#$&*
Phy 121
Your 'energy conversion 1' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** Energy Conversion 1_labelMessages **
** **
Note that the data program is in a continual state of revision and should be downloaded with every lab.
Most students report completion times between 2 and 3 hours, with some as short as 1 hour and some as long as 5 hours.
For part of this experiment you will use the calibrated rubber band you used in the preceding experiment 'Force vs. Displacement 1', as well as the results you noted for that experiment.
For this experiment you will need to use at least one rubber band in such a way as to make it useless for subsequent experiments. DO NOT USE ONE OF YOUR CALIBRATED RUBBER BANDS. Also note that you will use four of the thin rubber bands in a subsequent experiment, so DO NOT USE THOSE RUBBER BANDS HERE.
If your kit has extra rubber bands in addition to these, you may use one of them.
You are going to use the rubber band to bind three of your dominoes into a block. If you don't have extra rubber bands, you could use some of the thread that came with your kit, but rubber bands are easier to use.
The idea of binding the dominoes is very simple. Just set one domino on a tabletop so that it lies on one of its long edges. Then set another right next to it, so the faces of the two dominoes (the flat sides with the dots) are touching. Set a third domino in the same way, so you have a 'block' of three dominoes.
Bind the three dominoes together into a 'block' using a rubber band or several loops of thread, wrapping horizontally around the middle of the 'block', oriented in such a way that the block remains in contact with the table. The figure below shows three dominoes bound in this manner, resting on a tabletop.
Now place a piece of paper flat on the table, and place the block on the paper, with the block at one end of the paper.
Give the block a little push, hard enough that it slides about half the length of the paper.
Give it a harder push, so that it slides about the length of the paper, but not quite.
Give it a push that's hard enough to send it past the other end of the paper.
You might need to slide the block a little further than the length of one sheet, so add a second sheet of paper:
Place another piece of paper end-to-end with your first sheet.
Tuck the edge of one sheet slightly under the other, so that if the block slides across the first sheet it can slide smoothly onto the second.
You are going to use a calibrated rubber band to accelerate the blocks and make them slide across the table.
Tie two pieces of thread through to the rubber bands holding the blocks, at the two ends of the block, so that if you wanted you could pull the block along with the threads. One thread should be a couple feet long--long enough that if the block is at one edge of one paper, the other end of the thread extends beyond the edge of the other paper. The other thread needs to be only long enough that you can grasp it and pull the block back against a small resistance.
At the free end of the longer thread, tie a hook made from a paper clip.
Use the rubber band you used in the preceding experiment (the 'first rubber band' from your kit, the one for which you obtained the average force * distance results). Hook that rubber band to the hook at the free end of the longer thread.
Make another hook, and put it through the other end of the rubber band loop, so that when you pull on this hook the rubber band stretches slightly, the string becomes taut and the block slides across the tabletop.
You will need something to which to attach the last hook:
Now place on the tabletop some object, heavy enough and of appropriate shape, so that the last hook can in one way or another be fixed to that object, and the object is heavy enough to remain in place if the rubber band is stretched within its limits. That is, the object should be able so remain stationary if a few Newtons of force is applied. Any rigid object weighing, or being weighted by, about 5-10 pounds ought to be sufficient.
Your goal is to end up with a moderately massive object, to which the last hook is tied or attached, with the rubber band extending from the hook to another hook, a thread from that hook to the block (with a shorter thread trailing from the other end of the block)
With a slight tension in the system the block should be a few centimeters from the 'far' edge of the paper which is furthest from the massive object.
If the block is pulled back a little ways (not so much that the rubber band exceeds its maximum tolerated length) the rubber band will stretch but the last hook will remain in place, and if the block is then released the rubber band will snap back and pull the block across the tabletop until the rubber band goes slack and the block then coasts to rest.
The figure below shows the block resting on the paper, with the thread running from a hook to the rubber band at the far end, which is in turn hooked to the base of a flatscreen monitor.
At the far end the rubber band is ready to be stretched between two hooks. A measuring device is shown next to the rubber band; to get accurate measurements of rubber band length it is recommended that a piece of paper be placed beneath the rubber band, and two points carefully marked on the paper to indicate the positions of the ends. The separation of the points can later be measured. Alternatively the two points can be marked in advance at the desired separation and the system stretched accordingly.
