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course PHY201
9/26 10:20pm In this assignment, I got stuck a couple of times. I put those points as questions so please take a look and let me know where my thinking needs to point to. Thanks
For the toy car and magnets:Give the three positions you measured for each trial, one trial to each line. Each line should consists of three numbers, representing the position in cm of the fixed magnet, the position in cm of the magnet of the toy car, and the position in cm at which the car came to rest after being released. This is your raw data:
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0, 8, 23
0, 7, 26
0, 6, 30
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Give a brief explanation of what the data mean, including a statement of the units of the numbers:
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The 1st number is the fixed position of the magnet at 0cm.
The 2nd number is the distance between the front of the car with the magnet on it and the stationary magnet.
The 3rd number is the ending distance between the fixed magnet at 0cm and where the front of the car ended up at 23cm, 26cm, and 30cm respectively
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For each trial, give the distance of separation between the two magnets at the instant of release, and the distance the car traveled between release and coming to rest. Give in the form of two numbers to a line, separated by commas, with separation first and coasting distance second. After the last line, give a brief explanation of how your results were obtained and what the numbers mean, including a statement of the units of the numbers.
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8, 15
7, 19
6, 24
To obtain these results, I set the meter stick beside the toy car and magnet. I held 1 magnet stationary with my right hand and held the car with magnet on top with my left hand. I then released the car and magnet with my left hand while still holding the magnet stationary with my right hand.
For the 1st trial, the front of the magnet car was 8cm away from the stationary magnet and when I released it, it ended up being 23cm away from the stationary magnet for a difference of 15cm.
For the 2nd trial, the front of the magnet car was 7cm away from the stationary magnet and when I released it, the magnet car ended up being 26cm away from the stationary magnet for a difference of 19cm.
For the 3rd trial, the front of the magnet car was 6cm away from the stationary magnet and when I released it, the magnet car ended up being 30cm away from the stationary magnet for a difference of 24cm.
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Sketch a graph of distance traveled vs. initial separation. Describe your graph.
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The y-axis represents distance traveled and the x-axis represents initial separation. There are 2 points. The 1st point represents the distance of separation between the stationary magnet and the front of the magnet car. The 2nd point represents the total distance traveled away from the stationary magnet.
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Sketch the smooth curve you think best represents the actual behavior of distance traveled vs. initial separation for this system.
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The curve would spike upwards at the start and then slowly taper off from left to right ending up at point 15 on the vertical axis. In the trial, when the car was 1st released there was a “burst” of transfer of energy from potential to kinetic from the magnet on top of the car therefore representing the spike of the curve.
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Identify the point where initial separation is 8 cm.
· What coasting distance corresponds to this point?
· What is the slope of your smooth curve in the neighborhood of this point?
Give the coasting distance as a number in the first line, the slope of the graph as a number in the second line. Starting in the third line give the units of your quantities and explain what each quantity means, and how you obtained it.
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15cm
1.5
The coasting distance was 15cm and represents the distance traveled once the magnet was let go from 8cm apart from the stationary magnet. The total distance away from the magnet was 23cm but it only coasted 15cm. To get slope of graph I divided rise/run where rise is 23cm and run is 15cm. The rise represents the total cm away from the initial magnet and the run represents the total coasting distance.
rise and run represent changes in quantities; you don't indicate what changes with each
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Repeat for the point where initial separation is 5 cm.
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31.5cm
1.2
The coasting distance was 31.5cm and represents the distance traveled once the magnet was let go from 5cm apart from the stationary magnet. The total distance away from the magnet was 36.5cm but it only coasted 31.5cm. To get slope of graph I divided rise/run where rise is 36.5cm and run is 31.5cm. The rise represents the total cm away from the initial magnet and the run represents the total coasting distance.
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According to your graph, if the initial separation is doubled, what happens to the distance the car travels?
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The distance is tripled if the initial separation is doubled.
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According to your graph, if the initial separation is doubled, what happens to the slope of the graph?
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The slope would be 1/3 less if the initial separation is doubled.
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The distance the car travels is an indication of the energy it gained from the proximity of the magnets. Specifically, the frictional force slowing the car typically is about .01 Newton, or 10 milliNewtons. The force exerted on the car by friction is in the direction opposite the car's displacement, so when you calculate the work done by this force, your force and the displacement will have opposite signs (i.e., one will be positive and the other will be negative).
