course Phy 201 Here is the work for assignment 9. While going through it I noticed that the line for Video Experiment 9 said not assigned at this time. I took that to mean that I did not have to work on or complete it so I skipped it. However when I was doing the query I noticed that there were a few questions about it in there. Do I have to do it or not?
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18:38:56 `q001. Note that there are 10 questions in this set. .You have done the Introductory Force Experiment in which you used rubber bands and bags of water, and you understand that, at least in the vicinity of the Earth's surface, gravity exerts downward forces. You have also seen that forces can be measured in units called Newtons. However you were not given the meaning and definition of the Newton as a unit of force. You are also probably aware that mass is often measured in kilograms. Here we are going to develop in terms of an experiment the meaning of the Newton as a force unit. Suppose that a cart contains 25 equal masses. The cart is equal in mass to the combined total of the 25 masses, as indicated by balancing them at equal distances from a fulcrum. The cart is placed on a slight downward incline and a weight hanger is attached to the cart by a light string and suspended over a low-friction pulley at the end of the ramp. The incline is adjusted until the cart, when given a slight push in the direction of the hanging weight, is observed to move with unchanging, or constant, velocity (and therefore zero acceleration). The masses are then moved one at a time from the cart to the hanger, so that the system can be accelerated first by the action of gravity on one of the masses, then by the action of gravity onto of the masses, etc.. The time required for the system to accelerate from rest through a chosen displacement is observed with one, two, three, four, five, ... ten of the masses. The acceleration of the system is then determined from these data, and the acceleration is graphed vs. the proportion of the total mass of the system which is suspended over the pulley. It is noted that if the entire mass of the system, including the cart, is placed on the weight hanger, there will be no mass left on the incline and the entire weight will fall freely under the acceleration of gravity. Suppose the data points obtained for the 5 of the first 10 trials were (.04, 48 cm/s^2), (.08, 85 cm/s^2), (.12, 125 cm/s^2), (.16, 171 cm/s^2), (.20, 190 cm/s^2). Sketch these points on an accurate graph of acceleration vs. proportion of weight suspended and determine the slope and y-intercept of the line. What is your slope and what is the y intercept? What is the equation of the line?
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RESPONSE --> I come up with a slope of 1000 and my y intercept is 20 cm/s^2. The equation of the line that I come up with is x = .02 and y = 20 cm/s^2 so: 20 cm/s^2 / .02 = 1000 Correct me if I'm wrong.
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18:41:05 Since there are 25 equal masses and the mass of the cart is equivalent to another 25 of these masses, each mass is 1/50 = .02 of the total mass of the system. Thus the first 10 data points should might have been something like (.02, 21 cm/s^2), (.04, 48 cm/s^2), (.06, 55 cm/s^2), (.08, 85 cm/s^2), (.10, 101 cm/s^2), (.12, 125 cm/s^2), (.14, 141 cm/s^2), (.16, 171 cm/s^2), (.18, 183 cm/s^2), (.20, 190 cm/s^2). The data given in the problem would correspond to alternate data points. The slope of the best-fit line is 925 x + 12.8, indicating a slope of 925 and a y intercept of 12.8. The 967.5 is in units of rise / run, or for this graph cm/s^2. If you calculated the slope based on the points (.04, 48 cm/s^2) and (.20, 190 cm/s^2) you would have obtained 151 cm/s^2 / (.16) = 950 cm/s^2. Whether this is close to the best-fit value or not, this is not an appropriate calculation because it uses only the first and last data points, ignoring all data points between. The idea here is that you should sketch a line that fits the data as well as possible, then use the slope of this line, not the slope between data points.
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RESPONSE --> I used the data points for y and x where the line that I had drawn met the x and y axis. This is the prescribed method isn't it?
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18:42:24 `q002. Do the points seem to be randomly scattered around the straight line or does there seem to be some nonlinearity in your results?
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RESPONSE --> The points as far as I can see are pretty in line with the line. The only point that does not seem to be right on the line is the last point which is slightly below.
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18:45:00 The slope of your line should probably be somewhere between 900 cm/s^2 and 950 cm/s^2. The points should be pretty much randomly scattered about the best possible straight line. Careful experiments of this nature have shown that the acceleration of a system of this nature is to a very high degree of precision directly proportional to the proportion of the weight which is suspended.