Consult your previous results and determine the rubber band length required to support the weight of two dominoes. Pulling by the shorter piece of thread (the 'tail' of thread), pull the block back until the rubber band reaches this length, and on the paper mark the position of the center of the block (there might well be a mark at the center of the domino; if not, make one, being sure it is within 1 millimeter of the center, and mark the paper according to this mark). Release the thread and see whether or not the block moves. If it does, mark the position where it comes to rest as follows:
Make a mark on the paper where the center mark comes to rest by drawing a short line segment, perhaps 3 mm long, starting from the center mark and running perpendicular to the length of the block.
Make another mark about twice the length of the first, along the edge of the block centered at the center mark.
This will result in a mark that looks something like the following, with the longer line indicating the direction of the block and the two lines coming together at the center mark: __|__. In the first figure below the lowest two marks represent the positions of the center of the dominoes at initial point and at the pullback point. The mark next to the domino is the horizontal part of a mark that looks something like |- ; the vertical part of that mark is obscured by the blocks, and the mark it also tilted a bit to coincide with the slightly rotated orientation of the block. In the second figure most of the |- mark can be seen.
You will make a similar mark for the final position for each trial of the experiment, and from these marks you will later be able to tell where the center mark ended up for each trial, and the approximate orientation of the block at the end of each trial.
Based on this first mark, how far, in cm, did the block travel after being released, and through approximately how many degrees did it rotate before coming to rest?
If the block didn't move, your answers to both of these questions will be 0.
Answer in comma-delimited format in the first line below. Give a brief explanation of the meaning of your numbers starting in the second line.
Your answer (start in the next line):
8.3 cm, 18 degrees
This is how far the block of dominoes moved and its angle of rotation.
#$&* _ 2 rb tension how far and thru what angle
Tape the paper to the tabletop, or otherwise ensure that it doesn't move during subsequent trials.
Repeat the previous instruction until you have completed five trials with the rubber band at same length as before.
Report your results in the same format as before, in 5 lines. Starting in the sixth line give a brief description of the meaning of your numbers and how they were obtained:
Your answer (start in the next line):
8.5 cm, 18 degrees
9.4 cm, 19 degrees
9.5 cm, 20 degrees
10.1 cm, 24 degrees
8.8 cm, 19 degrees
This is the distance the block traveled and its angle of rotation for each trial.
#$&* _ trials on paper
Now, without making any marks, pull back a bit further and release.
Make sure the length of the rubber band doesn't exceed its original length by more than 30%, with within that restriction what rubber band length will cause the block to slide a total of 5 cm, then 10 cm, then 15 cm.
You don't need to measure anything with great precision, and you don't need to record more than one trial for each sliding distance, but for the trials you record:
The block should rotate as little as possible, through no more than about 30 degrees of total rotation, and
it should slide the whole distance, without skipping or bouncing along.
You can adjust the position of the rubber band that holds the block together, the angle at which you hold the 'tail', etc., to eliminate skipping and bouncing, and keep rotation to a minimum.
Indicate in the first comma-delimited line the rubber band lengths that resulted in 5 cm, 10 cm and 15 cm slides. If some of these distances were not possible within the 30% restriction on the stretch of the rubber band, indicate this in the second line. Starting in the third line give a brief description of the meaning of these numbers.
Your answer (start in the next line):
8.5, 9.4, 11.0
#$&* _ rb lengths for 5, 10, 15 cm slides
Now record 5 trials, but this time with the rubber band tension equal to that observed (in the preceding experiment) when supporting 4 dominoes. Mark and report only trials in which the block rotated through less than 30 degrees, and in which the block remained in sliding contact with the paper throughout.
Report your distance and rotation in the same format as before, in 5 lines. Briefly describe what your results mean, starting in the sixth line:
Your answer (start in the next line):
10.8, 1
10.6, 1
11.2, 1
9.7, 2
11.0, 0
These are the distances and angles of rotation of a block of four dominoes.
#$&* _ 5 trials 4 domino length
Repeat with the rubber band tension equal to that observed when supporting 6 dominoes and report in the same format below, with a brief description starting in the sixth line:
Your answer (start in the next line):
12.8, 0
12.6, 0
13.2, 0
11.5, 1
12.7, 0
These are the distances and rotation of a block of six dominoes.
#$&* _ 5 trials for 6 domino length
Repeat with the rubber band tension equal to that observed when supporting 8 dominoes and report in the same format below, including a brief description starting in the sixth line:
Your answer (start in the next line):
13.3, 0
13.1, 0
12.8, 0
13.4, 0
13.2, 0
These are the distances and rotations of a block with eight dominoes.
