course Phy 201 Hwx㐲͆assignment #008
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09:09:39 QUESTION FROM STUDENT--Please define the differnece between Fnet and Force. See if you can answer this question.
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RESPONSE --> All I know is that F=m*a confidence assessment: 1
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09:11:10 ** Net force is the sum of all forces acting on an object. If you're pushing your car you are exerting a force, friction is opposing you, and the let force is the sum of the two (noting that one is positive, the other negative so you end up with net force less than the force you are exerting). Your heart rate responds to the force you are exerting and the speed with which the car is moving; the accel of the car depends on the net force. **
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RESPONSE --> I understand. This includes all forces, not just the one in question like in the formula F=m*a. self critique assessment: 3
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詇~Zޱ~\͘ assignment #004 krq찶N⯔~ Physics I Class Notes 05-30-2007
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09:16:15 How do we reason out the process of determining acceleration from rest given displacement and time duration?
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09:16:18 ** COMMON ERROR: By first finding the average velocity. Also we must understand this gives us the change in velocity which gives us the acceleration. INSTRUCTOR COMMENT: Average acceleration is not related to average velocity. It is related to the change in velocity, but change in velocity is not the same thing as acceleration. Acceleration is the rate of change of velocity. ANSWER TO QUESTION: Average acceleration is defined to be the average rate at which velocity changes, which is change in velocity/change in clock time, or equivalently change in velocity/time duration. However we do not know the change in velocity, nor from the given information can we directly determine the change in velocity. So we have to look at what we can reason out from what we know. Given displacement from rest and time duration we calculate average velocity: vAve = `ds / `dt = displacement/time duration. Since the initial velocity is 0, the final velocity is double the average so we find the final velocity by doubling the average velocity. Now that we know the initial velocity 0 and the final velocity we can find change in velocity by subtracting initial vel from final vel. Dividing change in velocity by a time duration we finally obtain the average acceleration. **
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09:16:24 How do we obtain an approximate velocity vs. clock time table and graph from a series of position vs. clock time observations? How do we then obtain an approximate acceleration vs. clock timetable and graph?
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09:16:28 ** If we start with a position vs. clock time graph we do divide the graph into intervals. On each interval velocity, which is rate of change of position, is represented by the slope of the graph. So we have to calculate the slope of the graph on each interval. If we then graph the slopes vs. the midpoint times of the intervals we get a good approximation of the velocity vs. clock time graph. The velocity at a given instant is the slope of the position vs. clock time graph. STUDENT ANSWER: We first calculate the time interval and displacement between each pair of data points. We use these calculations to calculate the average velocity. To obtain an approximate accelerations. Clock timetable and graph by associating the average velocity on a time interval with the midpoint clock time on that interval. INSTRUCTOR CRITIQUE: Using your words and amplifying a bit: We first calculate the time interval and displacement between each pair of data points. We use these results to calculate the average velocity, dividing displacement by time interval for each interval. Then we make a table, showing the average velocity vs. the midpoint time for each time interval. To obtain approximate accelerations we use the table and graph obtained by associating the average velocity on a time interval with the midpoint clock time on that interval. We find the time interval between each pair of midpoint times, and the change in average velocities between those two midpoint times. Dividing velocity change by time interval we get the rate of velocity change, or acceleration. **
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09:16:40 For water flowing from a uniform cylinder through a hole in the bottom, with how much certainty can we infer that the acceleration of the water surface is uniform?
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09:16:49 ** If we first calculate velocities from the position vs. clock time data we get decreasing velocities. If we graph these velocities vs. midpoint clock times we get a graph which appears to be well-fit by a straight line. This is evidence that the acceleration of the water surface is uniform. If we calculate average accelerations based on average velocities and midpoint clock times we get a lot of variation in our results. However since acceleration results depend on velocities and changes in clock times, and since the velocities themselves were calculated based on changes in clock times, our results are doubly dependent on the accuracy of our clock times. So these fluctuating results don't contradict the linearity of the v vs. t graph. We also find that the position vs. clock time data are very well-fit by a quadratic function of clock time. If the position vs. clock time graph is quadratic then the velocity is a linear function of clock time (University Physics students note that the derivative of a quadratic function is a linear function) and acceleration is constant (University Physics students note that the second derivative of a quadratic function is constant). **
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09:16:55 How does the graph make it clear that an average velocity of 4 cm / s, and initial velocity 0, imply final velocity 8 cm / s?
