course Phy 201 Made a mistake by clicking on the next question/answer on the form. Sometimes I dont click in the box before typing ô~x¹¢xó°‡âɆѧìÂjž½æ¶Ðèè™ÕÙassignment #022
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01:31:23 `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 --> Since we have a constant net force of 5newtons and a mass of 4kg we can first find the accleration of the magnitude by using the equation F=ma. We try to find a so we have a=F/m we get, a=5newtons/4kg,a=1.25m/s^2. Since the force is in the direction of 210deg we can find the x and y component. 1.25m/s^2*cos(210deg), x=-1.08m/s^2 and 1.25m/s^2*sin(210deg), y=-.63m/s^2. Since the object is moving in the x direction at 10m/s against the acceleration of -1.08m/s^2 the object will stop in 10m/s/-1.08=9.3s. Since the y velocity is -.63m/s^2 and and it takes 9.2 s for the object to come to rest we can say `dv=a*`dt, we get `dv=-.63m/s^2*9.2s,`dv=-5.8m/s. Im assuming here that the intial velocity is 0m/s and the `dv=-5.8m/s so the final velocity would be -5.8m/s confidence assessment: 1
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01:35:16 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 --> ok self critique assessment: 3
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01:58:25 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 --> Since we know the accelerattion of the x-component to be -1.08 and `ds=30m and the intial v=10m/s we can find the average velocity. First we take vf=sqrt(10m/s+2(-1.08m/s^2)*30m,vf=5.9m/s, we then take vAve=(10m/s+5.9m/s),vAve=8.0m/s. We then take the 30m/8.0m/s=3.8s. Now that we know the time and the acceleration of y- component we can find the final velocity by taking -.63m/s^2*3.8s=-2.4m/sNow we have the two components we can find the magntude. v=sqrt((6m/s)^2+(-2.4m/s)^2,v=6.5m/s and we can find the direction by taking the tan^-1(-2.4m/s/5.9m/s)=-22deg+360deg=338deg self critique assessment: 3
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