course Phy 201 |zMDqٽ{`~Ğassignment #009
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19:34:48 `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 --> Slope = rise/run = 85 -48/ (0.08-0.04) = 925 y intercept = y = mx +b 48 = 925 (.04) + b b = 48 - 37 = 11 Equation: y = 925x + 11 confidence assessment: 2
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19:37:03 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 didn't alter the data points according to the masses. I understand my mistake now. self critique assessment: 2
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19:38:13 `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 seem to be fairly close to the straight line. confidence assessment: 2
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19:38:59 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 --> ok self critique assessment: 3
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19:39:54 `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 --> No the straight line isn't very close to the origin. confidence assessment: 2
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19:41:34 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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RESPONSE --> If my graph was precise to scale then maybe the line would have passed through the origin! self critique assessment: 2
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19:42:43 `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 --> Force on the object confidence assessment: 3
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19:43:22 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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RESPONSE --> ok self critique assessment: 2
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19:49:34 `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 --> These results shows that to achieve a given acceleration the amount of force necessary is proportional to the amount of mass because as more mass was added the force became greater and thus the acceleration was greater. confidence assessment: 2
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19:50:26 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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RESPONSE --> ok that was commonsense! oops! self critique assessment: 3
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19:56:06 `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 --> 1 m/s/s / 9.80 m/s/s = 0.102m/s/s F= .102 m/s/s * 1 mass unit = 0.102 confidence assessment: 2
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19:56:29 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 self critique assessment: 3
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19:57:36 `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 --> 9.80 N confidence assessment: 3
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19:57:56 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 --> ok self critique assessment: 3
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19:58:25 `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 --> 9.80 N confidence assessment: 3
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19:58:39 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 self critique assessment: 3
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19:59:29 `q008. How much force would gravity exert on a mass of 8 kg?
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RESPONSE --> F = 9.80 m/s/s * 8 kg = 78.4 N confidence assessment: 3
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20:00:04 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 --> ok self critique assessment: 3
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20:01:31 `q009. How much force would be required to accelerate a mass of 5 kg at 4 m/s^2?
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RESPONSE --> 20 N of force on a mass of 5 kg with an acceleration of 4 m/s^2. confidence assessment: 3
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20:12:31 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 --> ok self critique assessment: 3
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20:13:10 `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 --> F = 1200 kg * 2 m/s/s = 2400 N confidence assessment: 3
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20:13:18 This force would be F = m * a = 1200 kg * 2 m/s^2 = 2400 kg * m/s^2 = 2400 Newtons.
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RESPONSE --> ok self critique assessment: 3
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