course phy 121 yDZϐ|Tassignment #011
......!!!!!!!!...................................
09:00:47 `q001. A cart on a level, frictionless surface contains ten masses, each of mass 2 kg. The cart itself has a mass of 30 kg. A light weight hanger is attached to the cart by a light but strong rope and suspended over the light frictionless pulley at the end of the level surface. Ignoring the weights of the rope, hanger and pulley, what will be the acceleration of the cart if one of the 2 kg masses is transferred from the cart to the hanger?
......!!!!!!!!...................................
RESPONSE --> The weight on the hanger will produce a force of 2kg * 9.8m/s^2 or 19.6N First find the total mass of the cart 10mass units * 2kg = 20kg 20kg - 2 kg (going to the hanger) = 18kg 18kg + 30kg = 48kg The acceleration of the cart is F/m a = 19.6N / 48kg 1= 19.6 kg*m/s^2 / 48kg a = .41m/s^2 confidence assessment: 2
.................................................
......!!!!!!!!...................................
09:03:05 At the surface of the Earth gravity, which is observed here near the surface to accelerate freely falling objects at 9.8 m/s^2, must exert a force of 2 kg * 9.8 m/s^2 = 19.6 Newtons on the hanging 2 kg mass. This force will tend to accelerate the system consisting of the cart, rope, hanger and suspended mass, in the 'forward' direction--the direction in which the various components of the system must accelerate if the hanging mass is to accelerate in the downward direction. This is the only force accelerating the system in this direction. All other forces, including the force of gravity pulling the cart and the remaining masses downward and the normal force exerted by the level surface to prevent gravity from accelerating the cart downward, are in a direction perpendicular to the motion of these components of the system; furthermore these forces are balanced so that they add up to 0. The mass of the system is that of the 30 kg cart plus that of the ten 2 kg masses, a total of 50 kg. The net force of 19.6 Newtons exerted on a 50 kg mass therefore results in acceleration a = Fnet / m = 19.6 Newtons / (50 kg) = 19.6 kg m/s^2 / (50 kg) = .392 m/s^2.
......!!!!!!!!...................................
RESPONSE --> I think the total mass of the cart is 48kg, not 50kg because I think the problem stated that 2kg was moved from the cart to the hanger. Otherwise I followed the same procedure self critique assessment: 3
.................................................
......!!!!!!!!...................................
09:11:55 `q002. Two 1-kg masses are suspended over a pulley, one on either side. A 100-gram mass is added to the 1-kg mass on one side of the pulley. How much force does gravity exert on each side of the pulley, and what is the net force acting on the entire 2.1 kg system?
......!!!!!!!!...................................
RESPONSE --> Fnet = m*a a= 9.8m/s^2 m1 = 1kg m2 = 1.1kg therefore F1 = 9.8N F2 = 10.78N I am not sure if the Fnet is F1 + F2 or F2-F1. Gravity is pulling down on both sides so the net force could be F1+F2 or 2.1kg*9.8m/s^2 = 20.58N. But the force on F2 is greater and therefore would pull F2 down and F1 up. Since the two are acting in opposition I think I should subtract F1 from F2 to get the net force. 10.78N - 9.8N = .98N confidence assessment: 2
.................................................
......!!!!!!!!...................................
09:13:37 The 1-kg mass experiences a force of F = 1 kg * 9.8 m/s^2 = 9.8 Newtons. The other side has a total mass of 1 kg + 100 grams = 1 kg + .1 kg = 1.1 kg, so it experiences a force of F = 1.1 kg * 9.8 Newtons = 10.78 Newtons. {}{}Both of these forces are downward, so it might seem that the net force on the system would be 9.8 Newtons + 10.78 Newtons = 20.58 Newtons. However this doesn't seem quite right, because when one mass is pulled down the other is pulled up so in some sense the forces are opposing. It also doesn't make sense because if we had a 2.1 kg system with a net force of 20.58 Newtons its acceleration would be 9.8 m/s^2, and we know very well that two nearly equal masses suspended over a pulley won't both accelerate downward at the acceleration of gravity. So in this case we take note of the fact that the to forces are indeed opposing, with one tending to pull the system in one direction and the other in the opposite direction. {}{}We also see that we have to abandon the notion that the appropriate directions for motion of the system are 'up' and 'down'. We instead take the positive direction to be the direction in which the system moves when the 1.1 kg mass descends. {}{}We now see that the net force in the positive direction is 10.78 Newtons and that a force of 9.8 Newtons acts in the negative direction, so that the net force on the system is 10.78 Newtons - 9.8 Newtons =.98 Newtons. {}{}The net force of .98 Newtons on a system whose total mass is 2.1 kg results in an acceleration of .98 Newtons / (2.1 kg) = .45 m/s^2, approx.. Thus the system accelerates in the direction of the 1.1 kg mass at .45 m/s^2.
