#$&* course phy 201 There were alot towards the end I could not answer it had no graph and did not say where it was , maybe I misinterpreted it? 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): ------------------------------------------------ Self-critique 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): ------------------------------------------------ Self-critique Rating: ********************************************* 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: dPe = dW - dKe 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): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qGive a specific example of such a process. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I am not sure what process is being talked about 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): ------------------------------------------------ Self-critique Rating: ********************************************* 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Well in the first place the weight on both sides to be equal then for the cart to move up the incline then there must be more weight on the other side leading to more work on the washers because friction will remove some of the strength from the force of the weights so the work on the washers is the greatest Work removed by friction + work done on cart = work on the weights 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qWhat is our evidence that the acceleration of the cart is proportional to the net force on the cart? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Fnet = m * a a = Fnet/m it is proportional to the net force divided by mass 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). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: Calculate the displacement and velocity at times of (a) 0.500, (b) 1.00, (c) 1.50, and (d) 2.00 s for a ball thrown straight up with an initial velocity of 15.0 m/s. Take the point of release to be y_0 = 0 . With the number of significant figures specified for this problem we would need to use a 3-significant-figure acceleration. However the acceleration at different points on the surface of the Earth varies in the third significant figure. Since the location isn't specified, the solution given here will ignore significant figures and use the value 10 m/s^2 for the acceleration of gravity and time intervals .5, 1, 1.5 and 2 s. You should be aware of the significant-figure implications both of the given information and the assumptions used this given solution. The acceleration of gravity is 10 m/s^2 downward. We understand the meaning of this acceleration to be that the speed of the object changes by 10 m/s every second, and that the velocity change is in the downward direction. It isn't explicitly stated, but it can be taken as implicit that the given times are in fact the times as measured on a running clock which reads 0 at the instant of release. Directly reasoned solution for clock time 0.5 second: In 0.5 second the change in speed will therefore be 5 m/s (if the change is 10 m/s every second, then in half a second it will be half of 10 m/s or 5 m/s). The velocity change is in the downward direction, so an initial upward speed of 15 m/s will in this time reduce to 10 m/s. During the 0.5 second interval, then, the velocity changes from 15 m/s to 10 m/s. The average velocity, since acceleration is uniform (so that the v vs. t graph is a straight line) is therefore 12.5 m/s. In 0.5 second with average velocity 12.5 m/s, the object will therefore move through displacement `dy = 12.5 m/s * 0.5 second = 6.75 meters. So after 0.5 second, the object's vertical position will be y = y_0 + 6.75 meters = 0 + 6.75 meters = 6.75 meters. Similar reasoning for the other given clock times leads to the following conclusions. Each line corresponds to the interval from clock time 0 to the given clock time. clock time change in velocity initial velocity final velocity ave velocity position 0.5 -5 m/s 15 m/s 10 m/s 12.5 m/s 6.75 m 1 -10 m/s 15 m/s 5 m/s 10 m/s 10 m 1.5 -15 m/s 15 m/s 0 m/s 7.5 m/s 11.25 m 2 -20 m/s 15 m/s -5 m/s 5 m/s 10 m More generally we could reason out the position at clock time t as follows: Taking upward as the positive direction, the acceleration being downward is negative, with value -10 m/s^2. The change in velocity for the interval from clock time 0 to clock time t is therefore -10 m/s^2 * t. The final velocity for this interval is thus 15 m/s - 10 m/s^2 * t. The average velocity for the interval is the average of the initial and final velocities, so vAve = (15 m/s + (15 m/s - 10 m/s^2 * t) ) / 2 = 15 m/s - 10 m/s^2 * t / 2 = 15 m/s - 5 m/s^2 * t. The displacement during the interval is thus `dy = vAve * `dt = vAve * (t - 0 s) = (15 m/s - 5 m/s^2 * t) * t = 15 m/s * t - 5 m/s^2 * t^2. Evaluating this for t = .5 s, 1 s, 1.5 s and 2 s we obtain respective displacements 6.75 m, 10 m, 11.25 m and 10 m. The above solutions provide insight into the nature of the motion that can be bypassed by using the third equation of motion `ds = v0 `dt * 1/2 a `dt^2. It isn't a good idea to bypass insights, so you should be able to reason out the solution in the above ways, but it's also important to be able to use the equations. And once you're comfortable with the reasoning process, the equations often provide the quickest path to the solution: Plugging t = 0, 0.5, 1, 1.5 and 2 into the equation we obtain respective displacements 6.75 m, 10 m, 11.25 m and 10 m. If the displacements are added to the initial position y_0 = 0 we obtain the desired heights, which are not surprisingly identical to those obtained previously. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Was the solution in the wrong spot? at .5 sec the velocity is 10 m/s displacement of 6.75 m at 1 sec the velocity is 5 m/s displacement of 10 m at 1.5 sec velocity is 0 m/s displacement of 11.25 m at 2 sec the velocity is -5 m/s displacement if 10 m ds = v0 dt * .5 a dt^2 putting in the variables you get the displacement confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qprin phy and gen phy prob 34: Car rolls off edge of cliff; how long to reach 85 km/hr? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 85000 meters and 3600 seconds 85000/3600 = 23.6 m/s 23.6/9.8 = 2.408 seconds confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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.. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I dont know the info about gears so I looked that up in the answer but we don't know the acceleration so we can't solve it , it says graph but where?
