#$&* course phy 201 011. `query 11
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Given Solution: `a** A conservative force conserves energy--you can get your energy back. For example: Push something massive up a hill, then climb back down the hill. The object, by virtue of its position, has the potential to return most of your energy to you, after regaining it as it rolls back down. You will have done work against gravity as you move along a path up the hill, and gravity can return the energy as it follows its path back down the hill. In this sense gravity conserves energy, and we call it a conservative force. However, there is some friction involved--you do extra work against friction, which doesn't come back to you. And some of the energy returned by gravity also gets lost to friction as the object rolls back down the hill. This energy isn't conserved--it's nonconservative. ** Another more rigorous definition of a conservative force is that a force is conservative if the work done to get from one point to another independent of the path taken between those two points. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I didn’t know what these were, given answer helped. Took notes ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qIf a system does work W1 against a nonconservative force while conservative forces do work W2 on the system, what are the change in the KE and PE of the system? Explain your reasoning from a commonsense point of view, and include a simple example involving a rubber band, a weight, an incline and friction. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Nonconservative - PE is negative and KE is negative Conservative PE is positive KE is negative confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: `dKE F_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 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Got this completely wrong, given answers helped ------------------------------------------------ Self-critique Rating: 3 ********************************************* 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The PE will change a lot too 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 = 0 W_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). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): This is really confusing. I read over given answer multiply times. ------------------------------------------------ Self-critique Rating: 2 ********************************************* Question: `qGive a specific example of such a process. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Pushing a ball up a hill and letting it roll? 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 mathmatically figure 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 other 100J 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. &&&&& 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.
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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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I got these totally wrong. My thinking is off. " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Because of the friction that is present. Fiction should be consistent. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** Informally: • The more clips, the more gravitational force, and the more the clips descend the more work is done by that force. • The amount of work depends on how many clips, and on how far they descend. • The number of clips required is proportional to the slope (as long as the slope is small). More formally, the force exerted by gravity is the same on each clip, so the total gravitational force on the hanging clips is proportional to the number of clips. The work done is the product of the force and the displacement in the direction of the force, so the work done is proportional to product of the number of washers and the vertical displacement. To pull the cart up a slope at constant velocity the number of washers required is proportional to the slope (for small slopes), and the vertical distance through which the cart is raised by a given distance of descent is proportional to the slope, to the work done is proportional to the vertical distance thru which the cart is raised. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I got these totally wrong. My thinking is off. " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&! ********************************************* Question: `qClass notes #10. Why does it make sense that the work done by gravity on a set of identical hanging washers should be proportional to the product of the number of washers and the distance through which they fall? Why is this consistent with the idea that the work done on a given cart on an incline is proportional to the vertical distance through which the cart is raised? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Because of the friction that is present. Fiction should be consistent. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I got these totally wrong. My thinking is off. " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!#*&!