Asst 12

In answer to your question about testing, in case you didn't see the message in last night's posting:

You need to get me the email address of someone in the testing center so I can send them the information they will need. This process usually works very smoothly at BRCC.

As you note at the end it does take time to absorb this material and to develop your problem-solving techniques.

See my notes and let me know if you have additional questions.

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20:03:48 `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 --> I am drawing a blank on how to set it up but here is what I have. Total mass = 7 kg, v0 = 5 m/s, vf = 0 m/s, mass of the friction is 0.14 kg and the force of the friction is 1.372 Newtons.

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20:06:51 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 I am still having a little trouble setting things up.

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20:16:03 `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 --> First we have the v0 = 5 m/s the mass = 3 kg and the 'dt = 10 seconds. Also the vf will at the end = 0 m/s. We know that it will fall freely at a rate of 9.8 m/s^2. The vAve will = 0 m/s - 5 m/s = - 2.5 m/s. Then if moving at a constant rate the 'ds would be 10 s / - 2.5 m/s = 4 meters of displacement from its initial position.

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20:18:51 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 --> Ok I totally misunderstood that question. I was looking at it as if the motion was in the direction of the lighter side for the entire 10 seconds.

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20:46:46 `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 --> The force the car is moving at is Fnet = 1400 kg * 9.8 m/s^2 = 13720 Newtons. So the friction is .01 times the normal force which is .01 * 13720 = 137.2 Newtons. The v0 = 5m/s and the 'dt = 100 meters. So we need to find the acceleration of the car which if the friction is opposing the force of the car then it will be negative. So we have 13720 N - 137.2 N = 13582.8 Newtons. So then the acceleration is = 13582.8 Newtons / 1400 kg = 9.702 m/s^2. So with this we can say that : 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 + 19.404 m/s^2(100 m) vf^2 = 25 m^2/s^2 + 1940.4 m^2/s^2 vf^2 = 1965.4 m^2/s^2 vf = 44.332 m/s I would say that is how fast it will be going but I didn't take the 5% incline into account and don't know how to.

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20:50:58 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 --> Ok I see exactly what I did. If it is a 5% incline then it is not a free fall. So instead of using the full force I had to place the incline into the equation to get the Fnet. I thought I had the Fnet when I calculated the mass to the free fall acceleration. I do understand it better now and can see that with practice I will get this.

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21:11:28 `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 --> First off the v0 = 11.2 m/s, m = 1400 kg, friction = .01 the mass, and vf = 0 m/s. The way I see it is the car will have all of its mass, the friction and the incline opposing its incline. So this would be 13720 Newtons + 137.2 Newtons + 686 Newtons = 14543.2 Newtons and this would be a negative force so then you could say 14543.2 N / 1400 kg = - 10.388 m/s^2. Now you can say that : vf^2 = v0^2 + 2 * a * 'ds (0 m/s)^2 = (11.2 m/s)^2 + 2 ( - 10.388 m/s^2) * 'ds 'ds = 1.08 meters

That 13720 Newtons is not in the direction of motion; the 686 Newtons is the component of that force in the direction of motion. You need the 686 Newtons in the equation, but the 13720 Newtons shouldn't be there.

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21:12:20 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 --> I ran into trouble when I figured the mass of the car into the equation.

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���Ԙ�������vߕ�j����z�^�[���{� assignment #001 yE�܃���������݌��֭ Liberal Arts Mathematics I 11-21-2005

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19:01:38 Problem Number 1 A coasting cart has velocities 4.5 m/s, 10 m/s and 2.25 m/s at clock times t = 0 s, 9 s and 13 s. Approximately how far does it move and at what average rate does it accelerate over each of the two time intervals (i.e., 0- 9 seconds and 9 - 13 seconds)? Show how you could use a graph of velocity vs. time to obtain your results. Directly reason out your results using the concept of rate. If the automobile is coasting, and if air resistance is not a significant factor, then over which time interval do you think the average slope of the road is greatest?

