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Submitting Assignment: CQ_1_15.1
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The purpose and the process of answering 'seed' questions:
In cloud seeding small crystalline particles (the 'seeds') are scattered throughout a cloud, so that water vapor in the cloud will build up on the 'seed' and eventually fall in the form of rain. These questions are posed without any previous explanation. You are expected to use what you already know, along with common sense, to answer the questions. It is standard practice in many courses to an instructor to give explanations and examples before asking students to answer questions, and you will see plenty of examples and explanations in this course. However the goal here is to first experience and think about a situation. Whether you think correctly or incorrectly, your thinking gets you started on an idea and forms a 'seed' on which understanding can accumulate. You are expected to answer it to the best of your ability, based on what you know at the beginning of this assignment. You are not expected to research this question until after you have submitted your best response. You are not penalized based on whether your answer is 'right' or 'wrong', but you are expected to think as clearly and deeply as you can about the question. You are not, however, expected to spend hours thinking about the question or agonize unduly about your answers. A rule of thumb is to give it up to 20 minutes, half for thinking and half for typing in your answers (maybe a little more for the typing if you don't have good keyboard skills). Your answers should consist of your best attempt at a solution, and/or one or more questions about the situation. If you think you know the answer or can make a reasonable attempt to answer, then give your answer along with a concise outline of your reasoning. If you aren't sure what the question is asking, make your best attempt to interpret and answer it, and consider including one or more questions. If you are very sure you don't know what the question is asking, then break it down phrase-by-phrase or even word-by-word and explain what you think each key phrase or word might mean. A question consists of a complete but concise statement of what you do and do not understand about the situation. There are two ways you can spend an excessive amount of time explaining your solutions and/or asking questions. One is to type a lot more than what is necessary, and another is to spend a lot of time worrying about what is and is not necessary. Balance the two in the way that works best for you. Remember that the 'concise' part is more for your benefit than mine. I can read a lot more quickly than you can type, and don't mind reading through a lot of words to understand your meaning. You are invited but not required to include comments and/or discussion. You are welcome to use reasonable abbreviations in your work.
In cloud seeding small crystalline particles (the 'seeds') are scattered throughout a cloud, so that water vapor in the cloud will build up on the 'seed' and eventually fall in the form of rain.
These questions are posed without any previous explanation. You are expected to use what you already know, along with common sense, to answer the questions. It is standard practice in many courses to an instructor to give explanations and examples before asking students to answer questions, and you will see plenty of examples and explanations in this course. However the goal here is to first experience and think about a situation. Whether you think correctly or incorrectly, your thinking gets you started on an idea and forms a 'seed' on which understanding can accumulate.
You are expected to answer it to the best of your ability, based on what you know at the beginning of this assignment.
You are not expected to research this question until after you have submitted your best response.
You are not penalized based on whether your answer is 'right' or 'wrong', but you are expected to think as clearly and deeply as you can about the question.
You are not, however, expected to spend hours thinking about the question or agonize unduly about your answers. A rule of thumb is to give it up to 20 minutes, half for thinking and half for typing in your answers (maybe a little more for the typing if you don't have good keyboard skills).
Your answers should consist of your best attempt at a solution, and/or one or more questions about the situation.
If you think you know the answer or can make a reasonable attempt to answer, then give your answer along with a concise outline of your reasoning.
If you aren't sure what the question is asking, make your best attempt to interpret and answer it, and consider including one or more questions.
If you are very sure you don't know what the question is asking, then break it down phrase-by-phrase or even word-by-word and explain what you think each key phrase or word might mean.
A question consists of a complete but concise statement of what you do and do not understand about the situation.
There are two ways you can spend an excessive amount of time explaining your solutions and/or asking questions. One is to type a lot more than what is necessary, and another is to spend a lot of time worrying about what is and is not necessary. Balance the two in the way that works best for you.
Remember that the 'concise' part is more for your benefit than mine. I can read a lot more quickly than you can type, and don't mind reading through a lot of words to understand your meaning.
You are invited but not required to include comments and/or discussion.
You are welcome to use reasonable abbreviations in your work.