......!!!!!!!!...................................
RESPONSE --> I didn't calculate the acceleration, but I did come up with the fact that you subtract because of the opposing directions. self critique assessment: 3
.................................................
......!!!!!!!!...................................
09:24:17 `q003. If in the previous question there is friction in the pulley, as there must be in any real-world pulley, the system in the previous problem will not accelerate at the rate calculated there. Suppose that the pulley exerts a retarding frictional force on the system which is equal in magnitude to 1% of the weight of the system. In this case what will be the acceleration of the system, assuming that it is moving in the positive direction (as defined in the previous exercise)?
......!!!!!!!!...................................
RESPONSE --> Previously Fnet was .98 fFrict =1% of 2.1kg fFrict = .01*2.1 = .021 It is opposing the previous Fnet of .98 new Fnet = .98 - .021 = .959N a = Fnet/m a = .959kg * m/s^2 / 2.1kg a = .457m/s/s confidence assessment: 1
.................................................
......!!!!!!!!...................................
09:35:33 We first determine the force exerted a friction. The weight of the system is the force exerted by gravity on the mass of the system. The system has mass 2.1 kg, so the weight of the system must be 2.1 kg * 9.8 m/s^2 = 20.58 Newtons. 1% of this weight is .21 Newtons, rounded off to two significant figures. This force will be exerted in the direction opposite to that of the motion of the system; since the system is assume to be moving in the positive direction the force exerted by friction will be frictional force = -.21 Newtons. The net force exerted by the system will in this case be 10.78 Newtons - 9.8 Newtons - .21 Newtons = .77 Newtons, in contrast with the .98 Newton net force of the original exercise. The acceleration of the system will be .77 Newtons / (2.1 kg) = .35 m/s^2, approx..
......!!!!!!!!...................................
RESPONSE --> I didn't think about weight and mass being different and just used the mass of the system to find the frictional force. Is weight always the product of mass * acceleration of gravity? Weight = m*g Are Newtons and lbs related in that sense? Would a 200lb man also be considered to be 200Newtons? Does that man have a mass of 200lb/9.8m/s^2 or 20.41kg? self critique assessment: 2
.................................................
......!!!!!!!!...................................
09:37:37 `q004. Why was it necessary in the previous version of the exercise to specify that the system was moving in the direction of the 1.1 kg mass. Doesn't the system have to move in that direction?
......!!!!!!!!...................................
RESPONSE --> Its necessary because it means that the fFrict was negative and had to be subtracted from the Fnet. confidence assessment: 1
.................................................
......!!!!!!!!...................................
09:42:49 If the system is released from rest, since acceleration is in the direction of the 1.1 kg mass its velocity will certainly always be positive. However, the system doesn't have to be released from rest. We could give a push in the negative direction before releasing it, in which case it would continue moving in the negative direction until the positive acceleration brought it to rest for an instant, after which it would begin moving faster and faster in the positive direction.
......!!!!!!!!...................................
RESPONSE --> I hope we don't have to calculate how much faster and faster it is going to go. self critique assessment: 3
.................................................
......!!!!!!!!...................................
09:52:12 `q005. If the system of the preceding series of exercises is initially moving in the negative direction, then including friction in the calculation what is its acceleration?
......!!!!!!!!...................................
