open query 11

#$&*

course Phy 201

011.  `query 11 

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Self-critique Rating:

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Question:  `q set 3 problems 15-19.  Explain the difference between a conservative and a nonconservative force, and give an example of each.

 

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Your solution: 

Conservative forces increase the potential energy to be released as kinetic energy. An example is lifting a book and then dropping it.

@& Good. Note that conservative forces can be in or opposite the direction of motion, so they can decrease or increase the PE.*@

Nonconservative forces are forces that do not store energy. friction and air resistance are examples of nonconservative forces. 

 

confidence rating #$&*:

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Given Solution: 

`a** A conservative force conserves energy--you can get your energy back. 

 

For example: 

 

Push something massive up a hill, then climb back down the hill.  The object, by virtue of its position, has the potential to return most of your energy to you, after regaining it as it rolls back down.  You will have done work against gravity as you move along a path up the hill, and gravity can return the energy as it follows its path back down the hill.  In this sense gravity conserves energy, and we call it a conservative force.

 

However, there is some friction involved--you do extra work against friction, which doesn't come back to you.  And some of the energy returned by gravity also gets lost to friction as the object rolls back down the hill.  This energy isn't conserved--it's nonconservative. **

 

Another more rigorous definition of a conservative force is that a force is conservative if the work done to get from one point to another independent of the path taken between those two points. 

 

 

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Self-critique (if necessary): I think my answer was ok but i have a question. When you push the rock up the hill, how do you get your energy back? by the rock smashing into you when it rolls back down? what if the rock doesn't roll back down? what if it's airlifted by helicopter back down to the bottom of the hill?

 

 

@& Very good questions.

If the rock rolls down the hill it will gain KE, and will also typically slide across and/or deform objects (including the ground) on the way down. All that takes energy. So some of the PE would go into KE, while some would do work against friction, and some would do the work of deformation (which would result in an increase in the temperature of the deformed objects, including the rock). Some energy would also be required to make the inevitable noise. Except fo the KE, the energy would ultimately be dissipated in the form of thermal energy (heat).

If the helicopter picks it up, then the force exerted on it while it's rising would be in the direction of motion and would do positive work on the rock, increasing its PE. When it's lowered, it's still supported and the supporting force is in the direction opposite its motion, and therefore does negative work on the rock.

It's better to think of the work done by the gravitational force, which is negative when the rock is being raised (the gravitational force acts downward, opposite the direction of the upward displacement, and hence does negative work) and positive when its it lowered.

By the time the rock is back in its original position, the positive and negative work done in the interim by gravity will be equal and opposite. So during this interval gravity will have done zero work, and there will be no change in the rock's PE.*@

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Self-critique Rating: 2

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Question:  `qIf a system does work W1 against a nonconservative force while conservative forces do work W2 on the system, what are the change in the KE and PE of the system?  Explain your reasoning from a commonsense point of view, and include a simple example involving a rubber band, a weight, an incline and friction.

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Your solution: 

the KE is the total net work done on the system, minus work done on the system by nonconservative forces. the PE is the total work done by the system on conservative and nonconservative forces. because the system is doing work a nonconservative force and the conservative forces are doing work on the system then the only potential energy comes from the nonconservative force.

an example would be a weight being released from the top of an incline with a rubber band attached to it and a decent amount of friction between the weight and the incline.

confidence rating #$&*:

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Given Solution: 

`a** `dKE is equal to the NET work done ON the system.

 

The KE of a system changes by an amount equal to the net work done on a system.

 

If work W1 is done BY the system against a nonconservative force then work -W1 is done ON the system by that force.

 

`dPE is the work done BY the system AGAINST conservative forces, and so is the negative of the work done ON the system BY nonconservative forces. 

• In the present case W2 stands for the work done on the system by conservative forces, so

`dPE = - W2.  PE decreases, thereby tending to increase KE.

So work -W1 is done ON the system by nonconservative forces and work W2 is done ON the system by a conservative force.

• The NET work done ON the system is therefore `dW_net_on = -W1 + W2.

The KE of the system therefore changes by `dKE = -W1 + W2.

If the nonconservative force is friction and the conservative force is gravity, then since the system must do positive work against friction, W1 must be positive and hence the -W1 contribution to `dKE tends to decrease the KE. 

e.g., if the system does 50 J of work against friction, then there is 50 J less KE increase than if there was no friction.

If the work done by the conservative force on the system is positive,

e.g., gravity acting on an object which is falling downward, then since force and displacement in the same direction implies positive work, gravity does positive work and the tendency will be to increase the KE of the system and W2 would be positive. 

