. And I did not think that the table exerting force on the

mass took it out of the system. I understand the idea though.

INSTRUCTOR RESPONSE

the table doesn't take the mass out of the system, but it does counter the force exerted

by gravity on that mass

so the total mass of the system is still the total of the accelerating masses, but the

net force is just the force of gravity on the suspended mass, (since the system is said

to be frictionless, there is no frictional force to consider)

SYNOPSIS

The change in gravitational PE is equal and opposite to the work done by gravity. This

is the definition of change in gravitational PE.

If the mass m2 descends distance `dy then the gravitational force m * g and the

displacement `dy are both downward, so that gravity does positive work m g `dy on the

mass. The change in gravitational PE is therefore - m g `dy.

As you say,

`dw_noncons + `dPE + `dKE = 0

If `dW_noncons is zero, as is the case here (since there are no frictional or other

nonconservative forces present), then

`dPE + `dKE = 0

and

`dKE = - `dPE.

In this case `dPE = - m g `dy so

`dKE = - ( - m g `dy) = m g `dy.

The signs are confusing at first, but if you just remember that signs are important

these ideas will soon sort themselves out.

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Self-critique (if necessary):

I did not add mass 1 back into the equation when I figured acceleration. It didn't make

sense to me that I'd end up back where I started, but I went with it anyway. So,

acceleration should be m2(9.8 m/s^2) / (m1+m2). For the gravitational PE question, I

did not use g for the gravitational force. So, what I need is mass times gravity times

the movement/displacement, and it needs to be negative because potential energy is lost

when it converts to kinetic energy (?). If it's okay to think of it that way (assuming

it's correct), I may be able to understand it. So, the PE would be a loss, and the

answer is -m*g*`dy.

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Self-critique Rating: 2

@&

Your thinking is fine at this stage. You'll probably continue to refine and gradually reformulate your understanding, but there's nothing wrong with any of your statements.

*@

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Question: `qHow would friction change your answers to the preceding question?

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Your solution:

If friction were a part of the previous question, it would affect the net force. You

would have to subtract it from the m2*9.8m/s^2. That force minus the friction force

would give you the net force. F - f(frict)=Fnet.

confidence rating #$&*:

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Given Solution:

`a**Friction would act to oppose the motion of the mass m1 as it slides across the

table, so the net force would be m2 * g - frictional resistance. **

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Self-critique (if necessary):

Do I start calling 9.8m/s^2 ""g"" now?

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Self-critique Rating: 2

@&

When expressing a relationship in symbols, g is preferred.

When expressing numerical quantities, 9.8 m/s^2 is good.

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Question: `qExplain how you use a graph of force vs. stretch for a rubber band to

determine the elastic potential energy stored at a given stretch.

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Your solution:

The graph is linear, starting at the origin. It needs to start at point (0,0) because

with zero force, there is 0 displacement. The y axis measures Force in Newtons and the

x axis measures the stretch of the rubber band in centimeters. The slope shows how much

force is needed for each centimeter of stretch. The area beneath the line tells how

much force it takes to stretch the rubber band a given distance.

confidence rating #$&*:

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Given Solution:

`a** If we ignore thermal effects, which you should note are in fact significant with

rubber bands and cannot in practice be ignored if we want very accurate results, PE is

the work required to stretch the rubber band. This work is the sum of all F * `ds

contributions from small increments `ds from the initial to the final position. These

contributions are represented by the areas of narrow trapezoids on a graph of F vs.

stretch. As the trapezoids get thinner and thinner, the total area of these trapezoids

approaches, the area under the curve between the two stretches.

So the PE stored is the area under the graph of force vs. stretch. **

STUDENT QUESTION

I am still a little confused about if the work is done by the rubber bands, or if the

work is done one the rubber bands.

Would you explain the difference?

INSTRUCTOR RESPONSE

This example might be helpful:

If you pull the end of an anchored rubber band to the right, it exerts a force to the

left, in the direction opposite motion, so it does negative work during the process.

You, on the other hand, pull in the direction of motion and do positive work on the

rubber band.

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Self-critique (if necessary):

I need to remember PE is the work required to stretch the rubber band. I am accustomed

to thinking of potential energy as the ability to do work--a rock sitting at the top of

a hill, a bubble about to pop, a cat about to pounce. All of these are still at that

moment and have potential to move; they just haven't done it yet. I know that's an

elementary level understanding of potential energy. I need to see how these definitions

fit into that, if they do at all.

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Self-critique Rating: 2

@&

The change in an ideal rubber band's PE is the work required to stretch it. Having been stretched, it can then do work equal to that done by the stretching force.

Similarly the rock at the top of the hill required work (perhaps from geophysical sources) to attain its position, and if it rolls down the hill that work will be converted to other forms of energy.

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Question: `q Does the slope of the F vs stretch graph represent something? Does the

area under the curve represent the work done? If so, is it work done BY or work done ON

the rubber bands?