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RESPONSE --> For the 0-9 time interval the 'dv would be 5.5 m/s. So the aAve = 5.5 m/s / 9 s = 0.61 m/s^2. The 9-13 time intervals' 'dv = - 7.75 m/s. The aAve = - 7.75 m/s / 4 s = - 1.9375 m/s^2. As for the distance traveled in the 0-9 time interval the v0 = 4.5 m/s and the vf = 10 m/s. So the vAve = 7.25 m/s. The 'ds = 65.25 meters. The 9-13 time intervals v0 = 10 m/s and the vf = 2.25 m/s making the vAve = -3.875. So the 'ds = -0.96875 meters. Since it 'dt can't be negative the answer is 0.96875 meters.

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19:06:58 Problem Number 2 We set up a straight ramp and measure the time required to coast down the ramp, from rest, at various slopes. The differences in elevation between one end and the other, for the different slopes, are 1.8, 4.2 and 6.7 cm. The time required for a cart to coast 78 cm down the ramp, starting from rest, is 2.722297 seconds on the first incline, 2.518101 seconds on the next, and 2.6606 seconds on the last. How well do these results support the hypothesis that acceleration is linearly dependent on slope?

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RESPONSE --> This is not a real good representation of that hypothesis. The incline increases with each trial so the time intervals should decrease, but the time you have for the third one is longer than that of the second one which has a less of an incline. I would say that the time for the second is flawed or the third one is.

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19:07:44 STUDENT RESPONSE: 1.8cm 2.722297s 4.2cm 2.518101s 6.7cm 2.6606s 'ds=78. These results show that the smallest slope the time to coast is the slowest The middle ramp has the fastest time down the ramp These results show that acceleration is fastest down the middle ramp (4.2cm) INSTRUCTOR COMMENT ON STUDENT RESPONSE: ** To support the idea that acceleration is linearly dependent on slope you first have to calculate the accelerations, as you did in the Video Experiments. You then need to plot a graph of acceleration vs. ramp slope and see if it forms a straight line. **

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RESPONSE -->

Your answer was good, but to be safe on problems of this type you have to calculate the accelerations.

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19:18:32 Problem Number 3 A ball reaches a ramp of length 50 cm with an unknown initial velocity and accelerates uniformly along the ramp, reaching the other end in 3.8 seconds. Its velocity at the end of the ramp is determined to be 6.31579 cm/s. What is its acceleration on the ramp?

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RESPONSE --> First the vAve can be determined by 'ds = 50 cm, 'dt = 3.8 s. So the vAve = 50 cm / 3.8 s = 13.16 cm/s. Then we can say that 13.16 cm/s = 6.31579 cm/s + v0 / 2.

This should be 13.16 cm/s = (6.31579 cm/s + v0) / 2. The expression vf + v0 is the numerator. So when you multiply by 2 you get 26.32 cm/s = 6.32 cm/s + v0 and

(13.16 cm/s - 6.31759 cm/s) 2 = v0 v0 = 26.32 cm/s - 6.38 cm/s = 19.94 cm/s

etc.

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19:22:13 STUDENT RESPONSE: .'ds=50cm vf=6.31579cm/s 'dt = 3.8s The average velocity on the ramp is 50cm/3.8s =13.16m/s The initial velocity will be v0 = 2 'ds /dt, v0 = 2 *50 /3.8, v0 = 26.31579 -vf which is 6.31579, so the v0 is 20.00 ** You got it but this step isn?t correctly stated. v0 = 2 `ds/dt is not a correct statement. v0 = 2 vAve ? vf, or v0 = 2 `ds/`dt ? v0 (which is the same thing since vAve = `ds/`dt) is the correct equation. The reasoning is that ave velocity is the average of initial and final velocities, since acceleration is uniform. So you have vAve = (v0 + vf) / 2.. You know vAve from your previous calculation and vf is given. So you solve this equation for v0 and then substitute these values and simplify. You get v0 = 2 vAve ? vf, then substitute. This is actually what you did; just be careful you state it this way. If you don?t state it right there?s a chance you might not do it right. ** The accleration is 6.31579-20/3.8 = -3.601m/s/s If your v0 was correct this would be right **

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19:32:03 Problem Number 4 A projectile leaves the edge of a table at 18 cm/s, then falls freely a distance of 95 cm to the floor. How far does it travel in the horizontal direction during the fall?