A couple of numerical examples:

If W2 is 150 J and W1 is 50 J, then in terms of the above example of a falling object, this would mean that gravity tends to increase the KE by 150 J but friction dissipates 50 J of that energy, so the change in KE will be only 100 J.  This is consistent with `dW_net_ON = -W1 + W2 = -50 J + 150 J = 100 J.

The previous example was of a falling object.  If the object was rising (e.g., a ball having been thrown upward but not yet at its highest point), displacement and gravitational force would be in opposite directions, and the work done by gravity would be negative. 

In this case W2 might be, say, -150 J.  Then `dKE would be -150 J - 50 J = -200 J. 

The object would lose 200 J of KE.  This would of course only be possible if it had at least 200 J of KE to lose.  For example, in order to lose 200 J of KE, the ball thrown upward would have to be moving upward fast enough that it has 200 J of KE.

STUDENT COMMENT

I find this really confusing. Could this be laid out in another way?
INSTRUCTOR RESPONSE

If you find this confusing at this point, you will have a lot of company.  This is a challenge for most students, and these ideas will occupy us for a number of assignments.   There is light at the end of the tunnel: It takes awhile, but once you understand these ideas, the basic ideas become pretty simple and even obvious, and once understood they are usually (but not always) fairly easy to apply
This could be laid out differently, but would probably be equally confusing to any given student. Different students will require clarification of different aspects of the situation. 
If you tell me what you do and do not understand about the given solution, then I can clarify in a way that will make sense to you.
I also expect that in the process of answering subsequent questions, these ideas will become increasingly clear to you.
In any case feel free to insert your own interpretations, questions, etc. into a copy of this document (mark insertions with &&&& so I can locate them), and submit a copy.

STUDENT QUESTION

If the system goes against the force will this always make it negative?
INSTRUCTOR COMMENT

If a force and the displacement are in opposite directions, then the work done by that force is negative.
If the system moves in a direction opposite the force exerted BY the system, the work done BY the system is negative.
Note, however, that if this is the case then any equal and opposite force exerted ON the system will be in the direction of motion, so the force will do positive work ON the system.

A separate document related to this problem is located in the document

• work_on_vs_by_dKE_dPE_etc_questions_answers.htm

STUDENT COMMENT

This is a little confusing and I have read over the link that you gave. It will take some time to get use to the concepts. So, almost all of the factors are equal and opposite of each other?
INSTRUCTOR COMMENT

In terms only of forces acting ON an object or system, we have the following:
1. The object or system can be subjected to any number of forces acting ON the system. The net force F_net_ON is the sum, the net effect, of all those forces.
2. On any given interval the work done by the net force is equal to the change in the KE of the object or system.
3. This is summarized in the work-kinetic energy theorem
`dW_net_ON = `dKE
4. Each force acting on the object or system can be classified as some combination of conservative and nonconservative forces, so 
5. the net force can be expressed as the sum of a net conservative and a net nonconservative force:
F_net_ON = F_net_cons_ON + F_net_noncons_ON.
6. Thus `dW_net_ON = `dW_net_cons_ON + `dW_net_noncons_ON.
7. Change in PE can be defined to be equal and opposite the work done ON the system by conservative forces:
`dW_net_cons_ON = - `dPE
8. Since `dW_net_ON = `dW_net_cons_ON + `dW_net_noncons_ON, the work-kinetic energy theorem becomes
`dW_net_cons_ON + `dW_net_noncons_ON = `dKE.
9. Since `dW_net_noncons_ON = -`dPE this can be written
-`dPE + `dW_net_noncons_ON = `dKE.
10. This can be rearranged to
`dW_net_noncons_ON = `dKE + `dPE.
In the above we have explained the relationships among six quantities:

`dKEF_net_ON`dW_net_ON`dW_net_ON_cons`dW_net_ON_noncons`dPE

The main relationships are

`dW_net_ON = `dKE and`dW_net_ON_noncons = `dKE + `dPE.

If we replace the word ON by the word BY (indicating forces exerted and work done BY rather than ON the system), the force and therefore the work reverse sign.  In particular this gives us

`dKE + `dPE + `dW_net_BY_noncons = 0,

a form which is useful in understanding some problems.
For more practice, you may apply these principles to the suggested exercises at the link

query_11_suggested_exercise.htm 

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Self-critique (if necessary): i think i understand.

If a force is pushing ON a system but the system is moving the opposite direction the force is pushing then that force is negative and the force applied BY the system is positive. and vice versa.

The kinetic energy change is always the net work done ON the system.

The potential energy change is the work done BY the system minus work done on nonconservative forces.

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Self-critique Rating: 2

@& Which direction is positive and which is negative is completely arbitrary, and is your decision.

However botht the displacement and the force have directions, so each is either positive or negative.