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A rubber band begins exerting a tension force when its length is 8 cm. As it is stretched to a length of 10 cm its tension increases with length, more or less steadily, until at the 10 cm length the tension is 3 Newtons. Between the 8 cm and 10 cm length, what are the minimum and maximum tensions? The minimum tension occurs at a length of 8 cm, where the rubber band just begins exerting tension. It hasn't stretched to a length where it will exert significant tension, so the tension at this point is zero. The maximum tension occurs at the 10 cm length and, as indicated, is 3 Newtons. Assuming that the tension in the rubber band is 100% conservative (which is not actually the case) what is its elastic potential energy at the 10 cm length? When stretching the rubber band from the 8 cm length to the 10 cm length the force gradually increases from 0 N to 3 N. For a rubber band the increase isn't exactly linear, so the average force probably isn't exactly halfway between the initial 0 N and final 3 N force, but it's likely to be reasonably close. In any case, our best conjecture for the average tension is 1.5 N. The work done to stretch the rubber band is therefore `dW = F_ave * `ds where F_ave is 1.5 N and `ds is the 2 cm distance through which it is stretched. Thus work to stretch = F_ave * `ds = 1.5 N * 2 cm = 3 N * cm. Standard units of work are Joules (N * m) or ergs (dynes * cm). The 3 N * cm product is 3 N * .01 m = .03 N * m = .03 Joules (or if you prefer 300,000 ergs). If the force is conservative, this will be the elastic potential energy of the rubber band. (Note that rubber band forces aren't completely conservative; we get only about 50-70% of the energy back, with most of the rest dissipated as thermal energy. However for the present problem we'll treat the rubber band as an 'ideal' elastic object, exerting a completely conservative force). If all this potential energy is transferred to the kinetic energy of an initially stationary 20 gram domino, what will be the velocity of the domino? If a 20 gram domino has .03 Joules of kinetic energy, then since its KE is 1/2 m v^2 we have 1/2 m v^2 = KE so v = sqrt( 2 * KE / m) = sqrt( 2 * .03 Joules / (.02 kg) ) = 1.7 m/s^2, approx.. If instead the rubber band is used to 'shoot' the domino straight upward, then how high will it rise? The domino will rise until its gravitational potential energy has increased by .03 Joules. The increase in gravitational potential energy is equal and opposite to the work done by the gravitational force. A mass m experiences a downward gravitational force F = m g, which we call its weight. As the mass rises, gravity acts downward as the object moves upward, so the gravitational force and the displacement are in opposite directions. Gravity therefore does negative work on the mass, so the gravitational potential energy increases. If the mass rises through displacement `dy, the work done on it by gravity is force * displacement = - m g * `dy, so its gravitational PE changes by + m g `dy. The domino therefore rises until m g `dy = .03 Joules, so that m g `dy = .03 Joules `dy = .03 Joules / (m g) = .03 Joules / (.02 kg * 9.8 m/s^2) = .15 meters, approx. Note that it's possible to just set the formula m g h for gravitational potential energy equal to .03 Joules and solve for h. For many problems this will be sufficient, and it's a significant skill. However we gain a lot more insight into the nature of energy by the preceding line of reasoning, which proceeds from the definitions. If time permits, it's a good idea to work to understand that reasoning. University Physics students should certainly work to understand this reasoning.
A rubber band begins exerting a tension force when its length is 8 cm. As it is stretched to a length of 10 cm its tension increases with length, more or less steadily, until at the 10 cm length the tension is 3 Newtons.
Between the 8 cm and 10 cm length, what are the minimum and maximum tensions?
The minimum tension occurs at a length of 8 cm, where the rubber band just begins exerting tension. It hasn't stretched to a length where it will exert significant tension, so the tension at this point is zero. The maximum tension occurs at the 10 cm length and, as indicated, is 3 Newtons.
The minimum tension occurs at a length of 8 cm, where the rubber band just begins exerting tension. It hasn't stretched to a length where it will exert significant tension, so the tension at this point is zero.
The maximum tension occurs at the 10 cm length and, as indicated, is 3 Newtons.
Assuming that the tension in the rubber band is 100% conservative (which is not actually the case) what is its elastic potential energy at the 10 cm length?
When stretching the rubber band from the 8 cm length to the 10 cm length the force gradually increases from 0 N to 3 N. For a rubber band the increase isn't exactly linear, so the average force probably isn't exactly halfway between the initial 0 N and final 3 N force, but it's likely to be reasonably close. In any case, our best conjecture for the average tension is 1.5 N. The work done to stretch the rubber band is therefore `dW = F_ave * `ds where F_ave is 1.5 N and `ds is the 2 cm distance through which it is stretched. Thus work to stretch = F_ave * `ds = 1.5 N * 2 cm = 3 N * cm. Standard units of work are Joules (N * m) or ergs (dynes * cm). The 3 N * cm product is 3 N * .01 m = .03 N * m = .03 Joules (or if you prefer 300,000 ergs). If the force is conservative, this will be the elastic potential energy of the rubber band. (Note that rubber band forces aren't completely conservative; we get only about 50-70% of the energy back, with most of the rest dissipated as thermal energy. However for the present problem we'll treat the rubber band as an 'ideal' elastic object, exerting a completely conservative force).
