course phy 121 I skipped the lab in asst 21. I hope to be able to do it in a few days. ?f??F????{????assignment #022
......!!!!!!!!...................................
16:43:38 `q001. Note that this assignment contains 2 questions, which relate to a force-field experiment which is done using a computer simulation, and could for example represent the force on a spacecraft, where uphill and downhill are not relevant concepts. . An object with a mass of 4 kg is traveling in the x direction at 10 meters/second when it enters a region where it experiences a constant net force of 5 Newtons directed at 210 degrees, as measured in the counterclockwise direction from the positive x axis. How long will take before the velocity in the x direction decreases to 0? What will be the y velocity of the object at this instant?
......!!!!!!!!...................................
RESPONSE --> I have no confidence in this answer because it seems too simple... Fnet * dt = mvf - mvo 5N * dt = (4kg * 0) - (4kg * 10m/s) 5N * dt = 0 - 40kg m/s dt = - 8s Since it can't be negative, I'll go with 8 secs. I didn't take into account the direction of the force, and I feel certain that I should have. I also don't know how to actually deal with the negative. confidence assessment: 0
.................................................
......!!!!!!!!...................................
16:58:06 A constant net force of 5 Newtons on a 4 kg object will result in an acceleration of 5 Newtons/(4 kg) = 1.25 meters/second ^ 2. If the force is directed at 210 degrees then the acceleration will also be directed at 210 degrees, so that the acceleration has x component 1.25 meters/second ^ 2 * cosine (210 degrees) = -1.08 meters/second ^ 2, and a y component of 1.25 meters/second ^ 2 * sine (210 degrees) = -.63 meters/second ^ 2. We analyze the x motion first. The initial velocity in the x direction is given as 10 meters/second, we just found that the acceleration in the x direction is -1.08 meters/second ^ 2, and since we are trying to find the time required for the object to come to rest the final velocity will be zero. We easily see that the change in the next velocity is -10 meters/second. At a rate of negative -1.08 meters/second ^ 2, the time required for the -10 meters/second change in velocity is `dt = -10 meters/second / (-1.08 meters/second ^ 2) = 9.2 seconds. We next analyze the y motion. The initial velocity in the y direction is zero, since the object was initially moving solely in the x direction. The acceleration in the y direction is -.63 meters/second ^ 2. Therefore during the time interval `dt = 9.2 seconds, the y velocity changed by (-.63 meters/second ^ t) * (9.2 seconds) = -6 meters/second, approximately. Thus the y velocity changes from zero to -6 meters/second during the 9.2 seconds required for the x velocity to reach zero.
......!!!!!!!!...................................
RESPONSE --> I read it. I wrote it down. I followed each step. But I don't really understand it. self critique assessment: 1
.................................................
......!!!!!!!!...................................
17:10:46 `q002. Suppose that the mass in the preceding problem encounters a region in which the force was identical to that of the problem, but that this region extended for only 30 meters in the x direction (assume that there is the limit to the extent of the field in the y direction). What will be the magnitude and direction of the velocity of the mass as it exits this region?
......!!!!!!!!...................................
RESPONSE --> I really have no idea, well I have one... I might use vf2 =v0^2 + 2a*ds. I could plug in both the x and the y components for acceleration seperately and 30 m for the ds to find the vf for x and vf for y. Then I might use those in Pythagorean's theorem to find the magnitude and then find the angle using arctan, but I am unwilling to go actually go through all of those mathematical steps when I don't think it is right. confidence assessment: 0
.................................................
......!!!!!!!!...................................