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RESPONSE --> I see what you are saying and do understand that the line should be in the area between points and not actually try to pass through any one point. I am just saying that in this case the line seems to right on the points. Now maybe my scale on the graph is to small which can possibly take away from the accuracy of the graph points.
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18:46:05 `q003. If the acceleration of the system is indeed proportional to the net force on the system, then your straight line should come close to the origin of your coordinate system. Is this the case?
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RESPONSE --> Yes it does as was stated in the last couple of question critiques.
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18:46:44 If the acceleration of the system is proportional to the net force, then the y coordinate of the straight line representing the system will be a constant multiple of the x coordinate--that is, you can always find the y coordinate by multiplying the x coordinate by a certain number, and this 'certain number' is the same for all x coordinates. The since the x coordinate is zero, the y coordinate will be 0 times this number, or 0. Your graph might not actually pass through the origin, because data inevitably contains experimental errors. However, if experimental errors are not too great the line should pass very close to the origin. In the case of this experiment the y-intercept was 12.8. On the scale of the data used here this is reasonably small, and given the random fluctuations of the data points above and below the straight-line fit the amount of deviation is consistent with a situation in which precise measurements would reveal a straight line through the origin.
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18:48:13 `q003. What is it that causes the system to accelerate more when a greater proportion of the mass is suspended?
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RESPONSE --> The cause of this can be summed with the pull of the gravitational force on the object. The heavier the object the more gravitational pull on the object.
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18:48:53 The gravitational forces exerted on the system are exerted on the suspended masses and on the cart and the masses remaining in it. The supporting force exerted by the ramp counters the force of gravity on the cart and the masses remaining in it, and this part of the gravitational force therefore does not affect the acceleration of the system. However there is no force to counter the pull of gravity on the suspended masses, and this part of the gravitational force is therefore the net force acting on the mass of the entire cart-and-mass system. The force exerted by gravity on the suspended masses is proportional to the number of suspended masses--e.g, if there are twice as many masses there is twice the force. Thus it is the greater gravitational force on the suspended masses that causes the greater acceleration.
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18:54:24 `q004. This results of this sort of experiment, done with good precision, support the contention that for a given mass the acceleration of a system is indeed proportional to the net force exerted on the system. Other experiments can be done using rubber bands, springs, fans and other nongravitational sources of force, to further confirm this result. In another sort of experiment, we can see how much force is required on different masses to obtain a certain fixed acceleration. In such experiments we find for example that if the mass is doubled, it requires twice the force to achieve the same acceleration, and that in general the force required to achieve a given acceleration is proportional to the amount of mass being accelerated. In a certain experiment using the same cart and masses as before, plus several additional identical carts, a single cart is accelerated by a single suspended mass and found to accelerate at 18 cm/s^2. Then a second cart is placed on top of the first and the two carts are accelerated by two suspended masses, achieving an acceleration of 20 cm / s^2. Then a third cart is placed on top of the first to and the three carts are accelerated by three suspended masses, achieving and acceleration of 19 cm/s^2. A fourth cart and a fourth suspended mass are added and an acceleration of 18 cm/s^2 is obtained. Adding a fifth cart in the fifth suspended mass an acceleration of 19 cm/s^2 is obtained. All these accelerations are rounded to the nearest cm/s^2, and all measurements are subject to small but significant errors in measurement. How well do these results indicate that to achieve a given acceleration the amount of force necessary is in fact proportional to the amount of mass being accelerated?
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RESPONSE --> Because for all of the masses that you stated, they all came up with an acceleration of the basic same amount.
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18:55:12 The accelerations obtained are all about the same, with only about a 10% variation between the lowest and the highest. Given some errors in the observation process, it is certainly plausible that these variations are the result of such observation errors; however we would have to have more information about the nature of the observation process and the degree of error to be expected before drawing firm conclusions. If we do accept the conclusion that, within experimental error, these accelerations are the same then the fact that the second through the fifth systems had 2, 3, 4, and 5 times the mass of the first with 2, 3, 4, and 5 times the suspended mass and therefore with 2, 3, 4, and 5 times the net force does indeed indicate that the force needed to achieve this given acceleration is proportional to the mass of the system.