When stretching the rubber band from the 8 cm length to the 10 cm length the force gradually increases from 0 N to 3 N. For a rubber band the increase isn't exactly linear, so the average force probably isn't exactly halfway between the initial 0 N and final 3 N force, but it's likely to be reasonably close. In any case, our best conjecture for the average tension is 1.5 N.
The work done to stretch the rubber band is therefore
`dW = F_ave * `ds
where F_ave is 1.5 N and `ds is the 2 cm distance through which it is stretched. Thus
work to stretch = F_ave * `ds = 1.5 N * 2 cm = 3 N * cm.
Standard units of work are Joules (N * m) or ergs (dynes * cm). The 3 N * cm product is 3 N * .01 m = .03 N * m = .03 Joules (or if you prefer 300,000 ergs).
If the force is conservative, this will be the elastic potential energy of the rubber band. (Note that rubber band forces aren't completely conservative; we get only about 50-70% of the energy back, with most of the rest dissipated as thermal energy. However for the present problem we'll treat the rubber band as an 'ideal' elastic object, exerting a completely conservative force).
If all this potential energy is transferred to the kinetic energy of an initially stationary 20 gram domino, what will be the velocity of the domino?
If a 20 gram domino has .03 Joules of kinetic energy, then since its KE is 1/2 m v^2 we have 1/2 m v^2 = KE so v = sqrt( 2 * KE / m) = sqrt( 2 * .03 Joules / (.02 kg) ) = 1.7 m/s^2, approx..
If a 20 gram domino has .03 Joules of kinetic energy, then since its KE is 1/2 m v^2 we have
1/2 m v^2 = KE so
v = sqrt( 2 * KE / m) = sqrt( 2 * .03 Joules / (.02 kg) ) = 1.7 m/s^2, approx..
If instead the rubber band is used to 'shoot' the domino straight upward, then how high will it rise?
The domino will rise until its gravitational potential energy has increased by .03 Joules. The increase in gravitational potential energy is equal and opposite to the work done by the gravitational force. A mass m experiences a downward gravitational force F = m g, which we call its weight. As the mass rises, gravity acts downward as the object moves upward, so the gravitational force and the displacement are in opposite directions. Gravity therefore does negative work on the mass, so the gravitational potential energy increases. If the mass rises through displacement `dy, the work done on it by gravity is force * displacement = - m g * `dy, so its gravitational PE changes by + m g `dy. The domino therefore rises until m g `dy = .03 Joules, so that m g `dy = .03 Joules `dy = .03 Joules / (m g) = .03 Joules / (.02 kg * 9.8 m/s^2) = .15 meters, approx. Note that it's possible to just set the formula m g h for gravitational potential energy equal to .03 Joules and solve for h. For many problems this will be sufficient, and it's a significant skill. However we gain a lot more insight into the nature of energy by the preceding line of reasoning, which proceeds from the definitions. If time permits, it's a good idea to work to understand that reasoning. University Physics students should certainly work to understand this reasoning.
The domino will rise until its gravitational potential energy has increased by .03 Joules.
The increase in gravitational potential energy is equal and opposite to the work done by the gravitational force.
A mass m experiences a downward gravitational force F = m g, which we call its weight.
As the mass rises, gravity acts downward as the object moves upward, so the gravitational force and the displacement are in opposite directions. Gravity therefore does negative work on the mass, so the gravitational potential energy increases.
If the mass rises through displacement `dy, the work done on it by gravity is force * displacement = - m g * `dy, so its gravitational PE changes by + m g `dy.
The domino therefore rises until m g `dy = .03 Joules, so that
m g `dy = .03 Joules `dy = .03 Joules / (m g) = .03 Joules / (.02 kg * 9.8 m/s^2) = .15 meters, approx.
m g `dy = .03 Joules
`dy = .03 Joules / (m g) = .03 Joules / (.02 kg * 9.8 m/s^2) = .15 meters, approx.
Note that it's possible to just set the formula m g h for gravitational potential energy equal to .03 Joules and solve for h. For many problems this will be sufficient, and it's a significant skill.
However we gain a lot more insight into the nature of energy by the preceding line of reasoning, which proceeds from the definitions. If time permits, it's a good idea to work to understand that reasoning. University Physics students should certainly work to understand this reasoning.
