course Phy 201 |}}s¯ӟͱdassignment #012
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15:00:13 `questionNumber 120000 `q001. Note that there are 4 problems in this set. Two 3 kg masses are suspended over a pulley and a 1 kg mass is added to the mass on one side. Friction exerts a force equal to 2% of the total weight of the system. If the system is given an initial velocity of 5 m/s in the direction of the lighter side, how long will a take the system to come to rest and how far will it travel?
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RESPONSE --> Well, I know that to find the weight I have to multiply the total mass of the system by 9.8m/s^2. So, 9.8m/s^2*7kg=68.6N To find the friction, which is two percent, I multiply 68.6N * .02=1.172N 68.6-1.172=67.228N=Fnet I am a bit confused about what constitutes the mass of the system, whether it is the 7kg total or whether it is the 4 kg hanging on the heaviest side. I am going to go with the 7kg since the problem indicates that the Force is equal to the ""total weight of the system."" So, a=Fnet/m=67.228N/7kg=9.604 m/s^2 We know the initial velocity, which we are going to call -5m/s since it is moving in the opposite direction of the weight. The system has to make a change-over from going toward the lighter side to going toward the heavier side, and at this changeover the velocity will equal 0. So, vf=0. 0-(-5)=5m/s='dv aAve='dv/'dt 9.604m/s^2=5m/s/'dt 9.604m/s^2*'dt=5m/s 'dt=.52 seconds Now, vf+vo/2=0-5/2=-2.5=vAve -2.5m/s*.52seconds=-1.3m This is just a totally random, trying what little bit I know type of thing. confidence assessment: 1
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15:07:59 `questionNumber 120000 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 --> Oh no! I made my mistake first thing: I put 7kg instead of 4, when I should have KNOWN it was 4 since the force of the system was acting on the heavier of the sides. Since I didn't do this, I also didn't calculate the forces that countered or balanced out the other forces, for example, the 3kg balancing out 3 of the 4kg on the other side. I am so excited! Even though I didn't get the right answer, this is the closest I have come so far to setting up an equation like this by myself. Maybe it is starting to sink in... self critique assessment: 2
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15:26:46 `questionNumber 120000 `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 --> We already know that the acceleration is 1.6m/s^2. Since vo=-5m/s and vo, the turning aroudn point, is 0, I can calculate vAve, we we know from last time is -2.5m/s. To find starting points and stopping points, I think Imight have to use instantaneous velocity and solve backward. Since vo=-5 -5*lim->0='ds (I chose time zero since that is from when the time period 3.1seconds is started). Oh darn! I can't figure it out. Nothing seems to be making sense. confidence assessment: 0
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16:12:31 `questionNumber 120000 Since the acceleration of the system is different if it is moving in the positive direction than when it is moving in the negative direction, if we have both positive and negative velocities this problem must the separated into two parts. As seen in the previous example, in the first 3.2 seconds the system displaces -7.8 meters. This leaves 6.8 seconds after that instant during which the system may accelerate from rest in the positive direction. We therefore analyze the motion from the instant the system comes to rest until the remaining .8 seconds has elapsed. The frictional force during this time will be negative, as it must oppose the direction of motion. The net force on the system will therefore be + 39.2 Newtons -29.4 Newtons - 1.4 Newtons = 8.4 Newtons, and the acceleration will be 8.4 Newtons / (7 kg) = 1.2 m/s^2, approx.. The initial velocity during this phase is 0, the time interval is 6.8 sec and the acceleration is 1.2 m/s^2. We therefore conclude that the velocity will change by 1.2 m/s^2 * 6.8 sec = 8.2 m/s, approx, ending up at 8.2 m/s since this phase started at 0 m/s. This gives an average velocity of 4.1 m/s; during 6.8 sec the object therefore displaces `ds = 4.1 m/s * 6.8 sec = 28 meters approx.. These results could have also been easily obtained from equations. For the entire 10 seconds, the displacements were -7.8 meters and +28 meters, for a net displacement of approximately +20 meters. That is, the system is at this instant about 20 meters in the direction of the 4 kg mass from its initial position.