Using the .01 Newton = 10 milliNewton force and your displacement in meters (you likely calculated the displacement in centimeters, so be sure you use the equivalent displacement in meters) find the work done by this force on each of your trials. Give below the initial separation of the magnets, the work done by the frictional force acting on the car in Newtons * m, the work in milliNewtons * m, in the form of three numbers per line separated by commas. In the first subsequent line, explain your results and include a detailed sample calculation.
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Initial separation of magnets, Work done by frictional force, work in millNewtons
80mm, -10milliNewtons, -1500milliNewtons
70mm, -10milliNewtons, -1900milliNewtons
60mm, -10milliNewtons, -2400milliNewtons
50mm, -10milliNewtons, -3150milliNewtons
40mm, -10milliNewtons, -4500milliNewtons
I used the formula aW(fric)=F(fric)*`dx=-10milliN*150mm=-1500milliNewtons/cm to find the work done for the 8cm separation.
your work multiplie force in milliNewtons by displacement in millimeters, so the result would be in milliNewtons * mm.
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Friction does negative work on the coasting car, which progressively depletes its kinetic energy (recall that kinetic energy is energy of motion). In this situation the original kinetic energy of the car came from the configuration of the magnets (the closer the magnets, the greater the KE gained by the car). We say that the initial magnet configuration had potential energy, with closer magnets associated with more potential energy. The initial potential energy of the magnets was therefore converted into the initial kinetic energy of the car, which was then lost to friction as friction did work on the car equal and opposite to its initial kinetic energy. (actually the potential-to-kinetic-energy transition takes place over a significant interval of time and distance, so it would be more appropriate to speak of the work by friction on the car begin equal and opposite to the initial potential energy, but we won't really worry much about that just yet).
Explain this below in your own words, as best you understand it.
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The reason that the car does not continue traveling once the potential energy was transferred to kinetic energy is because friction acted against as well as opposite the kinetic energy. If this were not so, then the car would continue to move in the direction the transfer of energy was exerted.
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The car also exerts a frictional force on the table which is equal and opposite the frictional force exerted on it (if the table was on frictionless rollers the frictional force exerted by the car would cause the table to accelerated very slowly in the direction of the car's motion), so it does work against friction which is equal and opposite to the work friction does on it. So the car does positive work against friction. This work is done at the expense of its kinetic energy.
If you were to make a table showing work done by the car against friction vs. initial separation it would be the same as the table you gave previously, except that the work would be positive (you did remember to make the work negative on the previous table, didn't you?). I'm not going to ask you to give that table here, since except for the sign of the work it is the same as your previous table.
What we are going to want is a graph of the work done by the car against friction, vs. initial separation.
You already have a graph of distance vs. initial separation.
You can add a new labeling to your vertical scale to represent the corresponding work done by the car against the frictional force.
If you don't understand what this means, you can go ahead and create a separate graph of work done vs. initial separation.
Either way:
According to your new graph, or the new labeling of your original graph, what is the work done by the car against friction when the initial separation is 8 cm? Give the quantity in the first line, a brief explanation in the second.
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The graph would start at the 150milliNewton point of the y-axis and because the work is negative would come from top to bottom. It would not spike as much at the top because there is less PE being transferred to KE on the car. But the graph slowly tapers off to the bottom right part of the graph ending at the 23cm point of the x-axis.
you previously calculated work in milliNewton * mm; work isn't in milliNewtons
work totals were in thousands of milliNewton * mm; 150 milliNewtons does not correspond to a quantity measured on the vertical axis; however my guess is that the number is OK except that you left off a zero. The unit is still incorrect, but you are on the right track with the idea and the description.
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Repeat for initial separation 5 cm.
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The graph would start at the 3150milliNewton point of the y-axis and because the work is negative would come from top to bottom as it did for the 8cm separation. It would spike more than the 8 cm mark at the top because there is more PE being transferred to KE on the car than was on the 8cm mark therefore making the car’s displacement greater. The graph will slowly taper off to the bottom right part of the graph ending at the 36.5cm point of the x-axis.
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According to your graph, how much more work was done against friction when the initial separation was 5 cm, than was done when initial separation was 8 cm?
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There was -1650milliNewtons more work being done against friction when the initial separation was 5cm.
On this, I figured the difference in the work by the 8cm separation and the 5cm separation. The difference between -3150milliNewtons and -1500milliNewtons
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How much work was therefore done by the magnetic force between the 5 cm and 8 cm separation, while the cars were moving apart?