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19:33:01 STUDENT RESPONSE: .'ds=95cm a=980cm/s/s v0=0 First, we take the equation 'ds = v0'dt + .5(980) 'dt^2 95cm = .5 (980) 'dt^2 'dt = sqrt 95cm/.5(980)=.194 So, 'dt = sqrt .194 'dt=.44s Next take the 18cm/s and multiply by the time duration .44s = 7.92cm/s INSTRUCTOR COMMENT TO ALL STUDENTS: Remember the basic principle used in this solution. You find the time of fall for the vertical motion, then find the horizontal displacement using the time interval and the initial horizontal velocity.

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19:42:26 Problem Number 5 What are the average acceleration and final velocity of an object which accelerates uniformly from rest, ending up with velocity 13.9 cm/sec after traveling a distance of 40 cm from start to finish?

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RESPONSE --> The vf is given as 13.9 cm/s and the v0 = 0 cm/s. So the vAve is 13.9 cm/s + 0 cm/s / 2 = 6.95 cm/s. So the time interval would be 40 cm / 6.95 cm/s = 5.75 s. So the aAve = 13.9 cm/s / 5.75 s = 2.41 cm/s^2

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19:43:19 STUDENT RESPONSE: 'ds=40cm v0=0 vf=13.9cm/s If vf is 13.9cm/s then the using the equation a 'ds = vf^2 - v0^2)/2 will be , a * 40 cm = ( (13.9 cm/s)^2 ? 0^2) /2 40 * a = 193.21 cm^2/s^2 / 2 a = 96.61 cm^2/s^2 / (40 cm) a = 2.415cm/s^2 ** If vf^2 = v0^2 + 2 a `ds then a `ds = (vf^2 ? v0^2) / 2, as you say, so a = (vf^2 ? v0^2) / (2 `ds). This is what you did (good job) but be careful to state it this way. ** Then use the equation 'ds = v0 'dt + .5 a 'dt^2 ** I wouldn?t use this equation to solve for `dt, since the equation is quadratic in `dt. At this stage you know all variables except `dt. The equation vf = v0 + a `dt would be much simpler. Or you could just figure average velocity and divide into displacement. Either way would be fine. ** 40= +.5 2.415cm/s/s dt^2 dt = sqrt 40/ ** You could have used the fact that vAve = (13.9 cm/s + 0) / 2 = 6.95 cm/s to obtain `dt: Since vAve = `ds / `dt, we have `dt = `ds / vAve = 40 cm / (6.95 cm/s) = 5.8 sec, approx. **

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���DŽ����˿��Y����ٳK�~������ assignment #012 yE�܃���������݌��֭ Physics I 11-21-2005

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21:50:51 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 --> The force of the gravity on the mass hanging over the pulley is pulling on the system and the net force is 9 kg * 9.8 m/s^2 = 88.2 Newtons. The acceleration of the system would be 88.2 Newtons / 21 kg = 4.2 m/s^2. The PE change would be the flip side of the KE which would be 88.2 Newtons * 3.8 meters = 335.16 Joules so the PE would be - 335.16 Joules.