Question for University Physics Students
Why does it make sense to say that the PE change is equal to the integral of the force vs. position function over an appropriate interval, and what is the appropriate interval? (see discussion below)
NOTES FOR UNIVERSITY PHYSICS STUDENTS:
The following is the most common solution I get to this problem. The notes below are related to important properties of the integral, which you should understand. PE = 3N * 10 cm = 30 J COMMENTS ON THIS SOLUTION: 3 N * 10 cm would be 30 N * cm, not 30 J. Since a cm is .01 m, the unit N * cm is N * (.01 m) = .01 J, and 30 N * cm = ..30 J. However 3 N is the max force, not the ave force. Forces are exerted through the 2 cm interval from 8 cm to 10 cm, not through a 10 cm interval. The rubber band exerts no tension force if its length is shorter than another length at which the tension force is zero. For this reason lengths less than 8 cm are outside the domain of the force function, and the integral of force with respect to length is not defined for such lengths. This leads to a few very important calculus-based connections: The force vs. position function can be integrated to yield the work. In this case the force function can be considered to be linear, so that the result is equal to the result you get when you multiply the average of the two forces by displacement. You should find the linear function defined by the force vs. length ordered pairs (8 cm, 0 N) and (10 cm, 3 N). (The correct function is F = 1.5 N/cm * x - 12 N.) Recall that an integral of any function over an interval is equal to the average value of that function on the interval, multiplied by the length of the interval. This is a property of integrals you should review and thoroughly understand. (The correct integral for this problem would be the integral from 8 cm to 10 cm of the function 1.5 N / cm * x - 12 N. An antiderivative is 1.5 N / cm * x^2 / 2 - 12 N * x. At x = 8 cm the value of the antiderivative is 1.5 N / cm * (8 cm)^2 / 2 - 12 N * 8 cm = 48 N * cm - 96 N * cm = -48 N * cm. At x = 10 cm the results is -45 N * cm. The change in the antiderivative is therefore (final value - initial value) = -45 N * cm - (-48 N * cm) = 3 N * cm. That's the solution, in terms of the meaning of integration. The solution should be understood in those terms. The breif expression of this solution is int( 1.5 N / cm * x - 12 N, x, 8 cm, 10 cm), which when written out looks like this: An antiderivative is , which when simplified and expressed in decimal form is You should probably have already sketched this graph, in the process of finding your linear function. In any case, the figure below depicts the linear force vs. position function defined by the two given points. The short segment of this function which extends below the x axis is outside the domain of the actual force function, as discussed above.
The following is the most common solution I get to this problem. The notes below are related to important properties of the integral, which you should understand.
PE = 3N * 10 cm = 30 J
COMMENTS ON THIS SOLUTION:
3 N * 10 cm would be 30 N * cm, not 30 J. Since a cm is .01 m, the unit N * cm is N * (.01 m) = .01 J, and 30 N * cm = ..30 J.
However 3 N is the max force, not the ave force. Forces are exerted through the 2 cm interval from 8 cm to 10 cm, not through a 10 cm interval. The rubber band exerts no tension force if its length is shorter than another length at which the tension force is zero. For this reason lengths less than 8 cm are outside the domain of the force function, and the integral of force with respect to length is not defined for such lengths.
This leads to a few very important calculus-based connections: The force vs. position function can be integrated to yield the work. In this case the force function can be considered to be linear, so that the result is equal to the result you get when you multiply the average of the two forces by displacement.
You should find the linear function defined by the force vs. length ordered pairs (8 cm, 0 N) and (10 cm, 3 N). (The correct function is F = 1.5 N/cm * x - 12 N.) Recall that an integral of any function over an interval is equal to the average value of that function on the interval, multiplied by the length of the interval. This is a property of integrals you should review and thoroughly understand.
(The correct integral for this problem would be the integral from 8 cm to 10 cm of the function 1.5 N / cm * x - 12 N. An antiderivative is 1.5 N / cm * x^2 / 2 - 12 N * x. At x = 8 cm the value of the antiderivative is 1.5 N / cm * (8 cm)^2 / 2 - 12 N * 8 cm = 48 N * cm - 96 N * cm = -48 N * cm. At x = 10 cm the results is -45 N * cm. The change in the antiderivative is therefore (final value - initial value) = -45 N * cm - (-48 N * cm) = 3 N * cm.
That's the solution, in terms of the meaning of integration. The solution should be understood in those terms. The breif expression of this solution is
int( 1.5 N / cm * x - 12 N, x, 8 cm, 10 cm), which when written out looks like this:
An antiderivative is
, which when simplified and expressed in decimal form is
You should probably have already sketched this graph, in the process of finding your linear function. In any case, the figure below depicts the linear force vs. position function defined by the two given points. The short segment of this function which extends below the x axis is outside the domain of the actual force function, as discussed above.