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18:57:21 `q005. Now we note again that the force of gravity acts on the entire mass of the system when an entire system is simply released into free fall, and that this force results in an acceleration of 9.8 m/s^2. If we want our force unit to have the property that 1 force unit acting on 1 mass unit results in an acceleration of 1 m/s^2, then how many force units does gravity exert on one mass unit?
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RESPONSE --> I would say that they are proportional to each other so that would mean that it would be 1 m/s^2.
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18:58:41 Since gravity gives 1 mass unit an acceleration of 9.8 m/s^2, which is 9.8 times the 1 m/s^2 acceleration that would be experienced from 1 force unit, gravity must exerted force equal to 9.8 force units on one mass unit.
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RESPONSE --> Ok I see what I did wrong. This is true since 1 mass unit is equal to an acceleration of 9.8 m/s^2.
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18:59:54 `q006. If we call the force unit that accelerates 1 mass unit at 1 m/s^2 the Newton, then how many Newtons of force does gravity exert on one mass unit?
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RESPONSE --> It would be 9.8 m/s^2 as a Newton.
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19:00:49 Since gravity accelerates 1 mass unit at 9.8 m/s^2, which is 9.8 times the acceleration produced by a 1 Newton force, gravity must exert a force of 9.8 Newtons on a mass unit.
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RESPONSE --> Would this be expressed as 9.8 Newtons or 9.8 m/s^2 Newtons?
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19:02:04 `q007. The mass unit used here is the kilogram. How many Newtons of force does gravity exert on a 1 kg mass?
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RESPONSE --> That would have to be 9.8 kg Newtons of force.
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19:03:26 Gravity exerts a force of 9.8 Newtons on a mass unit and the kg is the mass unit, so gravity must exert a force of 9.8 Newtons on a mass of 1 kg.
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RESPONSE --> Ok I am beginning to see that the Newton is a unit of measurement much like the meter of kilometer. So when stated it should be said as the way that you have stated.
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19:04:39 `q008. How much force would gravity exert on a mass of 8 kg?
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RESPONSE --> The force of gravity on a mass of 8 kg would be 78.4 Newtons since a Newton is 9.8 times the acceleration.
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19:05:11 Gravity exerts 8 times the force on 8 kg as on 1 kg. The force exerted by gravity on a 1 kg mass is 9.8 Newtons. So gravity exerts a force of 8 * 9.8 Newtons on a mass of 8 kg.
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RESPONSE --> Which comes out to be 78.4 Newtons.
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19:07:40 `q009. How much force would be required to accelerate a mass of 5 kg at 4 m/s^2?
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RESPONSE --> Well first you would have to multiply the 5 kg by 9.8 to get the Newtons which would be 49 Newtons. The rest of it I don't know.
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19:11:10 Compared to the 1 Newton force which accelerates 1 kg at 1 m/s^2, 2e have here 5 times the mass and 4 times the acceleration so we have 5 * 4 = 20 times the force, or 20 Newtons. We can formalize this by saying that in order to give a mass m an acceleration a we must exert a force F = m * a, with the understanding that when m is in kg and a in m/s^2, F must be in Newtons. In this case the calculation would be F = m * a = 5 kg * 4 m/s^2 = 20 kg m/s^2 = 20 Newtons. The unit calculation shows us that the unit kg * m/s^2 is identified with the force unit Newtons.
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RESPONSE --> I don't understand why we didn't insert the 9.8 into the equation at all? Is it because the 4 m/s^2 takes it place?
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19:12:46 `q010. How much force would be required to accelerate the 1200 kg automobile at a rate of 2 m/s^2?
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RESPONSE --> Well given the last problem I would have to go with the equation of F = m* a. This would give us F = 1200 kg * 2 m/s^2 = 2400 Newtons.
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19:12:50 This force would be F = m * a = 1200 kg * 2 m/s^2 = 2400 kg * m/s^2 = 2400 Newtons.
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Gװ~G˳ assignment #009 yE܃֭ Physics I 11-13-2005
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19:44:07 Introductory prob set 3 #'s 1-6 If we know the distance an object is pushed and the work done by the pushing force how do we find the force exerted by the object?
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RESPONSE --> You would simply multiply the work done (Newtons) by the distance (meters) and you would have the product in Joules.
