asst 22 qanda

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01:01:11 `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 --> F=ma m=4kg x component= L * cos theta y component=L *sine theta ycomponent= 5N *sine(210degrees)= 5N * -.5=-2.5N x component=10m/s vf^2=v0^2 + 2a(`ds) vf(x)=0 `dt=? y velocity=? confidence assessment: 0

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01:07:40 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 --> 5Newtons=4kg*a a=1.25m/s^2 acceleration has x component 1.25m/s^2 *cosine(210degrees)=-1.08m/s^2 y component of 1.25m/s^2 *sine(210degrees)=-.63m/s^2 v0(x)=10m/s a=-1.08m/s^2 vf(x)=0 change in velocity =-10m/s `dt=-10m/s/-1.08m/s^2=9.2s v0(y)=0 a=-.63m/s^2 `dt=9.2s y velocity changed by -.63m/s^2*9.2s=-6m/s Change in y velocity=-6m/s self critique assessment: 2

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01:12:26 `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 --> Mass=4kg Force=5N a=1.25m/s^2 `ds=30m in x direction vave=`ds/`dt vave=5m/s 5m/s=30m/`dt 5m/s*`dt=30m `dt=6s confidence assessment: 0

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01:17:36 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 --> Should use the acceleration for x as before= -1.08m/s^2 initial velocity(x)=10m/s 30m in x direction vf=+- `sqrt((10m/s)^2 + 2*(-1.08m/s^2*30m=+-6m/s Choose vf=6m/s vave=8m/s `dt=30m/8m/s=3.75s y velocity changing by -.63m/s^2 y -.63m/s^2 *3.75s=-2.4m/s x component=6m/s y component=-2.4m/s x^2 + y^2=z^2 z=6.4m/s direction= arctan(-2.4m/s)/(6m/s)=-22 degrees y is negative so add 360 degrees 338degrees self critique assessment: 2

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asst 22 qanda

Your work has been received. Please scroll through the document to see any inserted notes (inserted at the appropriate place in the document, in boldface) and a note at the end. The note at the end of the file will confirm that the file has been reviewed; be sure to read that note. If there is no note at the end, notify the instructor through the Submit Work form, and include the date of the posting to your access page.

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Your work looks very good. Let me know if you have any questions. &#

end of document

Your work has not been reviewed.

Please notify your instructor of the error, using the Submit Work form, and be sure to include the date 11-26-2007.

Good work.

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Let me know if you have questions. &#