RESPONSE --> This is even worse than I had feared. I think I should have to know what the force was that pushed it in the negative direction to get an acccurate answer, but since I don't.... The 10.78 is negative because system is moving in opposition to it, The 9.8 is positive because the system is moving with it, and the fFrict is still negative because it is still an opposing force to the system. -10.78 + 9.8 - .21 = -1.19N a = Fnet * m a = -1.19N * 2.1kg = -2.49 = -2.5 m/s^2 confidence assessment: 0
.................................................
......!!!!!!!!...................................
10:01:14 `q006. If friction is neglected, what will be the result of adding 100 grams to a similar system which originally consists of two 10-kg masses, rather than the two 1-kg masses in the previous examples?
......!!!!!!!!...................................
RESPONSE --> Enter, as appropriate, an answer to the question, a critique of your answer in response to a given answer, your insights regarding the situation at this point, notes to yourself, or just an OK. Always critique your solutions by describing any insights you had or errors you makde, and by explaining how you can make use of the insight or how you now know how to avoid certain errors. Also pose for the instructor any question or questions that you have related to the problem or series of problems. I multiplied 1.19N * 2.1kg by accident. But I am not sure why the answer is positive instead of negative, see my previous answer for my reasoing. Q006 came up with the answer to q005 F1 = 10kg * 9.8m/s^2 = 98N F2 = 10.1kg * 9.8m/s^2 = 98.98N = 99N Fnet = 1N a = Fnet/m a = 1N/20.1kg = .0497 = .05m/s^2 confidence assessment: 2
.................................................
......!!!!!!!!...................................
10:06:45 In this case the masses will be 10.1 kg and 10 kg. The force on the 10.1 kg mass will be 10.1 kg * 9.8 m/s^2 = 98.98 Newtons and the force on the 10 kg mass will be 10 kg * 9.8 m/s^2 = 98 Newtons. The net force will therefore be .98 Newtons, as it was in the previous example where friction was neglected. We note that this.98 Newtons is the result of the additional 100 gram mass, which is the same in both examples. The mass in this case will be 20.1 kg, so that the acceleration of the system is a = .98 Newtons / 20.1 kg = .048 m/s^2, approx.. The same .98 Newton net force acting on the significantly greater mass results in a significantly smaller acceleration.
......!!!!!!!!...................................
RESPONSE --> I shouldn't have rounded the Newtons to 1, but I see that the same force acting on a greater mass results in smaller acceleration. self critique assessment: 3
.................................................
......!!!!!!!!...................................
10:13:58 `q007. If friction is not neglected, what will be the result for the system with the two 10-kg masses with .1 kg added to one side? Note that by following what has gone before you could, with no error and through no fault of your own, possibly get an absurd result here, which will be repeated in the explanation then resolved at the end of the explanation.
......!!!!!!!!...................................
RESPONSE --> weight = m*g 20.1kg * 9.8m/s^2 = 196.98 friction is 1% of the weight 196.98 * .01 = 1.9698 = 1.97 since friction is in opposition fFrict = -1.97N Fnet = F1 - F2 - fFrict Fnet = 98.98 - 98 - 1.97 = -99N a = Fnet / m a = -99N/20.1kg a = -.049m/s^2 confidence assessment: 2
.................................................
......!!!!!!!!...................................
10:15:46 If friction is still equal to 1% of the total weight of the system, which in this case is 20.1 kg * 9.8 m/s^2 = 197 Newtons, then the frictional force will be .01 * 197 Newtons = 1.97 Newtons. This frictional force will oppose the motion of the system. For the moment assume the motion of the system to be in the positive direction. This will result in a frictional force of -1.97 Newtons. The net force on the system is therefore 98.98 Newtons - 98 Newtons - 1.97 Newtons = -.99 Newtons. This net force is in the negative direction, opposite to the direction of the net gravitational force. If the system is moving this is perfectly all right--the frictional force being greater in magnitude than the net gravitational force, the system can slow down. Suppose the system is released from rest. Then we might expect that as a result of the greater weight on the positive side it will begin accelerating in the positive direction. However, if it moves at all the frictional force would result in a -.99 Newton net force, which would accelerate it in the negative direction and very quickly cause motion in that direction. Of course friction can't do this--its force is always exerted in a direction opposite to that of motion--so friction merely exerts just enough force to keep the object from moving at all. Friction acts as though it is quite willing to exert any force up to 1.97 Newtons to oppose motion, and up to this limit the frictional force can be used to keep motion from beginning. In fact, the force that friction can exert to keep motion from beginning is usually greater than the force it exerts to oppose motion once it is started.