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RESPONSE --> I was totally off. I am going to try to put in my own words what the procedure is. First we have to determine when the velocity is going in one direction versus when it is going in another before we can determine what distance takes place in which time periods. From before, we know that the system moves -7.8 m in 3.2 seconds. Of the 10 seconds given in the problem, 6.8 seconds are left for the system to turn around and move in the positive direction. Then, we recalculate the acceleration and use it to get the new vAve, from the remaining time period. We then find the new vf. From this and the vo, we find the vAve. We then take the displacement found from the new time period and the new vAve. This is confusing. By referencing this again, I think I could do it myself, though. self critique assessment: 2
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22:55:50 `questionNumber 120000 `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 --> 1400kg*9.8m/s^2=13720N=Fnet of gravity 13720N-137.2N=13582.8N Fnet=m*a a=Fnet/m a=13582.8N/1400kg=33.96m/s^2 5m/s='ds/'dt 5m/s=100m/'dt 5m/s*'dt=100m 'dt=20s a=vf-vo/'dt 33.96m/s^2=vf-5m/s/20s 679.2=vf-5m/s 684.2m/s=vf confidence assessment: 1
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23:11:47 `questionNumber 120000 We are given the initial velocity and displacement of the automobile, so we need only find the acceleration and we can analyzes problem as a standard uniform acceleration problem. The automobile experiences a net force equal to the component of its weight which is parallel to the incline plus the force of friction. If we regard the direction down the incline as positive, the parallel component of the weight will be positive and the frictional force, which must be in the direction opposite that of the velocity, will be negative. The weight of the automobile is 1400 kg * 9.8 m/s^2 = 13720 Newtons, so the component of the weight parallel to the incline is parallel weight component = 13720 Newtons * .05 = 686 Newtons. The normal force between the road and the tires is very nearly equal to the weight of the car because of the small slope, so the magnitude of the frictional force is approximately frictional force = 13720 Newtons * .01 = 137 Newtons, approx.. The frictional force is therefore -137 Newtons and the net force on the automobile is Fnet = 686 Newtons - 137 Newtons = 550 Newtons (approx.). It follows that the acceleration of the automobile must be a = Fnet / m = 550 Newtons / 1400 kg = .4 m/s^2 (approx.). We now have a uniform acceleration problem with initial velocity v0 = 5 meters/second, displacement `ds = 100 meters and acceleration a = .4 m/s^2. We can easily find the final velocity using the equation vf^2 = v0^2 + 2 a `ds, which gives us vf = +- `sqrt(v0^2 + 2 a `ds) = +- `sqrt( (5 m/s)^2 + 2 * .4 m/s^2 * 100 m) = +- `sqrt( 25 m^2 / s^2 + 80 m^2 / s^2) = +-`sqrt(105 m^2 / s^2) = +- 10.2 m/s. It is obvious that the final velocity in this problem is the positive solution +10.2 m/s, since the initial velocity and acceleration are both positive.
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RESPONSE --> Well, I made two major mistakes. First, my acceleration was wrong, because I forgot to convert the mass to Newtons (I dont' know how I managed that considering I did convert the net force, but...).
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RESPONSE --> Ffric=-137 N 686N in this case will be negative, since it is going in the direction opposite of the velocity, as will the friction. So, I guess the net force will be ""way"" negative since it will be -686N-137N= -820N Then, since we know the mass and the Net force, we can find acceleration by dividing the Fnet by the mass. a=Fnet/mass =-820N/1400kg=-.58m/s^2 Since we know that vo=11.2 m/s from the equatoin and that vf=0m/s from deduction, we can find 'dt by using the equation a=vf-vo/'dt by rearranging and filling in what we do know. 'dt=vf-vo/a =-11.2/-.58=19.31 seconds Since we can find the average velocity (vo + vf/2), we can use the time and the vAve to find 'ds. vAve=11.2/2=5.6m/s 5.6m/s='ds/19.31seconds 108.14meters='ds confidence assessment: 2
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