#$&* _ 5 trials for 8 domino length
Repeat with the rubber band tension equal to that observed when supporting 10 dominoes and report in the same format below, including your brief description as before:
Your answer (start in the next line):
13.9, 0
13.5, 0
14.0, 0
13.7, 0
13.0, 0
These are the distances and rotations for a block of ten dominoes.
#$&* _ 5 trials for 10 domino length
In the preceding experiment you calculated the energy associated with each of the stretches used in this experiment.
The question we wish to answer here is how that energy is related to the resulting sliding distance.
For each set of 5 trials, find the mean and standard deviation of the 5 distances. You may use the data analysis program or any other means you might prefer.
In the space below, report in five comma-delimited lines, one for each set of trials, the length of the rubber band, the number of dominoes supported at this length, the mean and the standard deviation of the sliding distance in cm, and the energy associated with the stretch.
You might choose to report energy here in Joules, in ergs, in Newton * cm or in Newton * mm. Any of these choices is acceptable.
Starting in the sixth line specify the units of your reported energy and a brief description of how your results were obtained. Include your detailed calculations and specific explanation for the third interval. Be sure to give a good description of how you obtained the energy associated with each stretch:
Your answer (start in the next line):
2 dominoes: 9.3, 9.26, 0.63, 2.4
4 dominoes: 9.2, 10.66, 0.58, 1.9
6 dominoes: 9.0, 12.56, 0.63, 2.5
8 dominoes: 9.0, 13.16, 0.23, 3.5
10 dominoes: 9.0, 13.62, 0.39, 3.8
The units are joules. I used the data program to get the means and standard deviations.
#$&* _ for each set of trials length, # dom, mean, std of sliding dist, energy _ describe how results obtained esp energy calculations
Sketch a graph of sliding distance vs. energy, as reported in the preceding space .
Fit the best possible straight line to your graph, and give in the first comma-delimited line the slope and vertical intercept of your line.
In the second line specify the units of the slope and the vertical intercept.
Starting in the third line describe how closely your data points cluster about the line, and whether the data points seem to indicate a straight-line relationship or whether they appear to indicate some sort of curvature.
If curvature is indicated, describe whether the curvature appears to indicate upward concavity (for this increasing graph, increasing at an increasing rate) or downward concavity (for this increasing graph, increasing at a decreasing rate).
Your answer (start in the next line):
0.65, 11.15
J/cm; cm
My line are somewhat off of the line, indicating curvature at an increasing rate.
#$&* _ sliding dist vs. energy slope, vert intercept of st line, how close to line, describe curvature if any
Now repeat the entire procedure and analysis, but add a second rubber band to the system, in series with the first.
For each trial, stretch until the first rubber band is at the length corresponding to the specified number of dominoes, then measure the second rubber band and record this length with your results.
When graphing mean sliding distance vs. energy, assume for now that the second rubber band contributes an amount of energy equal to that of the first. You will therefore use double the energy you did previously.
When you have completed the entire procedure report your results in the space es below, as indicated:
Report in comma-delimited format the length of the first rubber band when supporting the specified number of dominoes, and the length you measured in this experiment for second band. You will have a pair of lengths corresponding to two dominoes, four dominoes, ..., ten dominoes. Report in 5 lines:
Your answer (start in the next line):
11.1, 10.7
11.6, 10.7
11.9, 11.5
12.3, 11.2
12.8, 11.7
#$&* _ lengths of 1st and 2d rbs in series each of 5 trials
Report for each set of 5 trials your mean sliding distance and the corresponding standard deviation; you did five sets of 5 trials so you will report five lines of data, with two numbers in each line:
Your answer (start in the next line):
7.5, 0.53
13.8, 1.35
30.1, 7.4
42.4, 6.51
#$&* _ sliding dist and std dev each tension
Give the information from your graph:
Give in the first comma-delimited line the slope and vertical intercept of your line.
In the second line specify the units of the slope and the vertical intercept.
Starting in the third line describe how closely your data points cluster about the line, and whether the data points seem to indicate a straight-line relationship or whether they appear to indicate some sort of curvature.
If curvature is indicated, describe whether the curvature appears to indicate upward concavity (for this increasing graph, increasing at an increasing rate) or downward concavity (for this increasing graph, increasing at a decreasing rate).
