#$&*
course Phy 241
September 17 around 1:00pm.
For the car and paperclipsLet 1 unit of force correspond to the weight of 1 small paperclip. On this scale the weight of a large paperclip is 4 units of force.
If the paperclip is on the car, its weight is balanced by the upward force exerted by the table and it has no direct effect on the car's acceleration. If it is suspended, then its weight contributes to the accelerating force.
You should have obtained a count and a distance from rest for each trial, and on each trial there will be some number of force units suspended from the thread (1 unit for every small, 3 units for every large clip). Report in the first line the number of clips, the count and the distance from rest, separated by commas, for your first trial. Report subsequent trials in subsequent lines. After reporting the data for all your trials, give a brief explanation of your setup and how the trials were conducted. Include also the information about how many of your counts take how many seconds.
(TOTAL # of paperclips: 7 big/3 small)
(#on string)
1 big/1 little, no movement, 0cm
2 big/1 little, no movement, 0cm
3 big/1 little, no movement, 0cm
4 big/1 little, 8.6 seconds, 50 cm
5 big/1 little, 5.8 seconds, 74 cm
6 big/1 little, 3.8 seconds, 60 cm
Our string was 1 meter (100cm) in length. We then tied the string to the rubber bands on the car and then hooked paper clips, which we took off our car, to the end of the string. We recorded each trial, according to how far it went with account of the weight that was pulling it. The last trial we did, we clocked it to a certain position due to it hitting the meter stick, which almost made the whole thing stop.
#$&*
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.
4 force units, 0cm/s^2
7 force units, 0cm/s^2
10 force units, 0cm/s^2
13 force units, 1.35 cm/s^2
16 force units, 4.4 cm/s^2
19 force units, 8.3 cm/s^2
I found my force units by doing 3(# of big)+1(# of little). I found my accelerations by plugging what I know into the equation “’ds=v0’dt+ ½ a’dt^2” and solved for “a.” Since my v0 was 0, the “v0’dt” in the equation cancelled out and therefore made the equation very simple to solve.
#$&*
Sketch a graph of acceleration vs. number of force units, and describe your graph. Fit a straight line to your graph and determine its slope. Describe how the trend of your data either indicates a good straight-line fit, or how it deviates from a straight line.
My graph first starts out as a straight line upward, then starts curving out (in the shape of a lowercase “r”). I determined its slope to be approx. 0.86. Since my accelerations in the first 3 trials are 0cm/s^2, this is what makes the straight vertical line on the y-axis. As my acceleration exceeds, so does my force units. So, the graph is somewhat an increasing graph at a decreasing rate.
for accel vs. force the accel is on the vertical axis, so those points would be on the horizontal axis
#$&*
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:
0cm, 3cm, 47cm
0cm, 5cm, 42cm
0cm, 6cm, 29cm
0cm, 7cm, 26cm
0cm, 10cm, 17cm
0cm, 15cm, 18cm
0cm, 20cm, 21cm
#$&*
Give a brief explanation of what the data mean, including a statement of the units of the numbers:
According to the outcome of my data, this clearly shows that the closer you put the magnet on the car to the fixed magnet, the total distance traveled is going to be greater. The further you hold it away from the fixed magnet, the less it’s going to travel.
#$&*
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.
3cm, 44cm
5cm, 37cm
6cm, 23cm
7cm, 19cm
10cm, 7cm
15cm, 3cm
20cm, 1cm
#$&*
Sketch a graph of distance traveled vs. initial separation. Describe your graph.
It has a negative slope, and it is decreasing at a pretty steady rate.
#$&*
Sketch the smooth curve you think best represents the actual behavior of distance traveled vs. initial separation for this system.
#$&*
Identify the point where initial separation is 8 cm.
* What coasting distance corresponds to this point?
* What is the slope of your smooth curve in the neighborhood of this point?
Give the coasting distance as a number in the first line, the slope of the graph as a number in the second line. Starting in the third line give the units of your quantities and explain what each quantity means, and how you obtained it.
13cm
-4 (slope)
I got the coasting distance 13cm, by using my own data and seeing that at 7cm it was 19cm, and at 10cm it was 7cm. Therefore, 8cm had to be between 8 and 18. 13 is the median. I got my slope by taking (10, 7) and (7, 19) and finding the slope between those two points, which 8cm will lie on. This also shows that it goes down -4 cm for every cm further from the magnet.
