course Phy 231 6/21 5:15 012. Problems involving motion and force.
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Given Solution: 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. • Our first step is to choose the positive direction. The system can move in one of two directions, the direction in which the greater mass descends while the lesser ascends, or the direction in which the greater mass ascends while the lesser descends. • Either choice is fine, but once we make the choice all directional quantities will have positive or negative signs relative to this direction. • We arbitrarily choose the positive direction for the present system to be the direction in which the system moves as the greater of the two hanging masses descends. Thus the positive direction is the one in which the 4 kg mass is descending. 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.. The system is given an initial velocity in the direction which makes the lesser mass descend. This direction is opposite the direction we have chosen here to be positive, so the initial velocity is negative. The frictional force will tend to oppose the motion of the system. So if the system is moving in the negative direction, then the direction of the frictional force is positive. The frictional force is thus +1.4 Newtons. The net force on the system is therefore • F_net = +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. At this point we know 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. Self-critique (if necessary): OK Self-critique rating #$&* OK ********************************************* Question: `q002. If the system in the previous example was again given an initial velocity of 5 m/s in the direction of the 3 kg mass, and was allowed to move for 10 seconds, then what would be its final displacement relative to its initial position? Your solution: Position = -7.82m V0 = 0m/s Net force = 39.2N-29.4N-1.372N = 8.43N 8.43N = 7kg(a) A = 8.43m/s^2 kg /7kg = 1.2m/s^2 Dt = 10s-3.13s = 6.87s Dv = 1.2m/s^2 * 6.87s = 8.24m/s = final velocity vAve= 8.24m/s / 2 = 4.12m/s ds = 4.12m/s*6.87s = 23.8m from the initial position Confidence rating #$&* 2
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Given Solution: The acceleration of the system is different if it is moving in the positive direction than when it is moving in the negative direction. We only know how to analyze a uniform-acceleration situation. Since 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 6.8 seconds has elapsed. Recall our choice of positive direction, made in the preceding problem. To avoid having to reorient ourselves, we will continue to use this choice: • We arbitrarily chose the positive direction for the present system to be the direction in which the system moves as the greater of the two hanging masses descends. Thus the positive direction is the one in which the 4 kg mass is descending. During the final 6.8 seconds the system will be moving in the positive direction, with the 4 kg mass descending. The frictional force during this time is directed opposite the motion so the frictional force will be negative. The net force on the system will therefore be F_net = + 39.2 Newtons -29.4 Newtons - 1.4 Newtons = 8.4 Newtons, and the acceleration will be a = F_net / m = 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 `dv = 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 = v_Ave * `dt = 4.1 m/s * 6.8 sec = 28 meters approx.. These results are easily reasoned out from the definitions of velocity and acceleration. They could have also been obtained equally easily using the 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. The direction of the frictional force depends on the direction in which the system is moving. The system here moves in one direction, reverses, and moves in another. So the frictional force changes when the direction of motion reverses. Self-critique (if necessary): OK Self-critique rating #$&* OK ********************************************* Question: `q003. An automobile has a mass of 1400 kg, and as it rolls the force exerted by friction is .01 times the 'normal' force between its tires and the road. The automobile starts down a 5% incline (i.e., an incline with slope .05) at 5 m/s. How fast will it be moving when it reaches the bottom of the incline, which is 100 meters away (neglect air friction and other forces which are not part of the problem statement)? Your solution: Force = 1400kg(9.8m/s^2) = 13720N Friction = 13720N*.01 = -137.2N Parallel weight: 13720*.05 = 686N Net force = 686N-137.2N = 549N 549N = 1400kg(a) A = 548 m/s^2 kg / 1400kg = .391m/s^2 V0 = 5m/s Ds = 100m Vf^2 = (5m/s)^2 +2(.391m/s^2)(100m) Vf^2 = 25m^2/s^2 +78.2m^2/s^2 Vf^2 = 103.2m^2/s^2 Vf = 10.2m/s Confidence rating #$&* 3
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Given Solution: 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. On an incline the road doesn't push straight up--it pushes in a direction perpendicular to the road surface, so it doesn't completely balance the gravitational force. On a small incline it still balances most of the gravitational force, but not all. A component of the gravitational force remains. (These force components are analyzed in upcoming assignments using vectors; for right now we simply make the statement that the force component parallel to the (small) incline is approximately equal to the slope multiplied by the weight). Self-critique (if necessary): OK Self-critique rating #$&* OK ********************************************* Question: `q004. If the automobile in the previous example started at the bottom of the incline with velocity up the incline of 11.2 m/s, how far up the hill would it be able to coast? Your solution: Force = 1400kg(-9.8m/s^2) = -13720N Friction = -13720N*.01 = -137.2N Parallel weight: -13720*.05 = -686N Net force = -686N-137.2N =-823.2N -823.2N = 1400kg(a) A = -823m/s^2 kg / 1400kg = -.588m/s^2 Vf= 0m/s V0 = 11.2m/s 0m/s = 11.2m/s -.588m/s^2(dt) -11.2m/s = -.588m/s^2 dt Dt = 19.0s Ds = (11.2m/s+0m/s / 2)(19s) Ds = 5.6m/s * 19s = 106.4m Confidence rating #$&* 3
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Given Solution: 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 now take the upward direction as positive. The frictional force will still be 137 Newtons and will again be directed opposite the velocity, and 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 our chosen positive direction 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 now 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). Self-critique (if necessary): OK Self-critique rating #$&* OK