course Phy 121 |o}assignment #012
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14:06:38 We know the initial velocity of the system and, for the period until it comes to rest, we know that its final velocity will be 0 m/s. If we can find the acceleration, we will have three of the five necessary variables with which to analyze the motion and we can therefore determine the time interval `dt and displacement `ds corresponding to this period. We begin by analyzing the forces acting on the system. Before we do so we declare the positive direction of motion for this system to be the direction in which the system moves as the greater of the two hanging masses descends, i.e., the direction of the net force on the system Gravity exerts forces of 4 kg * 9.8 m/s^2 = 39.2 Newtons on the 4 kg mass and 3 kg * 9.8 m/s^2 = 29.4 Newtons on the 3 kg mass. Taking the positive direction to be the direction in which the system moves when the 4 kg mass descends, as stated earlier, then these forces would be +39.2 Newtons and -29.4 Newtons. The total mass of the system is 7 kg, so its total weight is 7 kg * 9.8 m/s^2 = 68.4 Newtons and the frictional force is therefore frictional force = .02 * 68.4 Newtons = 1.4 Newtons, approx.. If the system is moving in the negative direction, then the frictional force is opposed to this direction and therefore positive so the net force on the system is +39.2 Newtons - 29.4 Newtons + 1.4 Newtons = +11.2 Newtons. This results in an acceleration of +11.2 N / (7 kg) = 1.6 m/s^2. We now see that v0 = -5 m/s, vf = 0 and a = 1.6 m/s^2. From this we can easily reason out the desired conclusions. The change in velocity is +5 m/s and the average velocity is -2.5 m/s. At the rate of 1.6 m/s^2, the time interval to change the velocity by +5 m/s is `dt = +5 m/s / (1.6 m/s^2) = 3.1 sec, approx.. At an average velocity of -2.5 m/s, in 3.1 sec the system will be displaced `ds = -2.5 m/s * 3.1 s = -7.8 meters. These conclusions could also have been reached using equations: since vf = v0 + a `dt, `dt - (vf - v0) / a = (0 m/s - (-5 m/s) ) / (1.6 m/s^2) = 3.1 sec (appxox). Since `ds = .5 (v0 + vf) * `dt, `ds = .5 (-5 m/s + 0 m/s) * 3.1 s = -7.8 meters.
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RESPONSE --> once i figured how to get the accelration which is 1.6 m/s^2 then i could apply that to the formula to figure both the 'dt and the 'ds but i looked ahead to get it. self critique assessment: 3
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14:18:19 `q002. If the system in the previous example was again given an initial velocity of 5 m/s in the direction of the 3 kg mass, and was allowed to move for 10 seconds without the application of any external force, then what would be its final displacement relative to its initial position?
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RESPONSE --> i used the formula 'ds=v0'dt + .5a 'dt^2 and got the 'ds to be 110 m but it had to have gone forward and then reversed direction i set up a graph on the times based on the acceleration and got that the relative time would be -4m if the direction was initially positive confidence assessment: 1
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14:27:05 `q003. An automobile has a mass of 1400 kg, and as it rolls the force exerted by friction is .01 times the 'normal' force between its tires and the road. The automobile starts down a 5% incline (i.e., an incline with slope .05) at 5 m/s. How fast will it be moving when it reaches the bottom of the incline, which is 100 meters away (neglect air friction and other forces which are not part of the problem statement)?
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RESPONSE --> v0=5m/s m=1400 kg friction = 14 kg 'ds=100m force = 9.8m/s^2 * 1386kg f=13583 N a=679 m/s confidence assessment: 0
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14:35:13 `q004. If the automobile in the previous example started at the bottom of the incline with velocity up the incline of 11.2 m/s, how far up the hill would it be able to coast?
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RESPONSE --> v0=11.2m/s m=1400kg pw= -686 N? fnet= -823 N? a= -.6m/s^2 vf=0 .6m/s^2=11.2m/s/'dt 'dt=6.7s 'ds=11.2m/s * 6.7s 'ds= 75m confidence assessment: 2
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~іKǿg assignment #012 012. `query 12 Physics I 07-04-2007
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11:55:04 Query set 3 #'s 13-14 If an object of mass m1 rests on a frictionless tabletop and a mass m2 hangs over a good pulley by a string attached to the first object, then what forces act on the two-mass system and what is the net force on the system? What would be the acceleration of the system? How much would gravitational PE change if the hanging mass descended a distance `dy?
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RESPONSE --> Fnet would be 9.8 m/s * m2 a would by a = Fnet/m1+m2 a change of 'dy would mean the PE change would be 'PE=Fnet/'dy ? confidence assessment: 1
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11:56:11 How would friction change your answers to the preceding question?
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RESPONSE --> we would have to multiply friction, say 2%, by the weight which would change the Fnet confidence assessment: 2
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11:58:34 Explain how you use a graph of force vs. stretch for a rubber band to determine the elastic potential energy stored at a given stretch.
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RESPONSE --> if the force vs stretch produced a straight, not accelerating and not decelerating line confidence assessment: 1
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12:00:12 STUDENT QUESTIONS: Does the slope of the F vs stretch graph represent something? Does the area under the curve represent the work done? If so, is it work done BY or work done ON the rbber bands?
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RESPONSE --> it would be the work done on the system wouldn't it? confidence assessment: 1
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12:00:53 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> the only thing confusing me right now is how exactly to get PE in a given situation. i shall study that. self critique assessment: 3
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yc䅲 assignment #013 013. `query 13 Physics I 07-04-2007
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15:26:07 prin phy and gen phy problem 4.02 net force 265 N on bike and rider accelerates at 2.30 m/s^2, mass of bike and rider
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RESPONSE --> 2.30 m/s^2=265N/m mass=115 kg confidence assessment: 3
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15:30:58 prin phy and gen phy problem 4.07 force to accelerate 7 g pellet to 125 m/s in .7 m barrel
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RESPONSE --> mass=.0007kg 'ds=.7m v0=0 vf=125 m/s initial KE=0 final KE =5.5 J F * .7m = 5.5J Fnet = 7.8 N confidence assessment: 2
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15:32:48 gen phy 4.08. breaking strength 22 N, accel 2.5 m/s^2 breaks line. What can we say about the mass of the fish?
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RESPONSE --> 2.5=22/m the fish weighs more than 8.8 kg? that is a whopper if that is right confidence assessment: 2
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