If both force and displacement have the same sign (i.e., if both are positive, or if both are negative), their product is positive. Force and displacement are in the same direction and the force does positive work.

If the signs are opposite, then force and displacement have opposite signs and their product is negative. In this case the force and displacement are in opposite directions, and the force does negative work.*@

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Question:  `qIf the KE of an object changes by `dKE while the total nonconservative force does work `dW_nc on the object, by how much does the PE of the object change?

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Your solution: 

'dKE - (-'dW_nc) = PE 

 

confidence rating #$&*:

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Given Solution: 

`a** We have `dKE + `dPE + `dWbyNoncons = 0:  The total of KE change of the system, PE change of the system and work done by the system against nonconservative forces is zero.

Regarding the object at the system, if W_nc is the work done ON the object by nonconservative forces then work -W_nc is done BY the object against nonconservative forces, and therefore `dWnoncons_on = -W_nc.

We therefore have `dKE + `dPE - W_nc = 0 so that `dPE = -`dKE + W_nc. **

Equivalently, the work-energy theorem can be stated

• `dW_ON_nc = `dKE + `dPE

In this example the work done on the system by nonconservative forces is labeled W_nc, without the subscript ON and without the `d in front.  However it means the same thing, so the above becomes

W_nc = `dKE + `dPE

and we solve for `dPE to get

`dPE = -`dKE + W_nc

STUDENT COMMENT

 I’m still confused on how to understand when the energy is done on the object and when the energy is done against the object.
INSTRUCTOR RESPONSE

In an application, that can be the difficult question.
However in this case it is stated that W_nc is the work done by nonconservative forces ON the object.

STUDENT COMMENT:

I had the same logic as the given solution, however I got ‘dPE = -‘dKE - W_nc as the answer. I some how got an extra negative. Maybe Work can only be positive….??
INSTRUCTOR RESPONSE:

In this problem W_nc was specified as the work done on the object by nonconservative forces.

You have to be careful about whether W_nc is ON the system or BY the system.
You used the equation `dKE + `dPE + W_nc = 0; however that equation applies to the work done BY the system against nonconservative forces.  Written more specifically the equation you used would be

‘dKE + ‘dPE + W_nc_BY = 0 so

`dPE = - `dKE - `W_nc_BY.
W_nc_BY = - W_nc_ON so `dPE = - `dKE + W_nc_ON.

STUDENT RESPONSE WITH INSTRUCTOR'S COMMENTS (instructor comments in bold):

ok, so dke + dPe - W_nc = 0W_nc is total nonconservative forces doing work on the object,

Right up to here

this increases kinetic energy and decreases potential energy.  

there is no assumption about the sign of any of these quantities; any quantity could be positive or negative, as long as `dKE + `dPE - `dW_nc_on = 0
If `dW_nc_ON is positive then `dPE + `dKE is positive, but this could occur with positive `dKE and `dPE, or with a negative `dPE with lesser magnitude than a positive `dKE, or with a negative `dKE with lesser magnitude than a positive `dPE. All you would know is that `dKE + `dPE would be positive.
If `dW_nc_ON is negative then `dPE + `dKE is negative, but this doesn't tell you anything about the sign of either of the two quantities.
All we can say is that `dPE = `dW_nc_on - `dKE.

Since it is decreasing the potential energy it is negative. dKE is the kinetic energy which is positive since the potential energy is increased. 
If `dW_nc_on = 0, for example, an increase in either KE or PE implies a decrease in the other. KE would increase due to a decrease in PE (e.g., if you drop an object), while an increase in PE would be associated with a decrease in KE (e.g., an object thrown upward gains PE as it loses KE).
So an increase in KE tends to decrease PE, though `dW_nc_on can be such that KE and PE both increase.
In this problem we solve for PE. So that dPE= - dKE + W_nc. As potential energy increasess kinetic energy decreases and the non conservative work is positive because it is going with the direction of force more so than against it. 

An increase in PE could be the result of loss of KE and/or positive work done by nonconservative forces.
PE could also increase along with KE as long as `dW_nc_on is positive and large enough (e.g., a rocket increases both PE and KE due to nonconservative forces (the nonconservative forces result from ejecting fuel at high speed, i.e., from the rocket engines). 

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Self-critique (if necessary):

why couldn't you subtract 'dPE by both sides to get 'dKE - W_nc = -'dPE ? 

 

 

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Self-critique Rating: 2

@& `dPE = -`dKE + W_nc

is identical to

`dKe - W_nc = - `dPE,

which you get if you change the signs of every term (i.e., if you multiply every term by -1).

So that's a perfectly good way to express it.*@

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Question:  `qGive a specific example of such a process.