Your answer (start in the next line):
The graph seems to curve upward at an increasing rate, but my points are somewhat scattered towards the right side of the graph.
#$&* _ slope, vert intercept, describe curvature
In the space below, report in the first line, in comma-delimited format, the sliding distance with 1 rubber band under 2-domino tension, then the sliding distance with 2 rubber bands under the same 2-domino tension.
Then in the subsequent lines report the same information for 4-, 6-, 8- and 10-domino tensions.
You will have five lines with two numbers in each line:
Your answer (start in the next line):
#$&* _ 5 lines comparing 1 rb to 2 rb trials
Your preceding answers constitute a table of 2-rubber-band sliding distances vs. 1-rubber-band sliding distances.
Sketch a graph of this information, fit a straight line and determine its y-intercept, its slope, and other characteristics as specified:
Give in the first comma-delimited line the slope and vertical intercept of your line.
In the second line specify the units of the slope and the vertical intercept.
Starting in the third line describe how closely your data points cluster about the line, and whether the data points seem to indicate a straight-line relationship or whether they appear to indicate some sort of curvature.
If curvature is indicated, describe whether the curvature appears to indicate upward concavity (for this increasing graph, increasing at an increasing rate) or downward concavity (for this increasing graph, increasing at a decreasing rate).
#$&* _ graph 2 rb dist vs 1 rb dist _ slope and intercept _ describe any curvature
To what extent do you believe this experiment supports the following hypotheses:
The sliding distance is directly proportional to the amount of energy required to stretch the rubber band. If two rubber bands are used the sliding distance is determined by the total amount of energy required to stretch them.
Your answer (start in the next line):
I do not believe this hypothesis is fully supported by this experiment since the distance did not double using two rubber bands.
#$&* _to what extend is hypothesis of sliding dist prop stretching energy supported _ to what extent for 2 rb
Your instructor is trying to gauge the typical time spent by students on these experiments. Please answer the following question as accurately as you can, understanding that your answer will be used only for the stated purpose and has no bearing on your grades:
Approximately how long did it take you to complete this experiment?
Your answer (start in the next line):
2 hours
:
#$&*
*#&!
&#Your work on this lab exercise looks good. Let me know if you have any questions. &#
Phy 121
Your 'conservation of energy on an incline' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** Conservation of Energy on an Incline_labelMessages **
** **
This short version is currently a working draft. Please ask for clarification when that is necessary. Modifications will be made to this document in response to questions, which will also be answered in the usual manner.
Note that this is an alternative short version of the experiment intended for
Principles of Physics students
General College Physics students whose lab goal in only a passing lab average
This is not intended for University Physics students.
See also the short video at
http://www.vhcc.edu/dsmith/genInfo/qa_query_etc/EnergyConIncline_DialUp200.wmv
which demonstrates a finger delivering a quick impulse to the ball, which coasts to a stop as it travels up the ramp, then coasts with increasing speed down the ramp. The finger 'pokes' the ball at the very beginning of the video.
Note: You may assume for the purposes of this writeup that the ball has a mass of 40 grams. This probably isn't accurate, but it will serve the purpose of the experiment.
Report a preliminary run of the experiment
I suggest that before actually running the experiment you read through the instructions, set up the system, get a few preliminary timings and submit them with a brief description of what you did. I'll be able to tell you if your results make sense, and might make a suggestion or two. No need to do any calculations, and no need for a detailed description. 15 minutes should do it for the preliminary observations. I'll also be glad to clarify anything you think requires clarification.
Goal of the experiment
The experiment concerns a ball which coasts up a ramp, stops, and coasts back down. We are trying to detect the difference between the magnitude of the acceleration going up and the acceleration going down. It is suggested that to get an initial 'feel' for the system you take the
ball, the ramp and a domino, set the ramp up with a fairly small slope, and use your finger to 'bump' the ball in this manner. Again, the ball
just has to go up and come back. This will give you a point of reference for further instructions.
Basic instruction
The basic instruction is this:
There is a difference between the time required for a ball, given an initial velocity at the bottom of the ramp, to roll up a ramp and the
time required for it to roll back down to its initial position.
This is easiest to detect on a ramp whose slope is just enough that the ball will 'turn around' and roll back down.
It is necessary to start the ball with an initial impulse in the form of a 'bump' from your finger or from an object, as opposed to a prolonged push. This is because the presumably uniform acceleration up the ramp does not begin until the ball loses contact with the source of the impulse.