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+1650milliNewtons
Because the work is being done by the magnetic force against friction, then the magnetic force has to be positive as well as equal and opposite to the frictional force to move the car. Or we could look at it like is described below.
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For a system released at a separation of 5 cm or less, the magnets exerted a decreasing force between the 5 cm separation and the 8 cm separation. The force has an average somewhere between the forces exerted at the 5 cm and 8 cm separations. If you answered the preceding question correctly you know the work between the two positions.
Through what displacement did the magnetic force act between these two separations?
How can you calculate the average force given the displacement and the work?
What therefore was the average force?
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The displacement for the 5cm mark was 31.5cm and the displacement for the 8cm mark was 15cm. 315,150
To get the average force we would just add up the 2 forces and then divide by 2.
The average force then becomes (315milliNewtons+150milliNewtons)/2=232.5milliNewtons
I’m not at all confident on that answer I just gave.
Again good reasoning, but not clear on the intended meanings of the quantities or their units.
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What therefore is the average rate at which work is being done by the magnets, per unit of separation, between the 5 cm and 8 cm separations?
What aspect of the graph of work vs. separation is associated with this average rate?
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The slope
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What is your best estimate of the average rate at which work is being done by the magnets, per unit of separation, when the magnets are 6 cm apart?
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In general how can you use a graph of work vs. separation to this system to find the force exerted by the magnets at a given separation?
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On the questions above, I am unclear on. How do I figure these?
For the car and paperclips
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I need to gather more data on this to complete the questions below. I will gather the data and resubmit when complete.
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Let 1 unit of force correspond to the weight of 1 small paperclip. On this scale the weight of a large paperclip is 3 units of force.
If the paperclip is on the car, its weight is balanced by the upward force exerted by the table and it has no direct effect on the car's acceleration. If it is suspended, then its weight contributes to the accelerating force.
You should have obtained a count and a distance from rest for each trial, and on each trial there will be some number of force units suspended from the thread (1 unit for every small, 3 units for every large clip). Report in the first line the number of clips, the count and the distance from rest, separated by commas, for your first trial. Report subsequent trials in susequent lines. After reporting the data for all your trials, give a brief explanation of your setup and how the trials were conducted. Include also the information about how many of your counts take how many seconds.
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Determine the acceleration for each trial. You may use the 'count' as your unit of time, or if you prefer you can convert your counts to seconds and use seconds as your time unit. In each line below list the number of suspended force units and your acceleration. After reporting your results, give in the next line the units and your explanation of how your results were obtained.
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Sketch a graph of acceleration vs. number of force units, and describe your graph. Fit a straight line to your graph and determine its slope. Describe how the trend of your data either indicates a good straight-line fit, or how it deviates from a straight line.
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Brief at-home experiment
Is the magnitude of a ball's acceleration up an incline the same as the magnitude of its acceleration down the incline?
Test this.
One possible suggested method: Use the short ramp to get the ball started, and let it roll from the short ramp onto the longer ramp, with the longer ramp inclined so the ball rolls up, rather than down. Get the ball started on the short ramp, either by inclining it toward the long ramp or giving it a push (before it reaches the long ramp). Click the TIMER at the instant the ball hits the 'bump' between the two ramps, again when the ball comes to rest for an instant before accelerating back down the long ramp, and once more when the ball again hits the 'bump'.
Everyone will tend to anticipate their 'clicks', and to try to compensate for their anticipation. If the effect of anticipation is the same for all three timed events, then uncertainties in your results will all be due to the TIMER. If the degree of anticipation differs, as it inevitably will, then the uncertainties are compounded. If the degree of anticipation (and/or compensation) tends to be either greater or less for the second event (the ball stopping for an instant) than for the first and third (the 'clicks' made when the ball hits the 'bump'), then a systematic error is introduced (you will have a tendency to 'short' one interval while extending the other).
If there exists a difference between the times up and down the ramp, and if the difference is great enough to show through the uncertainties and systematic errors, then you might get a useful result (e.g., either the actual times are the same, or they differ).
Give a brief report of your data and your results:
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Displacement and force vector for a rubber band
The points (2, 1) and (4, 6) have the following characteristics:
· From the first point to the second the 'run' is 4 - 2 = 2 and the 'rise' is 6 - 1 = 5.
· The slope of the line segment connecting the points is therefore 5 / 2 = 2.5.