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09:16:58 ** If the graph is linear then the average velocity occurs at the midpoint clock time, and is halfway between the initial and final velocities. In this case 4 cm/s would be halfway between 0 and the final velocity, so the final velocity would have to be 8 cm/s (4 is halfway between 0 and 8). **
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09:17:04 Why does a linear velocity vs. time graph give a curved position vs. time graph?
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09:17:07 ** Assuming uniform time intervals, a linear v vs. t graph implies that over every time interval the average velocity will be different that over the previous time intervals, and that it will be changing by the same amount from one time interval to the next. The result is that the distance moved changes by the same amount from one time interval to the next. The distance moved is the rise of the position vs. clock time graph. If over uniform intervals the rise keeps changing, by the same amount with each new interval, the graph has to curve. **
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yj{Ɩ assignment #006 krq찶N⯔~ Physics I Class Notes 05-30-2007
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09:17:30 How do flow diagrams help us see the structure of our reasoning processes?
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09:17:33 ** They help us to visualize how all the variables are related. Flow diagrams can also help us to obtain formulas relating the basic kinematic quantities in terms of which we have been analyzing uniformly accelerated motion. **
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ӳĶSнN assignment #007 krq찶N⯔~ Physics I Class Notes 05-30-2007
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09:17:55 Why do we say that the first equation of uniformly accelerated motion expresses the definition of average velocity, while the second expresses the definition of acceleration?
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09:18:06 ** When acceleration is uniform average velocity is the average of initial and final velocities, (vf + v0) / 2. Average velocity is `ds / `dt (whether accel is uniform or not). Setting `ds / `dt = (vf + v0) / 2 we obtain `ds = (vf + v0) / 2 * `dt, which is the first equation of uniformly accelerated motion. So the definition of average velocity is equivalent to the definition of average velocity. Average acceleration is aAve = `dv / `dt = (vf - v0) / `dt. Since for uniform acceleration the acceleration is constant, we can just say that in this case a = (vf - v0) / `dt. This equation is easily rearranged to give the second equation of motion, vf = v0 + a `dt. **
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09:18:15 Why, for uniform acceleration, is vAve = (vf + v0) / 2, while this is not usually true for nonuniform acceleration?
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09:18:17 ** If acceleration is uniform then the v vs. t graph is linear, so that the average velocity over any time interval must be equal to the velocity at the midpoint of that interval. It follows that the average velocity must be midway between the initial and final velocities. (vf + v0) / 2 is the average of the initial and final velocities, and is therefore halfway between the v0 and vf. **
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09:18:21 In commonsense terms, why does change of velocity over a given distance, with a given uniform acceleration, differ with initial velocity?
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09:18:23 ** If the uniform acceleration is the same in both cases, then assuming that both initial velocity and acceleration are positive, a greater initial velocity will result in a shorter time interval to cover the given distance. A shorter time interval leaves less time for velocity to change, resulting in a smaller change in velocity. **
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mXvxLzH assignment #008 krq찶N⯔~ Physics I Class Notes 05-30-2007
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09:18:43 What is the meaning of the x intercept of the graph of force vs. slope?
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09:19:27 ** On this graph the x axis represents slope, the y axis represents acceleration. The x axis point occurs when y = 0, so this point represents a slope for which acceleration is zero. Assuming that 'down the incline' is the positive direction, everywhere to the right of this point the acceleration is positive and everywhere to the left is negative. The x intercept therefore represents the maximum slope for which friction is great enough to prevent the ball from accelerating down the slope. Since friction is resisting the weight component down the incline, the x intercept represents the slope when friction and the weight component tending to accelerate the cart down the slope are exactly in balance, equal and opposite. **
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09:19:36 When we obtain a linear relationship between force and acceleration, is it plausible that the constant term in the equation is, within experimental uncertainty, zero?
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09:19:46 ** The constant term in the equation will be zero only if the graph passes through the origin. This would imply that any nonzero force will produce a nonzero acceleration, and that for example there is either no friction present or friction has been compensated for in some way. **
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09:19:54 For the force vs. acceleration relationship for the car, why should we expect that the acceleration corresponding to a net force equal to the car's weight is the acceleration of gravity?