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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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): That still does not explain where we got the information ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q **** Gen phy what is the meaning of the slope of the graph and why should it have this meaning? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Still no graph 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qGen phy what is the meaning of the area under the curve, and why does it have this meaning? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Assuming it is a velocity over time graph then it will be the distance travelled confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The area under the curve is the distance traveled. This is so because 'ds = vAve*'dt. 'dt is equal to the width of the section under the curve and vAve is equal to the average height of the curve. The area of a trapezoid is width times average height. Although this is not a trapezoid it's close enough that we for the purpose of estimation can analyze it as such. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qGen phy what is the area of a rectangle on the graph and what does it represent? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I can't tell what the area is but it represents an distance travelled confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The area of a rectangle on the graph represents a distance. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* 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)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** In time interval `dt after leaving the building the falling student has fallen through displacement `ds = v0 `dt + .5 a `dt^2, where v0 = 0 and, choosing the downward direction to be positive, we have a = -9.8 m/s^2. If `ds = -180 m then we have `ds = .5 a `dt^2 and `dt = sqrt(2 * `ds / a) = sqrt(2 * -180 m / (-9.8 m/s^2)) = 6 sec, approx.. Superman starts 5 seconds later, and has 1 second to reach the person. Superman must therefore accelerate at -9.8 m/s^2 thru `ds = -180 m in 1 second, starting at velocity v0. Given `ds, `dt and a we find v0 by solving `ds = v0 `dt + .5 a `dt^2 for v0, obtaining v0 = (`ds - .5 a `dt^2) / `dt = (-180 m - .5 * -9.8 m/s^2 * (1 sec)^2 ) / (1 sec) = -175 m/s, approx. Note that Superman's velocity has only about 1 second to change, so changes by only about -9.8 m/s^2, or about -10 m/s^2. ** ``qsketch a graph of Superman's position vs. clock time, and on the same graph show the student's position vs. clock time, with clock time starting when the person begins falling YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* 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)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** In time interval `dt after leaving the building the falling student has fallen through displacement `ds = v0 `dt + .5 a `dt^2, where v0 = 0 and, choosing the downward direction to be positive, we have a = -9.8 m/s^2. If `ds = -180 m then we have `ds = .5 a `dt^2 and `dt = sqrt(2 * `ds / a) = sqrt(2 * -180 m / (-9.8 m/s^2)) = 6 sec, approx.. Superman starts 5 seconds later, and has 1 second to reach the person. Superman must therefore accelerate at -9.8 m/s^2 thru `ds = -180 m in 1 second, starting at velocity v0. Given `ds, `dt and a we find v0 by solving `ds = v0 `dt + .5 a `dt^2 for v0, obtaining v0 = (`ds - .5 a `dt^2) / `dt = (-180 m - .5 * -9.8 m/s^2 * (1 sec)^2 ) / (1 sec) = -175 m/s, approx. Note that Superman's velocity has only about 1 second to change, so changes by only about -9.8 m/s^2, or about -10 m/s^2. ** ``qsketch a graph of Superman's position vs. clock time, and on the same graph show the student's position vs. clock time, with clock time starting when the person begins falling YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: #*&!