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19:47:34 ** Knowing the distance `ds and the work `dW we note that `dW = F * `ds; we solve this equation and find that force is F=`dw/`ds **
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RESPONSE --> Ok I can see where I went wrong. The work done is actually the Joules. So therefore I was wrong in the way I explained it but I explained for the question if you had force and displacement instead of displacement and work as you had stated.
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19:49:06 If we know the net force exerted on an object and the distance through which the force acts how do we find the KE change of the object?
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RESPONSE --> You would simply multiply the force by the distance and this would also be equal to the work done which is equal to the kinetic energy.
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19:49:34 **`dW + `dKE = 0 applies to the work `dW done BY the system and the change `dKE in the KE OF the system. The given force acts ON the system so F `ds is work done ON the system. The work done BY the system against that force is `dW = -F * `ds. When you use the energy equation, this is the work you need--the work done BY the system. **
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19:50:49 Why is KE change equal to the product of net force and distance?
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RESPONSE --> I am not sure I can put it into words yet. I am following the formulas for now. I will read further and comment later.
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19:52:09 ** It comes from the equation vf^2 = v0^2 + 2 a `ds. Newton's 2d Law says that a = Fnet / m. So vf^2 = v0^2 + 2 Fnet / m `ds. Rearranging we get F `ds = 1/2 m vf^2 - 1/2 m v0^2. Defining KE as 1/2 m v^2 this is F `ds = KEf - KE0, which is change in KE. **
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RESPONSE --> Ok I see but it will take a while to soak in.
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19:53:15 When I push an object with a constant force, why is KE change not equal to the product of the force I exert and the distance?
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RESPONSE --> Ok I thought that it was?
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19:54:57 ** Change in KE is equal to the work done by the net force, not by the force I exert. When I push an object in the real world, with no other force 'helping' me, there is always at least a little force resisting my push. So the net force in this case is less than the force I exert, in which case the change in KE would be less than the product of the force I exert and the distance. If another force is 'helping' me then it's possible that the net force could be greater than the force I exert, in which case the change in KE would be greater than the product of the force I exert and the distance. It is actually possible for the 'helping' force to exactly balance the resisting force, but an exact balance would be nearly impossible to achieve. ANOTHER WAY OF LOOKING AT IT: If I push in the direction of motion then I do positive work on the system and the system does negative work on me. That should increase the KE of the system. However if I'm pushing an object in the real world and there is friction and perhaps other dissipative forces which tend to resist the motion. So not all the work I do ends up going into the KE of the object. **
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RESPONSE --> Ok I see what you are saying.
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19:56:32 Query pendulum force proportionalities (video exp #9) Describe how we conclude that for small displacements the force tending to restore a pendulum to equilibrium is proportional to the displacement
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RESPONSE --> I read that the experiment was not assigned at this time so I did not do it. Is it part of the assignment?
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19:56:41 ** Since displacement over length equals the ratio of the effective suspended mass over mass of pend., and this ratio is equal to Force over weight, then displacement is also proportionate to the force. **
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19:56:55 Explain how your common sense leads you to believe in the plausibility of the hypothesis that the force tending to restore a pendulum to its equilibrium position is in fact in the same proportion to the weight of the pendulum as the displacement to the length of the pendulum.
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19:56:59 ** Since weight is the measurement of gravity acting on mass, then the weight implies some force, and the length of pendulum implies some distance over which gravity works, then there should be a set proportion. **
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19:57:05 Describe how we conclude that for small displacements the force tending to restore a pendulum to equilibrium is in the same proportion to the weight of the pendulum as the displacement to the length of the pendulum.
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19:57:07 ** Intuitively, the further you pull back the greater the restoring force. As you pull back the ratio x / L increases. As you pull back the ratio F / (M g) = restoring force / pendulum weight also increases. The proportionality between the two says, basically, that if you pull the pendulum back a distance equal to 5% of its length then the restoring force is 5% of the pendulum's weight. This proportionality applies very well to small displacements. As displacement gets larger the linearity of the relationship deteriorates. More rigorously: The force m * g exerted on the suspended mass m is the force exerted by gravity, and with a small correction for friction is equal to the displacing force. This force is equal and opposite to the restoring force. The weight of the pendulum is the force M * g exerted by gravity on its mass M. If x / m = L / M then x / L = m / M = m * g / M * g = displacing force / weight. **
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