......!!!!!!!!...................................
RESPONSE --> self critique assessment: 3
.................................................
......!!!!!!!!...................................
16:13:42 `q008. A cart on an incline is subject to the force of gravity. Depending on the incline, some of the force of gravity is balanced by the incline. On a horizontal surface, the force of gravity is completely balanced by the upward force exerted by the incline. If the incline has a nonzero slope, the gravitational force (i.e., the weight of the object) can be thought of as having two components, one parallel to the incline and one perpendicular to the incline. The incline exerts a force perpendicular to itself, and thereby balances the weight component perpendicular to the incline. The weight component parallel to the incline is not balanced, and tends to accelerate the object down the incline. Frictional forces tend to resist this parallel component of the weight and reduce or eliminate the acceleration. A complete analysis of these forces is best done using the techniques of vectors, which will be encountered later in the course. For now you can safely assume that for small slopes (less than .1) the component of the gravitational force parallel to the incline is very close to the product of the slope and the weight of the object. [If you remember your trigonometry you might note that the exact value of the parallel weight component is the product of the weight and the sine of the angle of the incline, that for small angles the sine of the angle is equal to the tangent of the angle, and that the tangent of the angle of the incline is the slope. The product of slope and weight is therefore a good approximation for small angles or small slopes.] What therefore would be the component of the gravitational force acting parallel to an incline with slope .07 on a cart of mass 3 kg?
......!!!!!!!!...................................
RESPONSE --> .07 * 3kg = .21 I have to admit I don't really understand this very well. i just got that I should multiply the slope by the mass. I understood up until the Trigonometry stuff. I was 16 the last time I had Trigonometry. That was a LONG time ago! What is the unit for the .21? The slope is unitless, but gravitational force should be Newtons? confidence assessment: 2
.................................................
......!!!!!!!!...................................
16:19:00 The gravitational force on a 3 kg object is its weight and is equal to 3 kg * 9.8 m/s^2 = 29.4 Newtons. The weight component parallel to the incline is found approximately as (parallel weight component) = slope * weight = .07 * 29.4 Newtons = 2.1 Newtons (approx.).. STUDENT COMMENT: It's hard to think of using the acceleration of 9.8 m/s^2 in a situation where the object is not free falling. Is weight known as a force measured in Newtons? Once again I'm not used to using mass and weight differently. If I set a 3 kg object on a scale, it looks to me like the weight is 3 kg.INSTRUCTOR RESPONSE: kg is commonly used as if it is a unit of force, but it's not. Mass indicates resistance to acceleration, as in F = m a. {}{}eight is the force exerted by gravity. {}{}The weight of a given object changes as you move away from Earth and as you move into the proximity of other planets, stars, galaxies, etc.. As long as the object remains intact its mass remains the same, meaning it will require the same net force to give it a specified acceleration wherever it is.{}{}An object in free fall is subjected to the force of gravity and accelerates at 9.8 m/s^2. This tells us how much force gravity exerts on a given mass: F = mass * accel = mass * 9.8 m/s^2. This is the weight of the object. **
......!!!!!!!!...................................
RESPONSE --> I forgot again that weight is different than mass and I forgot to multiply the 3kg * 9.8m/s^2. That explains where the units come from also. self critique assessment: 3
.................................................
......!!!!!!!!...................................
16:23:00 `q009. What will be the acceleration of the cart in the previous example, assuming that it is free to accelerate down the incline and that frictional forces are negligible?
......!!!!!!!!...................................
RESPONSE --> a = Fnet/m Fnet = 2.1N m = 3kg a = 2.1N/3kg = .7m/s^2 confidence assessment: 2
.................................................
......!!!!!!!!...................................
16:23:38 The weight component perpendicular to the incline is balanced by the perpendicular force exerted by the incline. The only remaining force is the parallel component of the weight, which is therefore the net force. The acceleration will therefore be a = F / m = 2.1 Newtons / (3 kg) = .7 m/s^2.