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21:51:32 ** The net force on the system is the force of gravity on the suspended weight: Fnet = m2*9.8m/s/s Gravity also acts on m1 which is balanced by force of table on m1, so this force makes no contribution to Fnet. Acceleration=net force/total mass = 9.8 m/s^2 * m2 / (m1+m2). If the mass m2 descends distance `dy then gravitational PE decreases by - m2 g * `dy. COMMON MISCONCEPTIONS AND INSTRUCTOR COMMENTS: The forces acting on the system are the forces which keep the mass on the table, the tension in the string joining the two masses, and the weight of the suspended mass. The net force should be the suspended mass * accel due to gravity + Tension. INSTRUCTOR COMMENT: String tension shouldn't be counted among the forces contributing to the net force on the system. The string tension is internal to the two-mass system. It doesn't act on the system but within the system. Net force is therefore suspended mass * accel due to gravity only 'The forces which keep the mass on the table' is too vague and probably not appropriate in any case. Gravity pulls down, slightly bending the table, which response with an elastic force that exactly balances the gravitational force. **

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21:53:52 How would friction change your answers to the preceding question?

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RESPONSE --> You would have to consider friction as opposing the direction so it would be a negative figure to the net force. You would figure the percentage of the mass of the object to get the ffrict to subtract from the force net.

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21:54:21 **Friction would act to oppose the motion of the mass m1 as it slides across the table, so the net force would be m2 * g - frictional resistance. **

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21:55:56 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 --> I did not do the optional experiment so I will go on.

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21:57:08 ** If we ignore thermal effects, which you should note are in fact significant with rubber bands and cannot in practice be ignored if we want very accurate results, PE is the work required to stretch the rubber band. This work is the sum of all F * `ds contributions from small increments `ds from the initial to the final position. These contributions are represented by the areas of narrow trapezoids on a graph of F vs. stretch. As the trapezoids get thinner and thinner, the total area of these trapezoids approaches, the area under the curve between the two stretches. So the PE stored is the area under the graph of force vs. stretch. **

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RESPONSE --> Ok I see this is like the trapezoids that we did earlier in the course.

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21:57:39 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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21:58:12 ** Slope isn't directly related to any physical quantity. The area is indeed with work done (work is integral of force with respect to displacement). If the rubber band pulls against an object as is returns to equilibrium then the force it exerts is in the direction of motion and it therefore does positive work on the object as the object does negative work on it. If an object stretches the rubber band then it exerts a force on the rubber band in the direction of the rubber band's displacement, and the object does positive work on the rubber band, while the rubber band does negative work on it. **

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21:58:22 Query rubber bands supporting mass (experiment 11) When rubber bands support a mass, how can you determine the components of the force of one of the rubber bands?

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21:58:26 ** The length of the rubber band determines its force, which can be read from the calibration graph. The angle of the rubber bands can be used with the force to find x and y components of each, using standard vector methods. **

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21:58:30 When rubber bands support a mass, how do the horizontal components of their forces compare?

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21:58:32 ** If a mass is supported in a stationary position by two rubber bands, then the mass is in equilibrium. It follows that the net force on the mass is zero (if there was a nonzero net force it would have to be accelerating at rate a = F / m). So the net force in any direction must be zero. In particular the net horizontal force must be zero. In general both rubber bands will be pulling at nonzero angles with respect to vertical, so both will have both vertical and horizontal components. In the usual situation the horizontal components are the only forces exerted in the horizontal direction. Since the net force in this direction is zero, the horizontal components must be equal and opposite. The vertical forces acting on the object include the vertical components of the rubber band forces and the weight of the object. The total of the vertical components of the rubber band force is therefore equal and opposite to the downward pull of gravity. **

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21:58:36 When rubber bands support a masss, what can be said about the vertical components of the forces exerted and the force exerted by gravity on the mass?

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21:58:37 ** The vertical forces must add up to zero, since the object isn't accelerating in the horizontal direction. The vertical components of the forces exerted by the supporting rubber bands will be equal and opposite to the force exerted by gravity. **

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22:00:34 Query Add comments on any surprises or insights you experienced as a result of this assignment.

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RESPONSE --> I just find this part to be very difficult and hard to get a grasps of. I feel as if I need time to try to absorb it a little more but the clock is not on my side. So I will do the best I can and try to get as much as I can in a short period of time.

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