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09:31:07 In terms of the equations of motion why do we expect that a * `ds is proportional to the change in v^2, and why do we then expect that the change in v^2 is proportional to Fnet `ds?
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RESPONSE --> confidence assessment:
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09:31:34 ** It's very important in physics to be able to think in terms of proportionality. To say that y is proportional to x is to say that for some k, y = k x. That is, y is a constant multiple of x. To say that a * `ds is proportional to the change in v^2 is to say that for some k, a * `ds = k * ( change in v^2)--i.e., that a * `ds is a constant multiple of the change in v^2. In terms of the equations of motion, we know that vf^2 = v0^2 + 2 a `ds so a `ds = 1/2 (vf^2 - v0^2), which is 1/2 the change in v^2. So a `ds is a constant multiple (1/2) of the change in v^2. Formally we have a `ds = k ( change in v^2) for k = 1/2. Now since Fnet = m a we see that Fnet is proportional to a for a given mass m, and it follows that Fnet `ds = k * change in v^2, for the appropriate k (specifically for k = mass / 2. **
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RESPONSE --> self critique assessment:
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09:31:39 How do our experimental results confirm or refute this hypothesis?
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RESPONSE --> confidence assessment:
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09:31:59 ** We didn't actually do this part of the experiment, but on a ramp with fixed slope a `ds is simply proportional to `ds. When we measured `dt for different distances from rest down the same ramp, we were then able to determine the average and final velocities. The change in v^2 for each timing would be from 0 to vf^2. The change would therefore be just vf^2. If a `ds is proportional to the change in vf^2 then a graph of vf^2 vs. a `ds should be linear. Since a is constant we don't even need to determine it--a graph of vf^2 vs. `ds would be linear. This would confirm the hypothesis. **
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RESPONSE --> self critique assessment:
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緀Si assignment #008 008. `query 8 Physics I 05-30-2007
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09:33:01 QUESTION FROM STUDENT--Please define the differnece between Fnet and Force. See if you can answer this question.
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RESPONSE --> Fnet is the sum of all forces acting on an object. The only reason this is my guess is because any time I have ever seen ""net"" it means the ""sum of"" whatever entity, in this case, force. confidence assessment: 1
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09:33:11 ** Net force is the sum of all forces acting on an object. If you're pushing your car you are exerting a force, friction is opposing you, and the let force is the sum of the two (noting that one is positive, the other negative so you end up with net force less than the force you are exerting). Your heart rate responds to the force you are exerting and the speed with which the car is moving; the accel of the car depends on the net force. **
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RESPONSE --> I understand. self critique assessment: 3
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09:35:23 In terms of the equations of motion why do we expect that a * `ds is proportional to the change in v^2, and why do we then expect that the change in v^2 is proportional to Fnet `ds?
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RESPONSE --> If we use the formula vf^2=vo^2+2a'ds, we see that changes in a and 'ds could affect v^2 directly. I do not know why, though, that the change in v^2 is proportional to Fnet 'ds. confidence assessment: 2
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09:42:13 ** It's very important in physics to be able to think in terms of proportionality. To say that y is proportional to x is to say that for some k, y = k x. That is, y is a constant multiple of x. To say that a * `ds is proportional to the change in v^2 is to say that for some k, a * `ds = k * ( change in v^2)--i.e., that a * `ds is a constant multiple of the change in v^2. In terms of the equations of motion, we know that vf^2 = v0^2 + 2 a `ds so a `ds = 1/2 (vf^2 - v0^2), which is 1/2 the change in v^2. So a `ds is a constant multiple (1/2) of the change in v^2. Formally we have a `ds = k ( change in v^2) for k = 1/2. Now since Fnet = m a we see that Fnet is proportional to a for a given mass m, and it follows that Fnet `ds = k * change in v^2, for the appropriate k (specifically for k = mass / 2. **
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RESPONSE --> Oh wow, I understand how the formula was applied to show exactly how much change a'ds could cause for vf^2. As for Fnet 'ds, I sort of understand. All of this feels like it is way over my head. I am going to reread it and try to put it in my own words. I see that Fnet is proportional to a because of the formula F=m*a, because if a increases or decreases, the change will be somehow reflected proportionally in F. Now for Fnet='ds=k*change in v/^2, I still do not understand this. I understand from where the first 1/2 was derived (from the formula), but in terms of the Fnet equation, I do not understand how the ""specifically for k=mass/2"" comes into play. self critique assessment: 0
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09:43:18 How do our experimental results confirm or refute this hypothesis?