17:14:12 As we have seen in the preceding problem the object will have an acceleration of -1.08 meters/second ^ 2 in the x direction. Its initial x velocity is 10 meters/second and it will travel 30 meters in the x direction before exiting the region. Thus we have v0, a and `ds, so that you to the third or fourth equation of uniform accelerated motion will give us information. The fourth equation tells us that vf = +-`sqrt( (10 meters/second) ^ 2 + 2 * (-1.08 meters/second ^ 2) * (30 meters) ) = +-6 meters/second. Since we must exit the region in the positive x direction, we choose vf = + 6 meters/second. It follows that the average x velocity is the average of the initial 10 meters/second and the final 6 meters/second, or eight meters/second. Thus the time required to pass-through the region is 30 meters/(8 meters/second) = 3.75 seconds. During this time the y velocity is changing at -.63 meters/second ^ 2. Thus the change in the y velocity is (-.63 meters/second ^ 2) * (3.75 seconds) = -2.4 meters/second, approximately. Since the initial y velocity was zero, the y velocity upon exiting the region will be -2.4 meters/second. Thus when exiting the region the object has velocity components +6 meters/second in the x direction and -2.4 meters/second in the y direction. Its velocity therefore has magnitude `sqrt ( (6 meters/second) ^ 2 + (-2.4 meters/second) ^ 2) = 6.4 meters/second. The direction of velocity will be arctan ( (-2.4 meters/second) / (6 meters/second) ) = -22 degrees, approximately. Thus the object exits at 6.4 meters/second at an angle of 22 degrees below the positive x axis, or at angle -22 degrees + 360 degrees = 338 degrees.
......!!!!!!!!...................................
RESPONSE --> Well, I was amazingly on the right track. I'm sorry now that I was unwilling to pursue it. self critique assessment: 3
.................................................
??????;??????
......!!!!!!!!...................................
22:47:13 `q001. Note that this assignment contains 3 questions. . A chain 200 cm long has a density of 15 g/cm. Part of the chain lies on a tabletop, with which it has a coefficient of friction equal to .10. The other part of the chain hangs over the edge of the tabletop. If 50 cm of chain hang over the edge of the tabletop, what will be the acceleration of the chain?
......!!!!!!!!...................................
RESPONSE --> Did I miss something? I haven't worked with or read about density and length of an object, and I am not sure exactly what coefficient of friction means. I will attempt to make a guess... The part that lays over the edge is 50 cm at 15g/cm it is 750g or .750kg The part that lays on the table is 150cm at 15g/cm it is 2250g or 2.25kg Total mass = 200cm * 15g/cm = 3000g or 3kg Normal force = 2.25kg*9.8m/s^s = 22.05 Ffrict = .10*22.05=2.205N Fgrav = 3kg * 9.8 = 29.4N Fnet = 29.4N - 2.205N = 27.195N Fnet = m*a 27.195N = 3kg * a a = 9.065m/s^2 confidence assessment: 1
.................................................
......!!!!!!!!...................................
22:51:03 The part of the chain hanging over the edge of the table will experience an unbalanced force from gravity and will therefore tend to accelerate chain in the direction of the hanging portion. The remainder of the chain will also experience the gravitational force, but this force will be countered by the upward force exerted on the chain by the table. The force between the chain and the table will give rise to a frictional force which will resist motion toward the hanging portion of the chain. If 50 cm of chain hang over the edge of the tabletop, then we have 50 cm * (15 g/cm) = 750 grams = .75 kg of chain hanging over the edge. Gravity will exert a force of 9.8 meters/second ^ 2 * .75 kg = 7.3 Newtons of force on this mass, and this force will tend to accelerate the chain. The remaining 150 cm of chain lie on the tabletop. This portion of the chain has a mass which is easily shown to be 2.25 kg, so gravity exerts a force of approximately 21 Newtons on this portion of the chain. The tabletop pushes backup with a 21 Newton force, and this force between tabletop and chain results in a frictional force of .10 * 21 Newtons = 2.1 Newtons. We thus have the 7.3 Newton gravitational force on the hanging portion of the chain, resisted by the 2.1 Newton force of friction to give is a net force of 5.2 Newtons. Since the chain has a total mass of 3 kg, this net force results in an acceleration of 5.2 N / (3 kg) = 1.7 meters/second ^ 2, approximately.
......!!!!!!!!...................................
RESPONSE --> Why doesn't gravity cause an acceleration on the whole 3 kg. I went wrong with the force of gravity step. I multiplied 3kg by 9.8, instead of just .75kg * 9.8. I think I did everything else correctly. self critique assessment: 2\
.................................................
......!!!!!!!!...................................
23:08:22 `q002. What is the maximum length of chain that can hang over the edge before the chain begins accelerating?
......!!!!!!!!...................................
