course Phy 201 ?????????????}??Student Name: assignment #020
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23:03:27 `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 --> Th 5 kg block will not contribute to the acceleration because it is supported by the table. In order to find the acceleration, I must first find the gravitational force on the 2 kg block. W = 2kg * 9.8m/sec^2 = 19.6N Now I can use the equation F = m * a to solve for a. 19.6N = (2kg+5kg) * a a=2.8m/sec^2
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23:03:46 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 --> ok
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23:07:49 `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 --> I have to find the force between the tabletop and the block. W = 5kg * 9.8m/sec = 49N If friction is .10 times the force between the table and the block: friction = 49N*.1 = 4.9N Net force = gravitional force - frictional force net force = 19.6N - 4.9N = 14.7N Now, I will solve for acceleration using F = m * a 14.7N = 7kg * a a = 2.1m/sec^2
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23:08:23 `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 -->
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23:10:53 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 --> oops, somehow, I skipped the question part of this one. I did not mean to. I will look this problem over. Wow, that was a hard one anyways. This seems to combine all of the concepts together.
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a???????z????? assignment #020 ???????????Physics I 07-07-2006
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23:16:55 Explain how we get the components of the resultant of two vectors from the components of the original vectors.
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RESPONSE --> We get the resultant components of the two vectors by simply adding the x components together and the y components together.
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23:17:12 ** If we add the x components of the original two vectors we get the x component of the resultant. If we add the y components of the original two vectors we get the y component of the resultant. **
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RESPONSE --> ok
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23:18:21 Explain how we get the components of a vector from its angle and magnitude.
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RESPONSE --> x component = magnitude * cos(angle) y component = magnitude * sin(angle
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23:18:33 ** To get the y component we multiply the magnitude by the sine of the angle of the vector (as measured counterclockwise from the positive x axis). To get the x component we multiply the magnitude by the cosine of the angle of the vector (as measured counterclockwise from the positive x axis). **
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RESPONSE --> ok
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23:41:31 prin phy, gen phy 7.02. Const frict force of 25 N on a 65 kg skiier for 20 sec; what is change in vel?
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RESPONSE --> First I will find the impulse of the ofrce on the skiier. Impulse = -25N * 20sec = -500N Impulse = Change in momentum and change in momentum = m * dv -500N = 65kg * dv dv = -7.69m/sec
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23:41:39 If the direction of the velocity is taken to be positive, then the directio of the frictional force is negative. A constant frictional force of -25 N for 20 sec delivers an impulse of -25 N * 20 sec = -500 N sec in the direction opposite the velocity. By the impulse-momentum theorem we have impulse = change in momentum, so the change in momentum must be -500 N sec. The change in momentum is m * `dv, so we have m `dv = impulse and `dv = impulse / m = -500 N s / (65 kg) = -7.7 N s / kg = -7.7 kg m/s^2 * s / kg = -7.7 m/s.
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RESPONSE --> ok
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23:42:02 gen phy #7.12 23 g bullet 230 m/s 2-kg block emerges at 170 m/s speed of block
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RESPONSE --> i have no clue.
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23:46:02 **STUDENT SOLUTION: Momentum conservation gives us m1 v1 + m2 v2 = m1 v1' + m2 v2' so if we let v2' = v, we have (.023kg)(230m/s)+(2kg)(0m/s) = (.023kg)(170m/s)+(2kg)(v). Solving for v: (5.29kg m/s)+0 = (3.91 kg m/s)+(2kg)(v) .78kg m/s = 2kg * v v = 1.38 kg m/s / (2 kg) = .69 m/s. INSTRUCTOR COMMENT: It's probably easier to solve for the variable v2 ': Starting with m1 v1 + m2 v2 = m1 v1 ' + m2 v2 ' we add -m1 v1 ' to both sides to get m1 v1 + m2 v2 - m1 v1 ' = m2 v2 ', then multiply both sides by 1 / m2 to get v2 ` = (m1 v1 + m2 v2 - m1 v1 ' ) / m2. Substituting for m1, v1, m2, v2 we will get the result you obtained.**
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RESPONSE --> I know that I should have this mastered by now but can you just let me know exactly how this equation works. on the left side I think m1v1 corresponds to the mass and velocity of the forst object and m2v2 corresponds to the mass and velocity of the second object. I dont quite understand what goes where.
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