· A right triangle can be constructed whose hypotenuse is the line segment from (2, 1) to (4, 6), and whose legs are parallel to the coordinate axes. The leg parallel to the x axis has length 2, and the leg parallel to the y axis has length 5. The hypotenuse therefore has length sqrt(2^2 + 5^2) = sqrt(29) = 5.4, approx..
· The area of the trapezoid formed by projecting the two points onto the x axis, whose sides are the projection lines, the segment [2, 4] of the x axis and the line segment between the two points, has 'graph altitudes' 1 and 5, giving it average 'graph altitude' (1 + 6) / 2 = 3.5, and width 5 - 1 = 4, so its area is 14.
· The vector from the first point to the second has x component 2 and y component 5. Its length is sqrt(20) and its direction is specified by the ratio of the y to the x component.
Any time you see two points on a graph you should be aware that you can easily construct the right triangle and use it to calculate the slope of the segment joining them, the distance between the points, the area of the associated trapezoid and the components of the vector. Depending on the nature of the graph, some of these quantities will sense and be have useful interpretations, while others probably will not.
In the particular example of a rubber band stretched between the two points, with the coordinates in centimeters, the most important quantities are the components and length of the vector. The vector from the first point to the second is in this case a displacement vector. The displacement vector has x component 2 cm, y component 5 cm and length sqrt( 2 cm)^2 + (5 cm)^2 ) = sqrt( 29 cm^2) = sqrt(29) cm, about 5.4 cm.
· We denote the displacement vector from (2 cm, 1 cm) to (4 cm, 6 cm) as <2 cm, 5 cm>
If we divide the this vector 2 we get a vector of length 2.15 cm. The x component will be 1 cm (half of the original 2 cm) and the y component 5 cm /2 = 2.5 cm (half of the original 5 cm). The ratio of the components is the same, so this vector will be in the same direction as the original vector. If we multiply this vector by 2 we get a vector of length 8.6 cm. Its components will be double those of the original vector, and it will also be in the same direction as the original vector.
If we divide our vector by 5.4 cm, then the resulting vector has length 5.3 cm / (5.3 cm) = 1. The x and y components will be 2 cm / (5.4 cm) = .37 and 5 cm / (5.4 cm) = .92. As before, the ratio of the components is the same as for the original vector (accurate to 2 significant figures),
The vector <.37, .92> has magnitude 1 (calculated to 2 significant figure), as you can easily verify using the Pythagorean Theorem. Its direction is the same as that of the displacement vector <2 cm, 5 cm>.
Now we invoke a rule to find the tension in the rubber band. Assume that for this rubber band the tension is .5 Newtons for each centimeter in excess of its 'barely-exerting-force' length of 4.8 cm.
· The 5.4 cm length of the stretched rubber band is therefore .6 cm in excess of the 'barely-exerting-force' length of 4.8 cm.
· According to the rule, we conclude that the tension is therefore .6 cm * .5 N / cm = .3 N.
The vector <.37, .92> has magnitude 1, and its direction is the same as that of the displacement vector. A rubber band can exert a force only in the direction of its displacement vector.
· If we multiply our vector <.37, .92> , which we again mention has length 1, by .3 N, we will obtain a vector whose magnitude is .3 N. (whatever quantity we multiply by a vector of length 1, we end up with a vector whose length is equal to that quantity, or more correctly whose length represents that quantity)
· The direction of this vector will be the same as that of our original vector.
· <.37, .92> * .3 N = <0.11 N, 0.28 N>
· This is the tension vector. It tells us that the tension of this rubber band has x component 0.11 N and y component 0.28 N.
Using your data from the rubber band experiment, check your previous calculations of the displacement vector, the unit vector and the force vector for each rubber band. If your original work on this experiment was not done correctly, please correct it and resubmit. Please acknowledge that you have read this instruction and understand it.
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Acknowledged
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Linear force function and work\energy
If the graph below indicates the tension in Newtons vs. length in cm for a rubber band:
At what length does the rubber band begin exerting a force?
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For this problem, can we have a vector with points (5,0) and (7,1) even if the coordinates are not the same? If so, then the x and y coordinates would be 2cm and 1N respectfully? So, our displacement vector is (2, 1)? I’m not sure if that can be given the 2 points are not the same unit.
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What is the average force exerted between the lengths x = 5 cm and x = 7 cm?
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As the rubber band is stretched from length 5 cm to length 7 cm, through what distance is the force exerted?
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How much work is therefore done in stretching the rubber band from 5 cm to 7 cm length?
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How much work is done in stretching the rubber band from the 5 cm length to the 8 cm length?