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Your solution:

The slope tells how much force, in Newtons, it takes to stretch the rubber band 1

centimeter. The area tells the Newtons (Force) times centimeters, so it tells how much

force is used to move the rubber band. This is work done to move an object. The work

isn't done by the rubber bands because they can't move on their own. The work is done

on the rubber bands in the case by a person pulling them.

confidence rating #$&*:

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Given Solution:

`a** The rise of the graph is change in force, the run is change in stretch. So slope =

rise / run = change in force / change in stretch, which the the average rate at which

force changes with respect to stretch. This basically tells us how much additional

force is exerted per unit change in the length of the rubber band.

The area is indeed with work done (work is integral of force with respect to

displacement).

If the rubber band pulls against an object as is returns to equilibrium then the force

it exerts is in the direction of motion and it therefore does positive work on the

object as the object does negative work on it.

If an object stretches the rubber band then it exerts a force on the rubber band in the

direction of the rubber band's displacement, and the object does positive work on the

rubber band, while the rubber band does negative work on it. **

STUDENT QUESTION

Okay, so are you saying that the rubber band could either be doing work or getting work

done on it?

I believe I understand this, but just wanted to double check.

INSTRUCTOR RESPONSE

Yes, and that depends on whether the rubber band is being stretched, or contracting.

When it is being stretched positive work is being done on the rubber band.

After being released the rubber band does positive work on the object to which its force

is applied.

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Self-critique (if necessary):

It makes sense that as the rubber band is being stretched, work is being done on it, and

as it is let go, it is doing work on another object.

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Self-critique Rating: 2

PREVIOUS STUDENT RESPONSE TO REQUEST FOR COMMENTS

There is a whole lot of stuff concerning Newton’s laws of motion and there applications

to force and acceleration. They will take some serious application to master. I

understand what potential energy is, I understand that it is decreasing as kinetic

energy increase, but I don’t understand how to measure it. Its like an invisible force,

and the only relation to which I can apply it is in the context of gravity. If we have a

1kg object and we hold it 5meters off the ground, then according to the equation above

PE = m*g*`dy this would be

PE = 1kg * 9.8m/s^2 * 5m = 49 kg * m^2/s^2

my algebra is so bad but I still cant see this contributing to a useful measurement or

unit. I don’t know how to swing it so it’ll give me a newton, PE has to be measured in

newtons right because it is indeed a force?

Ohhh I get it now!! I remember, a kg times a m/s^2 is a newton, and a newton times a

meter is a Joule!!! So this is a valid measurement, which would make that equation

valid, the potential energy for the above circumstance would be 49 Joules then.

INSTRUCTOR RESPONSE

Very good.

Remember that F_net = m a

If you multiply mass m in kg by acceleration a in m/s^2, you get the force in Joules.

Of course when you multiply kg by m/s^2 you get kg m/s^2.

This is why a Newton is equal to a kg m / s^2.

Work being the product of a force and a displacement will therefore have units of

Newtons * meters, or kg * m/s^2 * m, which gives us kg m^2 / s^2.

Query Add comments on any surprises or insights you experienced as a result of this

assignment.

It started to make some sense in the video when you said the rail gets energy from the

rubber band at one rate that is dissipated. So 1/2Kx^2 = F`ds. (I'm still working on

what all that means, but.....) The potential energy = kinetic energy and so potential

energy is negative when it converts to kinetic energy. I hope I'm starting to get these

concepts down."

Self-critique (if necessary):

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Self-critique rating:

PREVIOUS STUDENT RESPONSE TO REQUEST FOR COMMENTS

There is a whole lot of stuff concerning Newton’s laws of motion and there applications

to force and acceleration. They will take some serious application to master. I

understand what potential energy is, I understand that it is decreasing as kinetic

energy increase, but I don’t understand how to measure it. Its like an invisible force,

and the only relation to which I can apply it is in the context of gravity. If we have a

1kg object and we hold it 5meters off the ground, then according to the equation above

PE = m*g*`dy this would be

PE = 1kg * 9.8m/s^2 * 5m = 49 kg * m^2/s^2

my algebra is so bad but I still cant see this contributing to a useful measurement or

unit. I don’t know how to swing it so it’ll give me a newton, PE has to be measured in

newtons right because it is indeed a force?

Ohhh I get it now!! I remember, a kg times a m/s^2 is a newton, and a newton times a

meter is a Joule!!! So this is a valid measurement, which would make that equation

valid, the potential energy for the above circumstance would be 49 Joules then.

INSTRUCTOR RESPONSE

Very good.

Remember that F_net = m a

If you multiply mass m in kg by acceleration a in m/s^2, you get the force in Joules.

Of course when you multiply kg by m/s^2 you get kg m/s^2.

This is why a Newton is equal to a kg m / s^2.

Work being the product of a force and a displacement will therefore have units of

Newtons * meters, or kg * m/s^2 * m, which gives us kg m^2 / s^2.

Query Add comments on any surprises or insights you experienced as a result of this

assignment.

It started to make some sense in the video when you said the rail gets energy from the

rubber band at one rate that is dissipated. So 1/2Kx^2 = F`ds. (I'm still working on

what all that means, but.....) The potential energy = kinetic energy and so potential

energy is negative when it converts to kinetic energy. I hope I'm starting to get these

concepts down."

Self-critique (if necessary):

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Self-critique rating:

#*&!

&#This looks good. See my notes. Let me know if you have any questions. &#