#$&*
Repeat for the point where initial separation is 5 cm.
37cm
#$&*
According to your graph, if the initial separation is doubled, what happens to the distance the car travels?
The distance the car travels is going to be a lot less. It will eventually get to be 0cm, according to my data.
#$&*
According to your graph, if the initial separation is doubled, what happens to the slope of the graph?
The slope is not going to be as “steep.” It will almost come out looking more like a horizontal line.
#$&*
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, the work in milliNewtons, 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.
3, -0.044, -4.4
5, -0.037, -3.7
6, -0.023, -2.3
7, -0.019, -1.9
10, -0.007, -0.7
15, -0.003, -0.3
20, -0.001, -0.1
I did these calculations by knowing the equation for WORK. I first, converted my “cm” into “m” for each distance. Then I plugged my numbers into the work formula. For example, on the first one:
W=F(change in position)
=0.1 N*0.44m
=0.044N*m
For milliNewtons:
=10mN*0.44m
=4.4 mN*m
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.
Since friction does negative work on the traveling car, it reduces its KE. On the other hand, we saw that the closer the magnets were, the greater the KE was gained by the car. The initial magnet distances we took had PE, but the closer they were the more. The initial PE was then “transformed” into the initial KE done, which friction had a hold of and did work that was equal and opposite of its original KE.
#$&*
... impulse ...
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.
... to cheerios ... add to table fraction of cheerio, then correct for my 15% efficiency
Now 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.
-1.3 milliNewtons
What I did was take the distance I found for 8cm, in the problems before this, which was 13cm and converted that to meters. I then used the work formula again to find its work in Newtons and milliNewtons. Since friction does negative work on the car, my answer is negative.
10 mN*0.13m=1.3 mN but NEGATIVE!
That's 1.3 mN * m, not 1.3 mN. You were mislead by my omission in the question. I should have asked for work in Newton * meters and milliNewtons * m.
The Newton is the standard unit of force, the Newton * meter the standard unit of work.
#$&*
Repeat for initial separation 5 cm.
- 3.7 milliNewtons
should be milliNewton * m
#$&*
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?
2.4 milliNewtons
I just subtracted 1.3mN by 3.7mN to get 2.4mN. Is this right?!
right, but should be milliNewton * m
#$&*
How much work was therefore done by the magnetic force between the 5 cm and 8 cm separation?
Approx. -2.5 milliNewtons
(-1.3mN) + (-3.7mN)= -5mN
-5mN/2= -2.5 mN
I found the average of the work done by the two distances. Is this right?!
The magnetic force provided the energy to overcome friction. It provided more energy when the car was released from the 5 cm position than from the 8 cm position. The difference, not the average, would be the energy provided between the 5 cm and 8 cm positions.
#$&*
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?
3cm
How can you calculate the average force given the displacement and the work?
I’m thinking you could convert your work to joules, and then divide the work by the distance over which the work was done.
What therefore was the average force?
-0.012 N*m/s^2
I did 0.03m/(-2.5J)= -0.012N*m/s^2
Did I use the right calculations for this?
Looks like your division was upside down.
It isn't necessary to convert units:
If the work was 2.5 mN * m, then ave force = work / displacement = 2.5 mN * m / (.03 m) = 800 mN, or .8 N.
This is not to say whether 2.5 mN * m is correct in the first place.
#$&*
What therefore is the average rate at which work is being done by the magnets, per unit of separation, between the 5 cm and 8 cm separations?
2.5 mN/cm
What aspect of the graph of work vs. separation is associated with this average rate?
The middle of the curve.
we'll leave this question open for awhile; give it some more thought
#$&*
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?
Since my personal data has 6cm as an initial point, I got -0.023N (-2.3 milliNewtons). However, I’m not sure of what this question is really asking for. Do I multiply (-0.023N) by the separation (0cm to 6cm)? Or do I divide? Or am I using the wrong calculations?!
#$&*
In general how can you use a graph of work vs. separation to this system to find the force exerted by the magnets at a given separation?
By having your intervals plotted on your graph, you can make the conclusion that the less separation between the magnets, the greater the work done. The more distance between the two magnets results in a less work done. You can always plug your points into your calculator and make a plot graph and then find a certain force at a certain separation by plugging it into the table. Or, just by using your graph, you could make a good estimation on what it would be according to what the initial was.
#$&*"
Very good. See my notes. We're going to leave those last few questions open for a little while.
#$&*