course Phy 231 ž´ˆÉó«èÙ¥ã•ÝóЂÓZðÕÛÁÙ’¸ôw¡²assignment #012
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21:02:26 `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 --> Frictional force = 7 kg * .02 * 9.8 m/s^2 = 1.372 Newtons Fnet on the system is force exerted by gravity - therefore: Fnet = m * a - Ffrict Fnet = 7 kg * 9.8 m/s^2 - 1.372 N = 67.228 Newtons acceleration of the system is: a = Fnet / m = 67.228 Newtons / 7 kg = 9.604 m/s^2 vf^2 = v0^2 + 2 a 'ds 'ds = (vf^2 - v0^2) / 2 a 'ds = (0 - (5 m/s)^2) / 2 * 9.604 m/s^2 'ds = -25 m^2/s^2 / 19.208 m/s^2 'ds = -1.302 m ***but why is it negative?? not sure about that one... 'dt = (vf - v0) / a 'dt = [0 m/s - 5 m/s] / 9.604 m/s^2 'dt = - 5 m/s / 9.604 m/s^2 = - 0.521 s *** again, not sure how I'm getting a negative or why - should the acceleration be negative, thereby eliminating the negative produced from the starting to ending velocities? confidence assessment: 1
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21:06:54 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 --> OK, interesting. So you separated the two masses and calculated the force for each. I started to do this, but after reading through Set 3, prob #13, I reversed course and used the Fnet for the entire mass. I was able to calculate the Frictional Force correctly, but naturally my acceleration for the system was WAY off because I did not calculate the separate Fnet's for each mass. I'm still not clear however on my the change in velocity is a positive integer - shouldn't this be negative if 'dv = vf - v0 ?? I understand how 'ds and 'dt were calculated, though my answers differed because of improper acceleration value.
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21:27:46 `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 --> Force = 3 kg * 9.8 m/s^2 = 29.4 Newtons Force = 4 kg * 9.8 m/s^2 = 39.2 Newtons Fnet = 39.2 Newtons - 29.4 Newtons = 9.8 Newtons (assuming no frictional force as stated in the question) Acceleration of the system would therefore be: a = Fnet / m = 9.8 Newtons / 7 kg = 1.4 m/s^2 given that v0 = 5 m/s, 'dt = 10 s, and a = 1.4 m/s^2, we can now find 'ds: 'ds = v0 'dt + 0.5 a * 'dt ^2 'ds = 5 m/s * 10 s + 0.5 * 1.4 m/s^2 * (10 s)^2 'ds = 50 m + 0.7 m/s^2 * 100 s^2 'ds = 50 m + 70 m = 120 m confidence assessment: 2
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21:34:19 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 --> OH I thought the problem stated there was no frictional force now... Both the calculation of Fnet and acceleration are apparent and appear easy, now that I've seen the correct #'s to plug in. I took note that v0 for this 2nd phase is 0 - that makes sense because the vf from the 1st stage was 0 m/s after 3.2 s. Having read through your solution, I seem to understand how you structured it, though prior to seeing it worked out, I was obviously way off in my calculation. I guess my real problem in solving these questions is in determining how best to set the problem up. The calculations themselves seem reasonably easy, but setting up the problem in the correct manner is throwing me off every time! I've read and taken notes in all of the ""ideas"" and ""problem sets"" and I try to draw diagrams and the such. Any suggestions for improvement here?
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21:43:35 `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 --> Ffrict = 1400kg * 9.8 m/s^2 * .01 = 137.2 Newtons Fnet = m * a - Ffrict Fnet = 1400 kg * 9.8 m/s^2 - 137.2 Newtons Fnet = 13,582.8 m/s^2 acceleration of system = Fnet / m a = 13,582.8 m/s^2 / 1400 kg = 9.702 m/s^2 v0 = 5 m/s 'ds = 100 m m = 1400 kg a = 9.702 m/s^2 find vf: vf^2 = v0^2 + 2 a 'ds vf^2 = ( 5 m/s )^2 + 2 * 9.702 m/s^2 * 100 m vf^2 = 25 m^2/s^2 + 1940.4 m^2/s^2 vf = sqrt( 1965.4 m^2 / s^2 ) vf = 44.33 m/s confidence assessment: 2
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21:49:49 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 --> Darn it..... I believe I understood everything, with the exception of how the slope of the incline played into it. ""Parallel weight component"" - OK, I have committed this to my notes. So in addition to the frictional force, we have this parallel weight component force, which is essentially the WEIGHT of the auto multiplied by the slope of the incline. Also interesting to note that the frictional force is unaffected the slope of the incline. However, the frictional force in the solution is a negative integer, which is actually contrary to what I would have originally thought. Parallel component force less frictional force (or plus a negative frictional force) = our Fnet for this problem. I understand how the accel was calculated and how the additional calculations for vf are calculated. self critique assessment: 2
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22:04:09 `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 = 11.2 m/s Fnet = 550 Newtons a = ? vf = 0 m/s m= 1400 kg weight = 1400kg * 9.8 m/s^2 = 13, 720 Newtons The parallel weight component and frictional force are now operating in the opposite direction. Fnet = -686 Newtons + 137 Newtons = -549 Newtons a = Fnet / m = -549 Newtons / 1400 kg = - 0.4 m/s^2 'dt = ( vf - v0 ) / a dt = - 11.2 m/s / - 0.4 m/s^2 = 28 s 'ds = v0 'dt + 0.5 a 'dt^2 'ds = 11.2 m/s * 28 s + 0.5 * -0.4 m/s^2 * (28 s)^2 'ds = 313.6 m - 156.8 m = 156.8 m confidence assessment: 2
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22:09:10 We again have a uniform acceleration situation where the initial velocity is known and we have the information to determine the acceleration. The other quantity we can deduce is the final velocity, which at the furthest point up the hill will be zero. Since the automobile is coasting up the incline, we will take the upward direction as positive. The frictional force will still be 137 Newtons and will again be directed opposite the velocity, so will therefore be negative. The parallel component of the weight will be the same as before, 686 Newtons, but being directed down the incline will be in the direction opposite to that if the velocity and will therefore also be negative. The net force on the automobile therefore be net force = -686 Newtons - 137 Newtons = -820 Newtons (approx.). Its acceleration will be a = Fnet / m = -820 Newtons / 1400 kg = -.6 m/s^2 (approx.). We see now that v0 = 11.2 m/s, vf = 0 and a = -.6 m/s^2. Either by direct reasoning or by using an equation we easily find that `dt = (-11.2 m/s) / (-.6 m/s^2) = 19 sec (approx) and `ds = (11.2 m/s + 0 m/s) / 2 * 19 sec = 5.6 m/s * 19 sec = 106 meters (approx).
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RESPONSE --> OK, so both frictional force AND parallel component force are negative? It is this key mistake that once again led me astray. Had I done that part correctly, I would have gotten the answer correct!! It seems there is always something about the set up of the problems that I've left out or done slightly incorrectly - perfect example is the Fnet of this problem. I would have suspected that the Ffrict was now in the opposite direction and would therefore be positive, but I guess it makes imperical sense that frictional force is always negative? I feel like I'm getting closer to a point where my understanding for the material is improved, but I can't quite seem to make it over the hill, where I'm understanding the questions/problems to the extent that I don't make key mistakes like in this problem. Maybe it's as simple as going to the text and doing additional problems until my comprehension is better. Do you have any suggestions?
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