course Phy 201 •™yÄÛšªÅ“wæxµûË©ê›çwÙ‹ÜÖû¿³assignment #020
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21:58:55 `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 --> For the 5 kg mass on the table, I could find the force, but overall the force will be zero between the force of the mass acting on the table and the force of the table acting on the mass. For the 2kg mass, however, it is not resting on anything; so, the only force acting on it is gravity. 2kg*9.8m/s^2=Fnet for the 2 kg mass 19.6 N=force So, for both masses, the total force is 19.6 N, and the total mass is 2 kg + 7kg=9 kg. A=Fnet/mass =19.6N/7kg =2.8m/s^2 confidence assessment: 3
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21:58:57 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 --> self critique assessment: 3
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21:59:12 `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 --> 5kg*9.8m/s^2=49N 49N*.10N=4.9N-frictional force Fnet=19.6N-4.9N=14.7N A=Fnet/mass =14.7N/7kg =2.1 m/s^2 confidence assessment: 3
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21:59:15 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: 3
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21:59:31 `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 --> Force=5kg*9.8m/s^2=49N We are given the 12 degree angle, but I am not sure how to find the other angle. I need the angle, I know, to plug into the equations for x and y (magnitude*cos/sin(angle)), but I don't know how to find it. confidence assessment:
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I am still confused on how the 270 degrees was derived. I understand that the 12 and 270 angles were added to get 282; however, how was 270 derived? The rest of this makes perfect sense… self critque assessment: 0
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