You can set the system up with too much slope, which will make the difference in time up and time down undetectable, or with too little slope, which will not result in uniform acceleration. It is up to you to determine the optimal slope, but if your time intervals are less than a couple of seconds you won't be able to time them with sufficient accuracy.
You also need to 'bump' the ball hard enough, but not too hard. You won't always get it right; simply disregard the trials that don't result in sufficient time intervals.
If the 'bump' isn't strong enough the ball won't go very far and you'll be timing a short interval.
If the 'bump' is too strong the ball will keep going right off the high end of the ramp.
More specific instructions
More specifically:
Too little slope causes problems:
If the ramp has too little slope it will be difficult to give the ball an initial impulse that causes it to travel most of the length of the
ramp without rolling off the 'high' end.
If you don't have quite enough ramp slope, once the ball comes to rest it might simply stay at rest.
Also, if there is too little slope, the small irregularities in the track interfere with the uniformity of the acceleration and throw off the results.
Too much slope causes problems:
There are unavoidable uncertainties in timing. If the slope is too great the time intervals will be short, and the resulting percent uncertainty will be too high to make an accurate distinction between the 'up' and 'down' intervals.
A ramp with a rise of a single domino is probably steeper than necessary. It is suggested that you use coins or shims. You won't need an accurate measurement of the slope.
Suggestions for experimental technique:
You want to start the ball rolling up the ramp, using a sudden impulse rather than a sustained push, giving it enough velocity that it travels
20 cm or more before coming to rest for an instant and then traveling back down.
At the instant of the 'bump' you need to start the TIMER,
and you need to operate the TIMER in such a way as to determine as accurately as possible the time up the ramp and the time back down.
You also need to observe, with reasonable accuracy, the point at which the ball comes to rest before rolling back down.
In the event of a mis-strike (e.g., too hard or too easy) it's simple enough to try again. After a couple of minutes' practice it's not difficult to do this at least well enough to get a good trial every minute or so.
To get good results it's important to avoid backspin and overspin; if the ball is struck just a little higher than the middle, the ball will start out with about the right amount of spin.
You have a good trial when you have data that allows you to determine the acceleration of the ball up the ramp, and back down.
You need at least half a dozen good trials.
You therefore need to get a reasonable number of trials, timing the ball from 'strike' to 'turnaround' then back to the original position. It's
important to try to detect and eliminate or correct for systematic errors in timing.
The goal is to try to detect the difference in acceleration between the ball as it travels up the incline and as it travels down. It is assumed
that this difference is independent of how far the ball travels, and also independent of the slope, as long as the slope is small.
It's up to you to find a slope that yields good results. As outlined above, too much slope is counterproductive, as is too little.
Analysis of data and interpretation
First give a synopsis of your setup and all relevant data.
I set up the ramp with a quarter under the right end and two quarters under the left end. The ramp was 30 cm long, and I used a large marble.
Motion Time up (seconds) Time down (seconds) Distance (cm) Acceleration (g, up, down)
Left to right 3.84 5.32 30 3.92 N, 0.25 N, 0.48 N
Right to left 3.28 2.73 30 3.92 N, 0.18 N, 0.12 N
Left to right 3.74 3.52 19 3.92 N, 0.37 N, 0.33 N
Right to left 3.59 4.48 28 3.92 N, 0.23 N, 0.36 N
Left to right 4.02 4.49 30 3.92 N, 0.27 N, 0.34 N
Right to left 4.60 3.14 10 3.92 N, 1.06 N, 0.49 N
@&
Your data are clear. However you need to show how you obtained your calculated results.
*@
Then, for each trial, determine the acceleration of the ball as it travels up the ramp, and as it travels down the ramp. Show, using a couple of representative sample calculations, how your results were obtained from your data.
I used the third uniform acceleration equation to find accelerations.
The acceleration of the ball results mainly from two forces, one being the component of the gravitational force parallel to the incline, the other the force of rolling friction between the ball and the ramp. The difference in the accelerations is due to two facts:
The gravitational component is in the direction opposite the ball's velocity during one phase of the motion, and in the same direction during the other phase.
The frictional force is in the direction opposite the ball's velocity.
In the absence of friction the only force would be that of the gravitational component parallel to the incline, which is the same for motion up the incline as for motion down the incline.
How then does rolling friction result in a difference in the two accelerations?
Friction would cause acceleration to be less because it would hinder the ball’s velocity.