#$&* course phy 201 There were alot towards the end I could not answer it had no graph and did not say where it was , maybe I misinterpreted it? 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): ------------------------------------------------ Self-critique 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): ------------------------------------------------ Self-critique Rating: ********************************************* 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: dPe = dW - dKe 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): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qGive a specific example of such a process. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I am not sure what process is being talked about 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): ------------------------------------------------ Self-critique Rating: ********************************************* 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Well in the first place the weight on both sides to be equal then for the cart to move up the incline then there must be more weight on the other side leading to more work on the washers because friction will remove some of the strength from the force of the weights so the work on the washers is the greatest Work removed by friction + work done on cart = work on the weights 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qWhat is our evidence that the acceleration of the cart is proportional to the net force on the cart? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Fnet = m * a a = Fnet/m it is proportional to the net force divided by mass 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). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: Calculate the displacement and velocity at times of (a) 0.500, (b) 1.00, (c) 1.50, and (d) 2.00 s for a ball thrown straight up with an initial velocity of 15.0 m/s. Take the point of release to be y_0 = 0 . With the number of significant figures specified for this problem we would need to use a 3-significant-figure acceleration. However the acceleration at different points on the surface of the Earth varies in the third significant figure. Since the location isn't specified, the solution given here will ignore significant figures and use the value 10 m/s^2 for the acceleration of gravity and time intervals .5, 1, 1.5 and 2 s. You should be aware of the significant-figure implications both of the given information and the assumptions used this given solution. The acceleration of gravity is 10 m/s^2 downward. We understand the meaning of this acceleration to be that the speed of the object changes by 10 m/s every second, and that the velocity change is in the downward direction. It isn't explicitly stated, but it can be taken as implicit that the given times are in fact the times as measured on a running clock which reads 0 at the instant of release. Directly reasoned solution for clock time 0.5 second: In 0.5 second the change in speed will therefore be 5 m/s (if the change is 10 m/s every second, then in half a second it will be half of 10 m/s or 5 m/s). The velocity change is in the downward direction, so an initial upward speed of 15 m/s will in this time reduce to 10 m/s. During the 0.5 second interval, then, the velocity changes from 15 m/s to 10 m/s. The average velocity, since acceleration is uniform (so that the v vs. t graph is a straight line) is therefore 12.5 m/s. In 0.5 second with average velocity 12.5 m/s, the object will therefore move through displacement `dy = 12.5 m/s * 0.5 second = 6.75 meters. So after 0.5 second, the object's vertical position will be y = y_0 + 6.75 meters = 0 + 6.75 meters = 6.75 meters. Similar reasoning for the other given clock times leads to the following conclusions. Each line corresponds to the interval from clock time 0 to the given clock time. clock time change in velocity initial velocity final velocity ave velocity position 0.5 -5 m/s 15 m/s 10 m/s 12.5 m/s 6.75 m 1 -10 m/s 15 m/s 5 m/s 10 m/s 10 m 1.5 -15 m/s 15 m/s 0 m/s 7.5 m/s 11.25 m 2 -20 m/s 15 m/s -5 m/s 5 m/s 10 m More generally we could reason out the position at clock time t as follows: Taking upward as the positive direction, the acceleration being downward is negative, with value -10 m/s^2. The change in velocity for the interval from clock time 0 to clock time t is therefore -10 m/s^2 * t. The final velocity for this interval is thus 15 m/s - 10 m/s^2 * t. The average velocity for the interval is the average of the initial and final velocities, so vAve = (15 m/s + (15 m/s - 10 m/s^2 * t) ) / 2 = 15 m/s - 10 m/s^2 * t / 2 = 15 m/s - 5 m/s^2 * t. The displacement during the interval is thus `dy = vAve * `dt = vAve * (t - 0 s) = (15 m/s - 5 m/s^2 * t) * t = 15 m/s * t - 5 m/s^2 * t^2. Evaluating this for t = .5 s, 1 s, 1.5 s and 2 s we obtain respective displacements 6.75 m, 10 m, 11.25 m and 10 m. The above solutions provide insight into the nature of the motion that can be bypassed by using the third equation of motion `ds = v0 `dt * 1/2 a `dt^2. It isn't a good idea to bypass insights, so you should be able to reason out the solution in the above ways, but it's also important to be able to use the equations. And once you're comfortable with the reasoning process, the equations often provide the quickest path to the solution: Plugging t = 0, 0.5, 1, 1.5 and 2 into the equation we obtain respective displacements 6.75 m, 10 m, 11.25 m and 10 m. If the displacements are added to the initial position y_0 = 0 we obtain the desired heights, which are not surprisingly identical to those obtained previously. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Was the solution in the wrong spot? at .5 sec the velocity is 10 m/s displacement of 6.75 m at 1 sec the velocity is 5 m/s displacement of 10 m at 1.5 sec velocity is 0 m/s displacement of 11.25 m at 2 sec the velocity is -5 m/s displacement if 10 m ds = v0 dt * .5 a dt^2 putting in the variables you get the displacement confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qprin phy and gen phy prob 34: Car rolls off edge of cliff; how long to reach 85 km/hr? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 85000 meters and 3600 seconds 85000/3600 = 23.6 m/s 23.6/9.8 = 2.408 seconds confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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.. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I dont know the info about gears so I looked that up in the answer but we don't know the acceleration so we can't solve it , it says graph but where?