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RESPONSE --> I honestly do not remember any experiment I have performed having much, if anything, to do with this. I am unable to answer this question. confidence assessment: 0
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09:45:34 ** We didn't actually do this part of the experiment, but on a ramp with fixed slope a `ds is simply proportional to `ds. When we measured `dt for different distances from rest down the same ramp, we were then able to determine the average and final velocities. The change in v^2 for each timing would be from 0 to vf^2. The change would therefore be just vf^2. If a `ds is proportional to the change in vf^2 then a graph of vf^2 vs. a `ds should be linear. Since a is constant we don't even need to determine it--a graph of vf^2 vs. `ds would be linear. This would confirm the hypothesis. **
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RESPONSE --> I understand this I think. If a remains constant, then 'ds is directly proportional to 'ds, not meaning that they are the same, but that a change in one will reflect a proportional change in the other, which will keep the slope constant. self critique assessment: 2
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ׂ assignment #009 009. `query 9 Physics I 05-30-2007
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13:09:52 Introductory prob set 3 #'s 1-6 If we know the distance an object is pushed and the work done by the pushing force how do we find the force exerted by the object?
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RESPONSE --> W=f*'ds W/'ds=f confidence assessment: 3
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13:10:30 ** Knowing the distance `ds and the work `dW we note that `dW = F * `ds; we solve this equation and find that force is F=`dw/`ds **
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RESPONSE --> Oops. I used the wrong symbol for Work. I will not make the mistake again. self critique assessment: 3
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13:11:22 If we know the net force exerted on an object and the distance through which the force acts how do we find the KE change of the object?
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RESPONSE --> The Kinetic Energy is basically the same as the work, I think, at least in terms of the formula. One would simply multiply the force by the displacement, yielding the kinetic energy or the work. confidence assessment: 2
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13:22:57 **`dW + `dKE = 0 applies to the work `dW done BY the system and the change `dKE in the KE OF the system. The given force acts ON the system so F `ds is work done ON the system. The work done BY the system against that force is `dW = -F * `ds. When you use the energy equation, this is the work you need--the work done BY the system. **
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RESPONSE --> The first part of the equation 'dw is the work done BY the system, whereas the other part of the equation F*'ds is the work done ON the system. That is the discrepancy and how to know where to fill in variables. self critique assessment: 2
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13:24:51 Why is KE change equal to the product of net force and distance?
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RESPONSE --> Because that is the work done on the system? confidence assessment: 1
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13:25:34 ** It comes from the equation vf^2 = v0^2 + 2 a `ds. Newton's 2d Law says that a = Fnet / m. So vf^2 = v0^2 + 2 Fnet / m `ds. Rearranging we get F `ds = 1/2 m vf^2 - 1/2 m v0^2. Defining KE as 1/2 m v^2 this is F `ds = KEf - KE0, which is change in KE. **
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RESPONSE --> I see, it is a derived equation. self critique assessment: 2
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13:32:23 When I push an object with a constant force, why is KE change not equal to the product of the force I exert and the distance?
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RESPONSE --> You must find the initial KE and the final KE to constitute change; otherwise, it is the average, just as with acceleration, velocity, etc. confidence assessment: 3
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13:42:24 ** Change in KE is equal to the work done by the net force, not by the force I exert. When I push an object in the real world, with no other force 'helping' me, there is always at least a little force resisting my push. So the net force in this case is less than the force I exert, in which case the change in KE would be less than the product of the force I exert and the distance. If another force is 'helping' me then it's possible that the net force could be greater than the force I exert, in which case the change in KE would be greater than the product of the force I exert and the distance. It is actually possible for the 'helping' force to exactly balance the resisting force, but an exact balance would be nearly impossible to achieve. ANOTHER WAY OF LOOKING AT IT: If I push in the direction of motion then I do positive work on the system and the system does negative work on me. That should increase the KE of the system. However if I'm pushing an object in the real world and there is friction and perhaps other dissipative forces which tend to resist the motion. So not all the work I do ends up going into the KE of the object. **
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RESPONSE --> I think I understand. Is this where the initial and final KE's come from? Or not?
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