course PHY201 ÍìˤQãw’z®vxõbäO“ ´êÒ|assignment #022
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14:34:07 `q001. Note that this assignment contains 2 questions, which relate to a force-field experiment which is done using a computer simulation, and could for example represent the force on a spacecraft, where uphill and downhill are not relevant concepts. . An object with a mass of 4 kg is traveling in the x direction at 10 meters/second when it enters a region where it experiences a constant net force of 5 Newtons directed at 210 degrees, as measured in the counterclockwise direction from the positive x axis. How long will take before the velocity in the x direction decreases to 0? What will be the y velocity of the object at this instant?
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RESPONSE --> With this problem we are looking for the time it will take to bring the object to zero, and what the final velocity of the y direction will be at this instant when a net force of 5N is applie in the direction of 210 deg. First we must calculate for the x and y forces. x = rcos(theta) x = 5N * cos(210 deg) = -4.33N y = rsin(theta) y = 5N * sin(210 deg) = -2.5N Now we will calculate for the acceleration in the those directions. ax = Fx / m ax = -4.33N / 4kg = -1.08m/s^2 ay = Fy / m ay = -2.5 / 4kg = -.63m/s^2 Now I will use the following formula to determine the time taken to bring the object to a velocity of 0m/s. dvx / ax = dt (0m/s - 10m/s) / -1.08m/s^2 = 9.26s dt = 9.26s Now I will use the formulas of uniform acceleration to determine the final velocity in the y-direction; but first I must calculate for the distance traveled. ds = Voy(dt) + .5a(dt^2) ds = 0m/s(9.26s) + .5(-.63m/s^2)(9.26^2) ds = 27m Vfy^2 = Voy^2 + 2a(ds) Vfy^2 = (0m/s)^2 + 2 * (-.63m/s^2)(27m) Vfy = sqrt(34.02m^2/s^2) = 5.83m/s Vfy = 5.83m/s confidence assessment: 2
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14:37:33 A constant net force of 5 Newtons on a 4 kg object will result in an acceleration of 5 Newtons/(4 kg) = 1.25 meters/second ^ 2. If the force is directed at 210 degrees then the acceleration will also be directed at 210 degrees, so that the acceleration has x component 1.25 meters/second ^ 2 * cosine (210 degrees) = -1.08 meters/second ^ 2, and a y component of 1.25 meters/second ^ 2 * sine (210 degrees) = -.63 meters/second ^ 2. We analyze the x motion first. The initial velocity in the x direction is given as 10 meters/second, we just found that the acceleration in the x direction is -1.08 meters/second ^ 2, and since we are trying to find the time required for the object to come to rest the final velocity will be zero. We easily see that the change in the next velocity is -10 meters/second. At a rate of negative -1.08 meters/second ^ 2, the time required for the -10 meters/second change in velocity is `dt = -10 meters/second / (-1.08 meters/second ^ 2) = 9.2 seconds. We next analyze the y motion. The initial velocity in the y direction is zero, since the object was initially moving solely in the x direction. The acceleration in the y direction is -.63 meters/second ^ 2. Therefore during the time interval `dt = 9.2 seconds, the y velocity changed by (-.63 meters/second ^ t) * (9.2 seconds) = -6 meters/second, approximately. Thus the y velocity changes from zero to -6 meters/second during the 9.2 seconds required for the x velocity to reach zero.
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RESPONSE --> I am glad I got it right! self critique assessment: 3
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15:17:00 `q002. Suppose that the mass in the preceding problem encounters a region in which the force was identical to that of the problem, but that this region extended for only 30 meters in the x direction (assume that there is the limit to the extent of the field in the y direction). What will be the magnitude and direction of the velocity of the mass as it exits this region?
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RESPONSE --> Using the values from the first problem... Vox = 10m/s ax = -1.08m/s^2 ay = -.63m/s^2 Fg = 5N @ 210 deg This time the region of the acting force is only 30m, so I will calculate the magnitude and direction of velocity for when the object leaves the region. First I will solve to final velocity in the x direction. Vf^2 = Vo^2 + 2a(ds) Vf^2 = (10m/s)^2 + 2(-.63m/s^2)(30m) Vf = sqrt(35.2m^2/s^2) Vfx = 5.93m/s Now we need to find final velocity in the y direction; but first we need to calculate the time it took to pass through the 30m region and the distance the object traveled in the y direction. dv / a = dt (5.93m/s - 10m/s) / -1.08m/s^2 = 3.77s dt = 3.77s ds = Voy(dt) + .5a(dt^2) ds = 0m/s(3.77s) + .5(-.63m/s^2)(3.77^2) ds = 4.48m Vfy^2 = Voy^2 + 2a(ds) Vfy^2 = (0m/s)^2 + 2(-.63m/s^2)(4.48m) Vfy = sqrt(5.6448m^2/s^2) Vfy = -2.38m/s Now we use Pythagorean Theorem to find the velocity in the z direction, and tan-1(y/x) to find the angle. a^2 + b^2 = c^2 (5.93m/s)^2 + (-2.38m/s)^2 = sqrt(40.8293m^2/s^2) c = 6.39m/s tan-1(y/x) = theta tan-1(-2.38m/s / 5.93m/s) theta = 338.13 deg Vf = 6.39m/s @ 338 deg confidence assessment: 3
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15:17:29 As we have seen in the preceding problem the object will have an acceleration of -1.08 meters/second ^ 2 in the x direction. Its initial x velocity is 10 meters/second and it will travel 30 meters in the x direction before exiting the region. Thus we have v0, a and `ds, so that you to the third or fourth equation of uniform accelerated motion will give us information. The fourth equation tells us that vf = +-`sqrt( (10 meters/second) ^ 2 + 2 * (-1.08 meters/second ^ 2) * (30 meters) ) = +-6 meters/second. Since we must exit the region in the positive x direction, we choose vf = + 6 meters/second. It follows that the average x velocity is the average of the initial 10 meters/second and the final 6 meters/second, or eight meters/second. Thus the time required to pass-through the region is 30 meters/(8 meters/second) = 3.75 seconds. During this time the y velocity is changing at -.63 meters/second ^ 2. Thus the change in the y velocity is (-.63 meters/second ^ 2) * (3.75 seconds) = -2.4 meters/second, approximately. Since the initial y velocity was zero, the y velocity upon exiting the region will be -2.4 meters/second. Thus when exiting the region the object has velocity components +6 meters/second in the x direction and -2.4 meters/second in the y direction. Its velocity therefore has magnitude `sqrt ( (6 meters/second) ^ 2 + (-2.4 meters/second) ^ 2) = 6.4 meters/second. The direction of velocity will be arctan ( (-2.4 meters/second) / (6 meters/second) ) = -22 degrees, approximately. Thus the object exits at 6.4 meters/second at an angle of 22 degrees below the positive x axis, or at angle -22 degrees + 360 degrees = 338 degrees.
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RESPONSE --> self critique assessment: 3
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