 

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Your solution: 

 i push a 20kg box at 5m/s while friction does 30joules of work on the box. the KE would equal 50 so to find the 'dPE it would be -50 + 30 = -20joules

confidence rating #$&*:

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Given Solution: 

`a** For example suppose I lift an object weighing 50 N and in the process the total nonconservative force (my force and friction) does +300 J of work on the object while its KE changes by +200 J.  

• The 300 J of work done by my force and friction is used to increase the KE by 200 J, leaving 100 J to be accounted for. 

More formally, `dW_noncons_ON = +300 J and `dKE = +200 J.  Since `dW_noncons_ON = `dKE + `dPE,  

So +300 J = +200 J + `dPE, and it follows that `dPE = +100 J.

• This 100 J goes into the PE of the object. **

STUDENT QUESTION (instructor responses inserted in bold)

&&&&&&I read your example first, and it makes no sense to me. Your force and the friction does 300J of work on the object....your force should be positive and friction should be negative right....so did you just add these two numbers together and get a positve number???

Right.  No numbers were assumed for the work I do and the work done by friction.  These two forces make up the nonconservative force on the system, and we just assumed a single total.  If you wish you can assume, say, that I do 350 J of work and friction does -50 J.  However the breakdown of the individual nonconservative forces isn't the point here.  All we really need is the total work done on the object by nonconservative forces.

How did you know that is KE changed by 200J.....did you just make that up or did you mathmaticallyfigure that out???

The phrase starts with 'for example, then reads

'suppose I lift an object weighing 50 N and in the process the total nonconservative force (my force and friction) does +300 J of work on the object while its KE changes by +200 J'

So all these quantities are simply assumed, for the sake of a numerical example.

How does your 300J increase the KE by 200J??? I understand that if 'dKE is only 200J then the other100J is 'dPE....I'm just not sure about the rest.

There is no specified connection between the 300 J of work I do and the 200 J of KE increase.  We just assume these quantities.  Once we have these quantities (in this case, simply by assumption; in other problems we will often find these quantities from other information), they dictate the PE increase.

There are a number of ways to think about this intuitively.  For example:

• If I do 300 J of work on an object, then if my force is the only force acting on it, the its KE will increase by 300 J.

• If I do more than 300 J of work, but friction reduces the net force on the object to 300 J, then KE will increase by 300 J.

• If all the nonconservative forces together (e.g., my force plus frictional force) do 300 J of work on the system, and if no other forces act, then the KE will increase by 300 J.

• If all nonconservative forces together do 300 J of work and the KE increases by only 200 J, then the nonconservative forces can't be the whole story, because the work done by the net force is equal to the change in KE.  The conclusion is that other forces must also be acting, and since they aren't nonconservative (we've assumed that all nonconservative forces together are accounted for in that 300 J), those forces must be conservative.  And they must do -100 J of work on the system, so that the net force does 300 J - 100 J = 200 J of work.

• Of course we can also resort to equations.  Since `dW_NC_on = `dPE + `dKE, it follows that `dPE = `dW_NC_on - `dKE = 300 J - 200 J = 100 J.

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STUDENT COMMENT

The problem didn’t seem to fit the equation we got earlier. This stuff is very confusing. I am going to read over it again. 


Your example is good.
You previously reasoned from the equation 
`dPE + `dKE + `dW_by_NC = 0,
obtaining `dPE = - `dKE - `dW_by_NC.
In your example you assumed that friction does 100 J of work on the box, and `dKE = 50 J. It would follow that the work done by the box against friction is -100 J; i.e., `dW_ON_nc = 100 J so `dW_by_nc = -100 J.
Thus in your example `dPE = - `dKE - `dW_by_NC. = - 50 J - (-100 J) = +50 J.

This is the result you obtained; however in your equation you used _by_ where you should have used _on_, so be sure you understand the distinction.  Since 'on' and 'by' are opposite, - `dKE - `dW_by_NC = -`dKE + `dW_on_NC.
You could as well have reasoned from the equation
`dW_nc_ON = `dPE + `dKE,
which leads to
`dPE = `dW_nc_ON - `dKE.
You assumed frictional work on the system to be 100 J, so `dW_nc_ON = 100 J. This leads to
`dPE = `dW_nc_ON - `dKE.= 100 J - 50 J = 50 J,
the same result obtained previously.
Since `dW_nc_ON = - `dW_nc_by, the two equations used in this instructor response are equivalent:
`dPE + `dKE + `dW_nc_by = 0 is completely equivalent to`dW_nc_ON = `dPE + `dKE.
Either equation, properly applied, leads to the correct result.

 

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Self-critique (if necessary):

did my example work? i can't tell

 

 

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Self-critique Rating:

@& You don't specify a distance, and you don't say how much force you exert on the box.