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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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): That still does not explain where we got the information ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q **** Gen phy what is the meaning of the slope of the graph and why should it have this meaning? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Still no graph 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qGen phy what is the meaning of the area under the curve, and why does it have this meaning? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Assuming it is a velocity over time graph then it will be the distance travelled confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The area under the curve is the distance traveled. This is so because 'ds = vAve*'dt. 'dt is equal to the width of the section under the curve and vAve is equal to the average height of the curve. The area of a trapezoid is width times average height. Although this is not a trapezoid it's close enough that we for the purpose of estimation can analyze it as such. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qGen phy what is the area of a rectangle on the graph and what does it represent? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I can't tell what the area is but it represents an distance travelled confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The area of a rectangle on the graph represents a distance. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* 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)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** In time interval `dt after leaving the building the falling student has fallen through displacement `ds = v0 `dt + .5 a `dt^2, where v0 = 0 and, choosing the downward direction to be positive, we have a = -9.8 m/s^2. If `ds = -180 m then we have `ds = .5 a `dt^2 and `dt = sqrt(2 * `ds / a) = sqrt(2 * -180 m / (-9.8 m/s^2)) = 6 sec, approx.. Superman starts 5 seconds later, and has 1 second to reach the person. Superman must therefore accelerate at -9.8 m/s^2 thru `ds = -180 m in 1 second, starting at velocity v0. Given `ds, `dt and a we find v0 by solving `ds = v0 `dt + .5 a `dt^2 for v0, obtaining v0 = (`ds - .5 a `dt^2) / `dt = (-180 m - .5 * -9.8 m/s^2 * (1 sec)^2 ) / (1 sec) = -175 m/s, approx. Note that Superman's velocity has only about 1 second to change, so changes by only about -9.8 m/s^2, or about -10 m/s^2. ** ``qsketch a graph of Superman's position vs. clock time, and on the same graph show the student's position vs. clock time, with clock time starting when the person begins falling YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* 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)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** In time interval `dt after leaving the building the falling student has fallen through displacement `ds = v0 `dt + .5 a `dt^2, where v0 = 0 and, choosing the downward direction to be positive, we have a = -9.8 m/s^2. If `ds = -180 m then we have `ds = .5 a `dt^2 and `dt = sqrt(2 * `ds / a) = sqrt(2 * -180 m / (-9.8 m/s^2)) = 6 sec, approx.. Superman starts 5 seconds later, and has 1 second to reach the person. Superman must therefore accelerate at -9.8 m/s^2 thru `ds = -180 m in 1 second, starting at velocity v0. Given `ds, `dt and a we find v0 by solving `ds = v0 `dt + .5 a `dt^2 for v0, obtaining v0 = (`ds - .5 a `dt^2) / `dt = (-180 m - .5 * -9.8 m/s^2 * (1 sec)^2 ) / (1 sec) = -175 m/s, approx. Note that Superman's velocity has only about 1 second to change, so changes by only about -9.8 m/s^2, or about -10 m/s^2. ** ``qsketch a graph of Superman's position vs. clock time, and on the same graph show the student's position vs. clock time, with clock time starting when the person begins falling YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: #*&!