According to your data, what is the magnitude of the acceleration due to the frictional force on the rolling ball, and what is this acceleration as a percent of the acceleration of gravity?
More appropriately, you might choose to give upper and lower bounds for the magnitude of the acceleration due to friction (e.g., the magnitude of the acceleration is at least ___, and at most ___).
Energy conservation states that `dW_NC_ON = `dKE + `dPE. How do your results illustrate this law?
My results show this because energy was not created nor destroyed. It was just transformed. If the ball went up only 19 centimeters, then it would only roll down the same distance. The ball could only travel as far as its energy would allow. Since it was moving down and up the ramp, kinetic and potential energy were both involved as it moved and stopped.
More detailed questions about interpretation
If you have answered the above questions with some degree of confidence you don't need to answer the following at this point. Depending on the answers you submit I might ask you to look at these questions.
You don't need to do any more calculations, except perhaps a couple of additions or subtractions, but answer the following. Don't spend hours thinking through your answers--just think about what is going on with this system and give the best answers you can in, say, 30 minutes. Many of the answers are pretty obvious. When I get your answers I'll be able, if necessary, to help clarify some of the more difficult points.
What forces act on the ball as it rolls freely up or down the ramp? List the forces.
Gravity and friction
Which of these listed forces are identical both for motion up the ramp and for motion down?
Gravity
Which of the listed forces change as the ball reverses direction?
Friction
How does your answer to these questions help explain why the magnitudes of the two accelerations should be different?
Since friction acts in opposition of the ball, it would create a hindrance on the acceleration.
What do you think the acceleration of the system would be in the absence of frictional forces?
Equal to the acceleration with only gravity
Don't actually calculate any of the quantities in the subsequent questions related to work and energy:
What happens to the gravitational PE of the ball as it goes up the incline, and what happens as it goes down the incline?
PE would increase as it goes up the incline and decrease as it goes down the incline.
What happens to the KE of the ball as it goes up the incline, and as it goes down?
KE would increase as it goes down the incline and decrease as it goes up the incline.
How does the PE change up the ramp compare to the PE change down the ramp?
PE would be converted into KE as the ball rolls down the incline.
As it rolls up the ramp, how does the PE change of the ball compare with the KE change?
PE increases as KE decreases.
As it rolls down the ramp, how does the PE change of the ball compare with the KE change?
PE decreases as KE increases.
In which case is the magnitude of the ratio of PE change to KE change greater, and why?
When the ball is going up the incline.
Do any nonconservative forces act along the line of the ball's motion as it rolls up the ramp, and as it rolls down the ramp?
How does the action of the nonconservative forces explain the answers to some of these questions?
Between the end of the 'bump' and the ball's return to the same position
Does its KE increase, decrease or remain the same?
Does it PE increase, decrease or remain the same?
Do nonconservative forces do positive work on the ball, negative work or no work?
In what ways do the results of this experiment and support the conservation-of-energy equation `dW_NC_ON = `dKE + `dPE?
@&
&#Please see my notes and submit a copy of this document with revisions, comments and/or questions, and mark your insertions with &&&& (please mark each insertion at the beginning and at the end).
Be sure to include the entire document, including my notes. &#
*@
course Phy 121
015. Impulse-Momentum
*********************************************
Question: `q001. Note that this assignment contains 5 questions.
. Suppose that a net force of 10 Newtons acts on a 2 kg mass for 3 seconds. By how much will the velocity of the mass change during these three seconds?
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
To find a:
Fnet = m * a
a = Fnet / m
= 10 Newtons / 2 kg = 5 m/s^2
To find `dv:
a = `dv / `dt
`dv = a * `dt = 5 m/s^2 * 3s = 15 m/s
confidence rating #$&*:#8232;
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution:
The acceleration of the object will be
accel = net force / mass = 10 Newtons / (2 kg) = 5 m/s^2.
In 3 seconds this implies a change of velocity
`dv = 5 m/s^2 * 3 s = 15 meters/second.
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&
Self-critique (if necessary): ok
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Self-critique rating: ok
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Question: `q002. By how much did the quantity m * v change during these three seconds?
What is the product Fnet * `dt of the net force and the time interval during which it acted?
How do these two quantities compare?
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
Fnet * `dt = 10 Newtons * 3 s = 30 kg m/s
m * v = 2kg * 15 m/s = 30 kg m/s
They changed the same.
confidence rating #$&*:#8232;
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution:
Since m remained constant at 2 kg and v changed by `dv = 15 meters/second, it follows that m * v changed by 2 kg * 15 meters/second = 30 kg meters/second.