However if `dKE is 50 Joules (e.g., if the 20 kg box increased its speed from rest to about 1.6 m/s) and friction does 30 Joules of work on the box, then your example would make sense.*@

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Question:  `qClass notes #10.

Why does it make sense that the work done by gravity on a set of identical hanging washers should be proportional to the product of the number of washers and the distance through which they fall? Why is this consistent with the idea that the work done on a given cart on an incline is proportional to the vertical distance through which the cart is raised?

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Your solution: 

 

gravity is the only force working on the washers and to find their potential energy we multiply their weight by the height of the fall by the acceleration of gravity.

the cart going up the incline uses the same principle

 

confidence rating #$&*:

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Given Solution: 

`a** Informally:

• The more clips, the more gravitational force, and the more the clips descend the more work is done by that force. 

• The amount of work depends on how many clips, and on how far they descend. 

• The number of clips required is proportional to the slope (as long as the slope is small).

More formally, the force exerted by gravity is the same on each clip, so the total gravitational force on the hanging clips is proportional to the number of clips.  The work done is the product of the force and the displacement in the direction of the force, so the work done is proportional to product of the number of washers and the vertical displacement.

 

To pull the cart up a slope at constant velocity the number of washers required is proportional to the slope (for small slopes), and the vertical distance through which the cart is raised by a given distance of descent is proportional to the slope, to the work done is proportional to the vertical distance thru which the cart is raised. **

 

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Self-critique (if necessary):

I think i overcomplicated things. it makes sense that the more clips the greater the gravitational force, but wouldn't the gravitational force and the weight of the washer added together be referred to as the normal force?

for the cart, i don't understand how the number of washers required and vertical distance is proportional to the slope. is that just something we have to know? 

 

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Self-critique Rating: 2

@& You need to view the experiment and the discussion in the Class Notes to fully understand the situation (and you might well have already done so).

We demonstrate that more washers will pull the cart at constant speed up a greater slope. And it's clear that greater slope implies a greater vertical displacement as the cart goes from one end of the ramp to the other.

The normal force is the force exerted by the incline on the cart, in the direction perpendicular to the incline. The weight of the cart is related to the normal force; the weight of the washers in this case is not (but would be if the string attached to the cart made a nonzero angle with the incline).*@

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Question:  `qHow does the work done against friction of the cart-incline-pulley-washer system compare with the work done by gravity on the washers and the work done to raise the cart?   Which is greatest?  What is the relationship among the three?

 

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Your solution: 

the friction of the system is less than the work done by gravity on the washers and the work done to raise the cart. the work done to raise the cart is the greatest because the work done by gravity on the washers is used to do work on the cart. if the friction force was greater than either of the other forces then the cart wouldn't move.

 

confidence rating #$&*:

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Given Solution: 

`a** The force exerted by gravity on the hanging weights tends to move the system up the incline.  The force exerted by gravity on the cart has a component perpendicular to the incline and a component down the incline, and the force exerted by friction is opposed to the motion of the system.

 

In order for the cart to move with constant velocity up the incline the net force must be zero (constant velocity implies zero accel implies zero net force) so the force exerted by gravity in the positive direction must be equal and opposite to the sum of the other two forces.  So the force exerted by gravity on the hanging weights is greater than either of the opposing forces.

 

So the force exerted by friction is less than that exerted by gravity on the washers, and since these forces act through the same distance the work done against friction is less than the work done by gravity on the washers.

 

The work done against gravity to raise the cart is also less than the work done by gravity on the washers. 

• Work done against friction + work against gravity to raise cart = work by gravity on the hanging weights. **

 

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Self-critique (if necessary):

I don't think my answer was right but i think i understand. so the work done by gravity on the washers has to be greater than both opposing forces for the cart to move?  

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Self-critique Rating: 1

@& Right idea, but you wouldn't want to compare work with force since work and force measure two different things.

The work done by gravity supplies the energy required to overcome friction and to raise the cart.*@

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Question:  `qWhat is our evidence  that the acceleration of the cart is proportional to the net force on the cart?

 

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Your solution: 

the fact that the cart doesn't start moving until the washers start going down  

 

confidence rating #$&*:

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Given Solution: 

The net force is the sum of the gravitational force on the weights, and the frictional force (one force being positive, the other negative).
The acceleration is the net force divided by total mass (mass of cart plus hanger plus washers).
Washers are progressively transferred from the cart to the hanger, which keeps the mass of the system constant while increasing the net force. So the acceleration increases with the number of washers on the hanger.

The gravitational force on the weights is therefore proportional to the number of washers on the hanger.

• With each added washer we get the same additional force, so we get the same additional acceleration.

• So the graph is linear.

However the acceleration is not proportional to the number of weights.