......!!!!!!!!...................................
RESPONSE --> self critique assessment: 3
.................................................
......!!!!!!!!...................................
16:29:34 `q010. What would be the acceleration of the cart in the previous example if friction exerted a force equal to 2% of the weight of the cart, assuming that the cart is moving down the incline? [Note that friction is in fact a percent of the perpendicular force exerted by the incline; however for small slopes the perpendicular force is very close to the total weight of the object].
......!!!!!!!!...................................
RESPONSE --> Fnet prior to friction = 2.1 m= 3kg fFrict = 2% of wt wt = 3kg * 9.8m/s^2 = 29.4N fFrict = 29.4 *.02 = .588 = .59N fFrict is in opposition to the movement so Fnet = 2.1 - .59 = 1.51 a = Fnet / m a = 1.51N / 3kg = .503R = .5 m/s^2 confidence assessment: 2
.................................................
......!!!!!!!!...................................
16:29:54 The weight of the cart was found to be 29.4 Newtons. The frictional force will therefore be .02 * 29.4 Newtons = .59 Newtons approx.. This frictional force will oppose the motion of the cart, which is down the incline. If the downward direction along the incline is taken as positive, the frictional force will be negative and the 2.1 Newton parallel component of the weight will be positive. The net force on the object will therefore be net force = 2.1 Newtons - .59 Newtons = 1.5 Newtons (approx.). This will result in an acceleration of a = Fnet / m = 1.5 Newtons / 3 kg = .5 m/s^2.
......!!!!!!!!...................................
RESPONSE --> self critique assessment: 3
.................................................
......!!!!!!!!...................................
16:34:55 `q011. Given the conditions of the previous question, what would be the acceleration of the cart if it was moving up the incline?
......!!!!!!!!...................................
RESPONSE --> I don't see how it could move up the incline without an additional force. But the gravitational force of 2.1 would become negative because the force is in opposition to the movement. The fFrict is also negative because it is opposition to the movement so Fnet would be -2.1 - .59 = -2.69 a = -2.69N / 3kg = -.896R = -.9 m/s^2 confidence assessment: 1
.................................................
......!!!!!!!!...................................
16:39:51 In this case the frictional force would still have magnitude .59 Newtons, but would be directed opposite to the motion, or down the incline. If the direction down the incline is still taken as positive, the net force must be net force = 2.1 Newtons + .59 Newtons = 2.7 Newtons (approx). The cart would then have acceleration a = Fnet / m = 2.7 Newtons / 3 kg = .9 m/s^2.
......!!!!!!!!...................................
RESPONSE --> I am still having trouble with the negatives. Why wouldn't the net force be negative? The force is in opposition to the movement. I can't see how the cart could have a positive acceleration when it is moving up hill against the force of gravity and friction and without any other force acting on it to give it a positive force. self critique assessment: 1
.................................................
......!!!!!!!!...................................
16:41:03 `q012. Assuming a very long incline, describe the motion of the cart which is given an initial velocity up the incline from a point a few meters up from the lower end of the incline. Be sure to include any acceleration experienced by the cart.
......!!!!!!!!...................................
RESPONSE --> It is going to go uphill, stop and then head back downhill. confidence assessment: 2
.................................................
......!!!!!!!!...................................
16:51:27 The cart begins with a velocity up the incline, which we still taken to be the negative direction, and an acceleration of +.9 m/s^2. This positive acceleration tends to slow the cart while it is moving in the negative direction, and the cart slows by .9 m/s every second it spends moving up the incline. Eventually its velocity will be 0 for an instant, immediately after which it begins moving down the incline as result of the acceleration provided by the weight component parallel to the incline. {}As soon as it starts moving down the incline its acceleration decreases to +.5 m/s^2, but since the acceleration and velocity are now parallel the cart speeds up, increasing its velocity by .5 m/s every second, until it reaches the lower end of the incline.
......!!!!!!!!...................................
RESPONSE --> How does a positive acceleration slow an object? It eventually slows down to a stop, which I intuitively understand, and then it accelerates based on the parallel component of gravity divided by the mass. Increasing velocity as it goes. Got it. self critique assessment: 2
.................................................