RESPONSE --> That's weird... That is exactly the question I wondered as I was working the last problem. I would rather solve this by trial and error, but since I don't have a 3kg chain... For a to equal 0 Fnet has to equal 0 Fnet = Fgrav - Ffrict Fnet = (downward mass * 9.8) - (.10 * norm force) Fnet = (dwn mass*9.8) - (.10 * tbl mass * 9.8) therefore down mass * 9.8 = .10 * tbl mass * 9.8 the 9.8 cancel each other out so down mass = .10 * tble mass How to find masses? Total mass = down mass + (.10*tble mass) 3kg = dmass + .1tmass my algebra skills are lacking. I don't know if I can solve this. I think the downward mass would be the same as .1* the mass on the table in order for the chain to stay still. Once I have the mass I can change it to grams and divide by 15 to get the length of the chain hanging over the edge.
.................................................
......!!!!!!!!...................................
23:09:34 The maximum length that can hang over is the length for which the frictional force opposing motion is precisely equal to the gravitational force on the hanging portion of the chain. If x stands for the length in cm of the portion of chain hanging over the edge of the table, then the mass of the length is x * .015 kg / cm and it experiences a gravitational force of (x * .015 kg / cm) * 9.8 m/s^2 = x * .147 N / cm. The portion of chain remaining on the tabletop is 200 cm - x. The mass of this portion is (200 cm - x) * .015 kg / cm and gravity exerts a force of (200 cm - x) * .015 kg / cm * 9.8 meters/second ^ 2 = .147 N / cm * (200 cm - x) on this portion. This will result in a frictional force of .10 * .147 N / cm * (200 cm - x) = .0147 N / cm * (200 cm - x). Since the maximum length that can hang over is the length for which the frictional force opposing motion is precisely equal to the gravitational force on the hanging portion of the chain, we set the to forces equal and solve for x. Our equation is .0147 N / cm * (200 cm - x) = .147 N/cm * x. Dividing both sides by .0147 N/cm we obtain 200 cm - x = 10 * x. Adding x to both sides we obtain 200 cm = 11 x so that x = 200 cm / 11 = 18 cm, approx..
......!!!!!!!!...................................
RESPONSE --> I have no idea if this is the same as what I was trying to do. Please comment. self critique assessment: 1
.................................................
......!!!!!!!!...................................
23:25:41 `q003. The air resistance encountered by a certain falling object of mass 5 kg is given in Newtons by the formula F = .125 v^2, where the force F is in Newtons when the velocity v is in meters/second. As the object falls its velocity increases, and keeps increasing as it approaches its terminal velocity at which the net force on the falling object is zero, which by Newton's Second Law results in zero acceleration and hence in constant velocity. What is the terminal velocity of this object?
......!!!!!!!!...................................
RESPONSE --> First off, there is the answer to a previous question I had. Yes, the acceleration can get to zero, but it won't stop the object, it will just maintain a constant velocity - duh! I should have known that! Second, I have to reason this one out too. I want to use Fnet = m*a and plug in 5kg and 0 acceleration , but then I get an Fnet of 0 also and I am unsure how to deal with this, because I need a force beside 0 to plug into F = .125 v^2 to get a velocity... Fnet = Fgrav - Fairresistance perhaps? 0 = Fgrav - F air resis Fgrav would equal 5kg * 9.8 = 49N So air resist force would also have to equal 49N? Then plug that into F = .125 v^2 49N = .125 * v^2 v^2 = 392 v = 19.8m/s confidence assessment: 1
.................................................
......!!!!!!!!...................................
23:27:33 Only two forces act on this object, the downward force exerted on it by gravity and the upward force exerted by air resistance. The downward force exerted by gravity remains constant at 5 kg * 9.8 meters/second ^ 2 = 49 Newtons. When this force is equal to the .125 v^2 Newton force of friction the object will be at terminal velocity. Setting .125 v^2 Newtons = 49 Newtons, we divide both sides by .125 Newtons to obtain v^2 = 49 Newtons/(.125 Newtons) = 392. Taking square roots we obtain v = `sqrt (392) = 19.8, which represents 19.8 meters/second.
......!!!!!!!!...................................
RESPONSE --> WoW! I love it when I reason out something right! It just takes me SO long. I wish I could do it faster. self critique assessment: 3
.................................................