• The net force on the system is equal to the gravitational force on the weights, plus the frictional force (which is of opposite sign, so while we are in fact adding quantities of opposite signs, it 'feels' like we're subtracting frictional force from gravitational force).

• The gravitational force on the weights is proportional to the number of washers, but when we add in the effect of the frictional force, our force is no longer proportional to the number of weights. Still linear, but not proportional to ... .

STUDENT COMMENT

I did not go into this great of detail. I simply used the one equation. Is this necessary to go in this thoroughly or was this just for our knowledge? 
INSTRUCTOR RESPONSE

The question is asking about 'our evidence'.
The fact that the law has been thoroughly tested by centuries of engineering and physics is what makes it a law, but the question here is whether the evidence obtained in the experiment (in the Class Notes) indicates that the acceleration of this particular cart is proportional to the net force acting on it.
Of course our experiments had better agree with the law, within their limits of precision. Our results will be more a test of our experimental design, and our execution of that design, than a test of the established law.
The law itself turns out to be valid only within certain restrictions, becoming invalid at relativistic velocities and at the quantum level of matter (realms far beyond our everyday experience and perception of the world).

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Self-critique (if necessary):

 i'm gonna fail and not be able to get my degree

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Self-critique Rating:

@& Not if you keep working hard.

This sequence of questions is challenging for almost everyone.*@

@& If you keep working hard that shouldn't happen.

This sequence of questions is challenging for almost everyone.*@

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Question:  `qprin phy and gen phy prob 34:  Car rolls off edge of cliff; how long to reach 85 km/hr?

 

 

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Your solution: 

23.61m/s / 9.8 = 2.41sec 

 

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Given Solution: 

`aWe know that the acceleration of gravity is 9.8 m/s^2, and this is the rate at which the velocity of the car changes.  The units of 85 km/hr are not compatible with the units m/s^2, so we convert this velocity to m/s, obtaining velocity

• 85 km/hr ( 1000 m/km) ( 1 hr / 3600 sec) = 23.6 m/s.

Common sense tells us that with velocity changing at 9.8 m/s every second, it will take between 2 and 3 seconds to reach 23.6 m/s. 

More precisely, the car's initial vertical velocity is zero, so using the downward direction as positive, its change in velocity is `dv = 23.6 m/s. 

• Its acceleration is a = `dv / `dt, so

• `dt = `dv / a = 23.6 m/s / (9.8 m/s^2) = 2.4 sec, approx..

 

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Self-critique (if necessary): ok

 

 

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Question:  `q****   prin phy and gen phy  problem 2.52 car 0-50 m/s in 50 s by graph

How far did the car travel while in 4 th gear and how did you get the result?

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Your solution: 

i don't understand. how many gears does the car have? did it end in 4th gear? are supposed to assume it travels an equal distance in each gear?

 

confidence rating #$&*:

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Given Solution: 

`a**  In 4th gear the car's velocity goes from about 36.5 m/s to 45 m/s, between clock times 16 s and 27.5 s.

 

Its average velocity on that interval will therefore be

 

vAve = (36.5 m/s  +  45 m/s) / 2 = 40.75 m/s and the time interval is

'dt =  (27.5s - 16s) = 11.5 s.

 

We therefore have

'ds = vAve * `dt = 40.75 m/s * 11.5 s = 468.63 m.

 

The area under the curve is the displacement of the car, since vAve is represented by the average height of the graph and `dt by its width.  It follows that the area is vAve*'dt, which is the displacement `ds.

 

The slope of the graph is the acceleration of the car. This is because slope is rise/run, in this case that is 'dv/'dt, which is the ave rate of change of velocity or acceleration.

 

We already know `dt, and we have `dv = 45 m/s - 36.5 m/s = 8.5 m/s.

 

The acceleration is therefore

 

a = `dv / `dt = (8.5 m/s) / (11.5 s) = .77 m/s^2, approx. **

 

 

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Self-critique (if necessary):

 

 

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Self-critique Rating:

@& This question requires that you analyze the graph given in the text, which clearly indicates what happens in each gear.*@

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Question:  `q ****   Gen phy   what is the meaning of the slope of the graph and why should it have this meaning?

 

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Your solution: 

 No idea

 

 

confidence rating #$&*:

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Given Solution: 

`a** The graph is of velocity vs. clock time, so the rise will be change in velocity and the run will be change in clock time.  So the slope = rise/run represents change in vel / change in clock time, which is acceleration. **

 

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Self-critique (if necessary):

 

 

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Self-critique Rating:

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Question:  `qGen phy   what is the meaning of the area under the curve, and why does it have this meaning?

 

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Your solution: 

 the displacement? i don't know

 

 

confidence rating #$&*:

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Given Solution: 

`a** The area under the curve is the distance traveled. This is so because 'ds = vAve*'dt.

