course phy201 ???va??l?z????assignment #020020. Forces (inclines, friction)
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23:19:58 `q001. Note that this assignment contains 3 questions. . A 5 kg block rests on a tabletop. A string runs horizontally from the block over a pulley of negligible mass and with negligible friction at the edge of the table. There is a 2 kg block hanging from the string. If there is no friction between the block in the tabletop, what should be the acceleration of the system after its release?
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RESPONSE --> F =ma of the hanging mass F = 2kg * 9.8m/s^2 F = 19.6N F=ma of the system a = F /m a = 19.6 kg/m/s/s / 5kg a = 3.92m/s/s confidence assessment: 3
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23:21:36 Gravity exerts a force of 5 kg * 9.8 meters/second = 49 Newtons on the block, but presumably the tabletop is strong enough to support the block and so exerts exactly enough force, 49 Newtons upward, to support the block. The total of this supporting force and the gravitational force is zero. The gravitational force of 2 kg * 9.8 meters/second = 19.6 Newtons is not balanced by any force acting on the two mass system, so we have a system of total mass 7 kg subject to a net force of 19.6 Newtons. The acceleration of this system will therefore be 19.6 Newtons/(7 kg) = 2.8 meters/second ^ 2.
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RESPONSE --> the forces of the block on the table and the upward force = 0 I assumed that without explaination Mass = mass of the system! self critique assessment: 2
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23:25:24 `q002. Answer the same question as that of the previous problem, except this time take into account friction between the block in the tabletop, which exerts a force opposed to motion which is .10 times as great as the force between the tabletop and the block. Assume that the system slides in the direction in which it is accelerated by gravity.
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RESPONSE --> Ffr = .10 * FN = 4.9N FN = ma Fn = 5 * 9.8 = 49 a = Fnet /m a = 19.6 - 4.9 /7 a = 2.1m/s/s confidence assessment: 3
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23:25:43 Again the weight of the object is exactly balance by the upward force of the table on the block. This force has a magnitude of 49 Newtons. Thus friction exerts a force of .10 * 49 Newtons = 4.9 Newtons. This force will act in the direction opposite that of the motion of the system. It will therefore be opposed to the 19.6 Newton force exerted by gravity on the 2 kg object. The net force on the system is therefore 19.6 Newtons -4.9 Newtons = 14.7 Newtons. The system will therefore accelerate at rate a = 14.7 Newtons/(7 kg) = 2.1 meters/second ^ 2.
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RESPONSE --> self critique assessment:
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23:41:07 `q003. Answer the same question as that of the preceding problem, but this time assume that the 5 kg object is not on a level tabletop but on an incline at an angle of 12 degrees, and with the incline descending in the direction of the pulley.
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RESPONSE --> Fn = 5 * 9.8 = 49 Ffr = .1 * 49 = 4. 9 pully F = 2 * 9.8 = 19.6 Fparallel = 49cos258 = -10.2 Fperpindic.= 49sin258 = -48 Fnet = .10.2 - 4.9 + 19.6 Fnet = 4.5 a = 4.5N / 7kg a = .64m/s/s This answer does not seem right I think my angle is incorrect confidence assessment: 2
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23:47:46 In this case you should have drawn the incline with the x axis pointing down the incline and the y axis perpendicular to the incline. Thus the x axis is directed 12 degrees below horizontal. As a result the weight vector, rather than being directed along the negative y axis, lies in the fourth quadrant of the coordinate system at an angle of 12 degrees with the negative y axis. So the weight vector makes an angle of 270 degrees + 12 degrees = 282 degrees with the positive x axis. The weight vector, which has magnitude 5 kg * 9.8 meters/second ^ 2 = 49 Newtons, therefore has x component 49 Newtons * cosine (282 degrees) = 10 Newtons approximately. Its y component is 49 Newtons * sine (282 degrees) = -48 Newtons, approximately. The incline exerts sufficient force that the net y component of the force on the block is zero. The incline therefore exerts a force of + 48 Newtons. Friction exerts a force which is .10 of this force, or .10 * 48 Newtons = 4.8 Newtons opposed to the direction of motion. Assuming that the direction of motion is down the incline, frictional force is therefore -4.8 Newtons in the x direction. The weight component in the x direction and the frictional force in this direction therefore total 10 Newtons + (-4.8 Newtons) = + 5.2 Newtons. This force tends to accelerate the system in the same direction as does the weight of the 2 kg mass. This results in a net force of 5.2 Newtons + 19.6 Newtons = 24.8 Newtons on the 7 kg system. The system therefore accelerates at rate {} a = (24.8 Newtons) / (7 kg) = 3.5 meters/second ^ 2, approximately.
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RESPONSE --> i knew my angle was wrong I considered 180-12 but that did not appear correct either. It is obvious my mistake could have been easily reasoned out. Fparallel = 49cos282 = 10.2 Fperpind = 49sin282 = 48 Fnet = 10 - 4.8 + 19.6 a = 24.1 /7 = 3.44 I knew that the accel was not correct because not only was the block moving down the incline but also in the direction of the F exerted by the pulley therefore common sense tells me the block would be moving faster. self critique assessment: 3
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