Fnet *`dt is 10 Newtons * 3 seconds = 30 Newton * seconds = 30 kg meters/second^2 * seconds = 30 kg meters/second.
The two quantities m * `dv and Fnet * `dt are identical.
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Self-critique (if necessary): ok
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Self-critique rating: ok
*********************************************
Question: `q003. The quantity m * v is called the momentum of the object.
The quantity Fnet * `dt is called the impulse of the net force.
The Impulse-Momentum Theorem states that the change in the momentum of an object during a time interval `dt must be equal to the impulse of the average net force during that time interval. Note that it is possible for an impulse to be delivered to a changing mass, so that the change in momentum is not always simply m * `dv; however in non-calculus-based physics courses the effective changing mass will not be considered.
If an average net force of 2000 N is applied to a 1200 kg vehicle for 1.5 seconds, what will be the impulse of the force?
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Your solution:
Impulse = Fnet * `dt
= 2000 N * 1.5 s = 3000 kg m/s
confidence rating #$&*:#8232;
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Given Solution:
The impulse of the force will be Fnet * `dt = 2000 Newtons * 1.5 seconds = 3000 Newton*seconds = 3000 kg meters/second. Note that the 1200 kg mass has nothing to do with the magnitude of the impulse.STUDENT COMMENT: That's a little confusing. Would it work to take the answer I got of 3234 N and add back in the weight of the person at 647 N to get 3881?
INSTRUCTOR RESPONSE: Not a good idea, though it works in this case.
Net force = mass * acceleration.
That's where you need to start with problems of this nature.Then write an expression for the net force, which will typically include but not be limited to the force you are looking for. *&*&
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Self-critique (if necessary): ok
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Self-critique rating: ok
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Question: `q004. If an average net force of 2000 N is applied to a 1200 kg vehicle for 1.5 seconds, what will be change in the velocity of the vehicle?
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Your solution:
a = Fnet / m = 2000 Newtons / 1200 kg = 1.7 m/s^2
1.7 m/s^2 * 1.5 s = 2.55 m/s
confidence rating #$&*:#8232;
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Given Solution:
The impulse of the 2000 Newton force is equal to the change in the momentum of the vehicle. The impulse is
impulse = Fnet * `dt = 2000 Newtons * 1.5 seconds = 3000 Newton*seconds = 3000 kg meters/second.
The change in momentum is m * `dv = 1200 kg * `dv.
Thus
1200 kg * `dv = 3000 kg m/s, so
`dv = 3000 kg m/s / (1200 kg) = 2.5 m/s.
In symbols we have Fnet * `dt = m `dv so that
`dv = Fnet * `dt / m.
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Self-critique (if necessary):
I derived the answer from Newtons 2nd by finding the acceleration and then multiplying that by the time interval. Since F = m a and a = `dv / `dt, then we have the same information both ways, so I'm ok? The given solution mentions an impulse and as of now I have that tied up tight to circular motion(pendulum) for some reason.
@&
Impulse is force * time interval.
Sort of analogous to work, which is force * displacement.
Work by the net force results in a change in KE.
Impulse by the net force results in change in momentum.
*@
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Self-critique rating: ok
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Question: `q005. Use the Impulse-Momentum Theorem to determine the average force required to change the velocity of a 1600 kg vehicle from 20 m/s to 25 m/s in 2 seconds.
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Your solution:
momentum = m *`dv
impulse = Fnet * `dt
m * `dv = Fnet * `dt
1600 kg * 5 m/s = Fnet * 2 s
Fnet = 4000 Newtons
confidence rating #$&*:#8232;
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Given Solution:
The vehicle changes velocity by 5 meters/second so the change in its momentum is m * `dv = 1600 kg * 5 meters/second = 8000 kg meters/second. This change in momentum is equal to the impulse Fnet * `dt, so
Fnet * 2 sec = 8000 kg meters/second and so
Fnet = 8000 kg meters/second / (2 seconds) = 4000 kg meters/second^2 = 4000 Newtons.
In symbols we have Fnet * `dt = m * `dv so that Fnet = m * `dv / `dt = 1600 kg * 5 m/s / ( 2 s) = 4000 kg m/s^2 = 4000 N.
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Self-critique (if necessary): ok
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Self-critique rating: ok
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&#Your work looks good. See my notes. Let me know if you have any questions. &#