 

'dt is equal to the width of the section under the curve and vAve is equal to the average height of the curve. The area of a trapezoid is width times average height. Although this is not a trapezoid it's close enough that we for the purpose of estimation can analyze it as such.

 

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Self-critique (if necessary):

 

 

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Question:  `qGen phy   what is the area of a rectangle on the graph and what does it represent?

 

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Your solution: 

is the graph of acceleration vs. time?

 

confidence rating #$&*:

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Given Solution: 

`a** The area of a rectangle on the graph represents a distance. **

 

 

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Self-critique (if necessary):

 why is this the answer?

 

 

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Self-critique Rating:

@& The graph shows the velocity of the car vs. clock time. If you haven't seen the graph these questions will be difficult to answer, but knowing that the graph is of velocity vs. clock time you should now understand.*@

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Question:  `quniv phy  problem 2.90 from 10th edition (University Physics students should solve this problem now).  Superman stands on the top of a skyscraper 180 m high.  A student with a stopwatch, determined to test the acceleration of gravity for himself, steps off the top of the building but Superman can't start after him for 5 seconds.  If Superman then propels himself downward with some init vel v0 and after that falls freely, what is the minimum value of v0 so that he catches the student before that person strikes the ground?

`quniv phy   what is Superman's initial velocity, and what does the graph look like (be specific)?

 

 

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Your Solution:

 

 

confidence rating #$&*:

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Given Solution: 

`a** In time interval `dt after leaving the building the falling student has fallen through displacement `ds = v0 `dt + .5 a `dt^2, where v0 = 0 and, choosing the downward direction to be positive, we have a = -9.8 m/s^2.  If `ds = -180 m then we have `ds = .5 a `dt^2 and `dt = sqrt(2 * `ds / a) = sqrt(2 * -180 m / (-9.8 m/s^2)) = 6 sec, approx..

 

Superman starts 5 seconds later, and has 1 second to reach the person. Superman must therefore accelerate at -9.8 m/s^2 thru `ds = -180 m in 1 second, starting at velocity v0.

 

Given `ds, `dt and a we find v0 by solving `ds = v0 `dt + .5 a `dt^2 for v0, obtaining v0 = (`ds - .5 a `dt^2) / `dt = (-180 m - .5 * -9.8 m/s^2 * (1 sec)^2 ) / (1 sec) = -175 m/s, approx.

 

Note that Superman's velocity has only about 1 second to change, so changes by only about -9.8 m/s^2, or about -10 m/s^2. **

 

``qsketch a graph of Superman's position vs. clock time, and on the same graph show the student's position vs. clock time, with clock time starting when the person begins falling

 

 

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Your solution: 

 

 

confidence rating #$&*:

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Given Solution: 

`a** If we start our clock at t = 0 at the instant the student leaves the top of the building then at clock time t the student's `dt will be just equal to t and his position will be x = x0 + v0 t + .5 a t^2 = .5 a t^2, with x0 = 180 m and a = -9.8 m/s^2.  A graph of x vs. t will be a parabola with vertex at (0,180), intercepting the t axis at about t = 6 sec.

 

For Superman the time of fall will be `dt = t - 5 sec and his position will be x = x0 + v0 (t-5sec) + .5 a (t-5sec)^2, another parabola with an unspecified vertex.

 

A graph of altitude vs. t shows the student's position as a parabola with vertex (0, 180), concave downward to intercept the t axis at (6,0).  Superman's graph starts at (5,180) and forms a nearly straight line, intercepting the t axis also at (6,0).  Superman's graph is in fact slightly concave downward, starting with slope -175 and ending with slope -185, approx. **

 

 

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Self-critique (if necessary):

 

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Self-critique Rating:

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Question:  `quniv phy  problem 2.90 from 10th edition (University Physics students should solve this problem now).  Superman stands on the top of a skyscraper 180 m high.  A student with a stopwatch, determined to test the acceleration of gravity for himself, steps off the top of the building but Superman can't start after him for 5 seconds.  If Superman then propels himself downward with some init vel v0 and after that falls freely, what is the minimum value of v0 so that he catches the student before that person strikes the ground?

`quniv phy   what is Superman's initial velocity, and what does the graph look like (be specific)?

 

 

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Your Solution:

 

 

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution: 

`a** In time interval `dt after leaving the building the falling student has fallen through displacement `ds = v0 `dt + .5 a `dt^2, where v0 = 0 and, choosing the downward direction to be positive, we have a = -9.8 m/s^2.  If `ds = -180 m then we have `ds = .5 a `dt^2 and `dt = sqrt(2 * `ds / a) = sqrt(2 * -180 m / (-9.8 m/s^2)) = 6 sec, approx..