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How much work is done in stretching the rubber band from the 6 cm length to the 8 cm length?
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What are the areas beneath the graph between each of the following pairs of lengths:
· x = 5 cm and x = 7 cm
· x = 5 cm and x = 8 cm
· x = 6 cm and x = 8 cm
· x = 7 cm and x = 8 cm
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Verify that the equation of the straight line in the given graph is F(x) = (x - 5) * .5, where F(x) is force in Newtons when x is position in centimeters.
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University Physics Students: Confirm your results for graph areas by integrating the force function over appropriate intervals. Suggested method: Integrate the function symbolically from x = a to x = b, then use the resulting expression for appropriate values of a and b.
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Nonlinear force function and work\energy
The graph given previously was linear. That graph would be realistic for a well-made spring, but not for a rubber band.
The graph given below is more realistic.
Estimate the average force exerted by this rubber band between the x = 5 cm and x = 6 cm lengths. Give your estimate in the first line, and explain how you made the estimate in the second:
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Repeat, for the x = 6 cm to x = 7 cm length interval.
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Repeat, for the x = 5 cm to x = 5.5 cm length interval.
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Repeat, for the x = 5.5 cm to x = 6 cm length interval.
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Estimate the work done in stretching the rubber band for each of these intervals. Give your four estimates in the first line below, separated by commas. Starting in the second line explain how you got your estimates:
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How much work do you estimate is done between the x = 5 cm length and the x = 7 cm length? Give your estimate in the first line, your explanation starting in the second.
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If a trapezoid was constructed by projecting the x = 5 and x = 7 points of the graph, what would be its area, and would this result be an overestimate or an underestimate of the actual area beneath the graph between these two points?
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If a trapezoid was constructed by projecting the x = 5 and x = 8 points of the graph, what would be its area, and would this result be an overestimate or an underestimate of the actual area beneath the graph between these two points?
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University Physics Students Only:
Note that these questions are easy to answer, if you understand what to do. Understanding what to do is fairly challenging at this point. I expect that some will get this, and some will not, and I can't predict who will fall into which category. Of course I'd love it if you would make it easy on me, and everyone would get these:
Suppose the function for the tension is T(L) = (L-6.5)^3*.15 + .5 + (L-5)*.2. How much work is done between length L = 5 cm and L = 7 cm, according to this function?
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How much work is done between lengths L = L_1 and L = L_2? Apply your expression to the work done over each of the following intervals:
· L = 5 cm to L = 6 cm
· L = 6 cm to L = 7 cm
· L = 5.5 cm to L = 6 cm
· L = 5.5 cm to L = 7 cm
Give your four results in the first line below, separated by commas. Starting in the second line explain how you got your results.
Directions of force and displacement vectors matter for work\energy
If force F and displacement `ds are both along the x axis, then what is the sign of F * `ds in each of the following cases:
· F is in the positive direction and `ds is in the positive direction.
· F is in the positive direction and `ds is in the negative direction.
· F is in the negative direction and `ds is in the positive direction.
· F is in the negative direction and `ds is in the negative direction.
Give your answers in the given order, in the first line below, separated by commas. Starting in the second line explain how you got your results.
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If the x axis in the car-and-magnet experiment was your meter stick, then what were the directions of F and `dx, in each of the following situations:
· F is the force exerted on the 'car magnet' by the 'fixed magnet', `dx is the displacement of the car for 10 cm after its release.
· F is the force exerted on the 'car magnet' by the 'fixed magnet', `dx is the displacement of the car as you move it toward the last 10 cm toward the 'fixed magnet'.
· F is the force exerted by the 'car magnet' on the 'fixed magnet', `dx is the displacement of the car for 10 cm after its release.
· F is the force exerted by the 'car magnet' by the 'fixed magnet', `dx is the displacement of the car as you move it toward the last 10 cm toward the 'fixed magnet'.
· F is the force exerted on the car by friction, `dx is the displacement of the car after being released.
· F is the force exerted by the car against friction, `dx is the displacement of the car after being released.
Give your six answers in the first line below, separated by commas. Starting in the second line explain how you reasoned out your answers.
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The figure below represents two plausible force vs. separation models for two of the ceramic magnets used in the experiment. Force is in Newtons while separation is in centimeters.
If the two magnets were touching before release, which model appears to predict the greater kinetic energy at the x = 2 cm position, and which at the x = 9 cm position?
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Good overall. You're on the right track with the questions you answered, though you have a step or two to go; see my notes
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