 

Superman starts 5 seconds later, and has 1 second to reach the person. Superman must therefore accelerate at -9.8 m/s^2 thru `ds = -180 m in 1 second, starting at velocity v0.

 

Given `ds, `dt and a we find v0 by solving `ds = v0 `dt + .5 a `dt^2 for v0, obtaining v0 = (`ds - .5 a `dt^2) / `dt = (-180 m - .5 * -9.8 m/s^2 * (1 sec)^2 ) / (1 sec) = -175 m/s, approx.

 

Note that Superman's velocity has only about 1 second to change, so changes by only about -9.8 m/s^2, or about -10 m/s^2. **

 

``qsketch a graph of Superman's position vs. clock time, and on the same graph show the student's position vs. clock time, with clock time starting when the person begins falling

 

 

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Your solution: 

 

 

confidence rating #$&*:

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Given Solution: 

`a** If we start our clock at t = 0 at the instant the student leaves the top of the building then at clock time t the student's `dt will be just equal to t and his position will be x = x0 + v0 t + .5 a t^2 = .5 a t^2, with x0 = 180 m and a = -9.8 m/s^2.  A graph of x vs. t will be a parabola with vertex at (0,180), intercepting the t axis at about t = 6 sec.

 

For Superman the time of fall will be `dt = t - 5 sec and his position will be x = x0 + v0 (t-5sec) + .5 a (t-5sec)^2, another parabola with an unspecified vertex.

 

A graph of altitude vs. t shows the student's position as a parabola with vertex (0, 180), concave downward to intercept the t axis at (6,0).  Superman's graph starts at (5,180) and forms a nearly straight line, intercepting the t axis also at (6,0).  Superman's graph is in fact slightly concave downward, starting with slope -175 and ending with slope -185, approx. **

 

 

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Self-critique (if necessary):

 

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Self-critique Rating:

#*&!

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Question:  `quniv phy  problem 2.90 from 10th edition (University Physics students should solve this problem now).  Superman stands on the top of a skyscraper 180 m high.  A student with a stopwatch, determined to test the acceleration of gravity for himself, steps off the top of the building but Superman can't start after him for 5 seconds.  If Superman then propels himself downward with some init vel v0 and after that falls freely, what is the minimum value of v0 so that he catches the student before that person strikes the ground?

`quniv phy   what is Superman's initial velocity, and what does the graph look like (be specific)?

 

 

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Your Solution:

 

 

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution: 

`a** In time interval `dt after leaving the building the falling student has fallen through displacement `ds = v0 `dt + .5 a `dt^2, where v0 = 0 and, choosing the downward direction to be positive, we have a = -9.8 m/s^2.  If `ds = -180 m then we have `ds = .5 a `dt^2 and `dt = sqrt(2 * `ds / a) = sqrt(2 * -180 m / (-9.8 m/s^2)) = 6 sec, approx..

 

Superman starts 5 seconds later, and has 1 second to reach the person. Superman must therefore accelerate at -9.8 m/s^2 thru `ds = -180 m in 1 second, starting at velocity v0.

 

Given `ds, `dt and a we find v0 by solving `ds = v0 `dt + .5 a `dt^2 for v0, obtaining v0 = (`ds - .5 a `dt^2) / `dt = (-180 m - .5 * -9.8 m/s^2 * (1 sec)^2 ) / (1 sec) = -175 m/s, approx.

 

Note that Superman's velocity has only about 1 second to change, so changes by only about -9.8 m/s^2, or about -10 m/s^2. **

 

``qsketch a graph of Superman's position vs. clock time, and on the same graph show the student's position vs. clock time, with clock time starting when the person begins falling

 

 

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Your solution: 

 

 

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution: 

`a** If we start our clock at t = 0 at the instant the student leaves the top of the building then at clock time t the student's `dt will be just equal to t and his position will be x = x0 + v0 t + .5 a t^2 = .5 a t^2, with x0 = 180 m and a = -9.8 m/s^2.  A graph of x vs. t will be a parabola with vertex at (0,180), intercepting the t axis at about t = 6 sec.

 

For Superman the time of fall will be `dt = t - 5 sec and his position will be x = x0 + v0 (t-5sec) + .5 a (t-5sec)^2, another parabola with an unspecified vertex.

 

A graph of altitude vs. t shows the student's position as a parabola with vertex (0, 180), concave downward to intercept the t axis at (6,0).  Superman's graph starts at (5,180) and forms a nearly straight line, intercepting the t axis also at (6,0).  Superman's graph is in fact slightly concave downward, starting with slope -175 and ending with slope -185, approx. **

 

 

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Self-critique (if necessary):

 

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Self-critique Rating:

#*&!#*&!

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