#$&* course phy201 3/161:02 011. Note that there are 12 questions in this set.
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Given Solution: 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. STUDENT QUESTION: Our answers are close, but wouldn’t the 50kg actually by 48kg since the 2kg weight was taken out of the kart? INSTRUCTOR RESPONSE: The 2 kg weight is still part of the mass that's being accelerated. It's been moved from the cart to the hanger, but it's still there. All 50 kg are being accelerated by that net force. STUDENT QUESTION Am I allowed to divide Newtons by Kg? Or do I have to change the Newtons to kg*m/s^2? INSTRUCTOR RESPONSE You need to reduce everything to fundamental units. How would you know what N / kg is unless they are both expressed in compatible units? You could of course memorize the fact that N / kg gives you m/s^2, along with about 200 other shortcuts, but that would be a waste of time and wouldn't contribute much to your insight or your ability to work out units in unfamiliar situations. You can count the number of fundamental units in all of physics on one hand. If you know the definitions of the quantities, this makes it very easy to deal with questions of units. Unit calculations come down to the simple algebra of multiplying and dividing fractions, whose numerators and denominators are just products and powers of a few simple units. STUDENT QUESTION Does it matter if you define aGravity as 9.8 m/s^2 or -9.8 m/s^2? Would -.39m/s^2 still be correct since we just know it's perpendicular, and not which perpendicular direction is positive? INSTRUCTOR RESPONSE: In your solution you said 'Fnet = 2kg * -9.8m/s^2 = -19.6 N in the downward direction' You weren't completely specific about what the - sign meant and what direction is positive. If the upward direction is considered positive, then we would simply say that the gravitational force is -19.6 N. We wouldn't add 'in the downward direction' because having declared upward as positive, the negative sign already tells us that the force is downward. If we were to say that the force is 19.6 N in the downward direction, this would be a true statement, without the use of the -sign. However this system doesn't move in just the up-down direction, and we have to be careful how we define our positive direction: The signs of the displacement, velocity and acceleration depend on the direction we choose as positive. We consider the system to consist of the cart and the hanging weights. Parts of this system are moving in the vertical direction and parts are moving in the horizontal direction, so neither vertical nor horizontal can be regarded as the positive direction. Let's assume that we are oriented so that from our position the cart moves to the right as the hanging weights descend. If the cart moves to the right, the hanging weights move downward; if the cart moves to the left the hanging weight move upward. We can describe these motions as 'right-down' and 'left-up'. We have to declare our choice of positive direction. We can choose either 'right-down' or 'left-up' to be positive, whichever we find more convenient; having made this choice the opposite direction will be regarded as negative. Now gravity will pull the system in the 'right-down' direction. You might prefer to regard the gravitational force as negative, in which case you will choose 'left-up' as the positive direction. You have implicitly done so by making your acceleration negative. So choosing 'left-up' as the positive direction the force exerted by gravity on the hanging mass, which pulls the system in the 'right-down' direction, will be negative. The resulting acceleration will in this case be a = F_net / m = -19.6 N / (50 kg) = -.392 m/s^2. This means that acceleration will be opposite the direction of motion--i.e., in the 'right-down' direction. It would have been equally valid to choose 'right-down' as the positive direction. The gravitational force on the hanging mass would be +19.6 N and the acceleration +.392 N. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I must not have understood what the question was asking or maybe I don’t understand the answer. I understood it as a pulley with 48 kg on one side and 2 kg on the other. ------------------------------------------------ Self-critique rating: 2 ********************************************* Question: `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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Mass 1: 1 kg + 100 g = 1.1 kg Mass 2: 1 kg Total mass: 2.1 kg Force on mass 1 = 9.8N Force on mass 2 = 10.78N Fnet = (1kg * 9.8) + (1.1kg * 9.8) = 20.58 N confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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, since 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 two 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. Additional note on + and - directions: One force tends to accelerate the system in one direction, the other tends to accelerate it in the opposite direction. • So you need to choose a positive direction and put a + or - sign on each force, consistent with your chosen positive direction. The positive direction can't be 'up' or 'down', since part of the system moves up while another part moves down. • The easiest way to specify a positive direction is to specify the direction of one of the masses. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: `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)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 2.1 kg * .01 = .021N
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Given Solution: 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) = .37 m/s^2, approx.. STUDENT SOLUTION (this solution and the instructor's commentary address a common error in expression and in thinking this is worth a look the main topic is why it's not appropriate to write an expression like .98N-.21N=.77N/2.1kg=.37m/s^2 Using my numbers from the previous problem, we had .98N pulling down, so I will subtract an additional .21 from that to get the Newtons pulling down w/ friction. .98N-.21N=.77N/2.1kg=.37m/s^2. INSTRUCTOR COMMENTARY It's clear what you mean by .98N-.21N=.77N/2.1kg=.37m/s^2, and everything you said up to this point is very good and correct. However as a mathematical statement .98N-.21N=.77N/2.1kg=.37m/s^2 is incorrect. If .98N-.21N=.77N/2.1kg=.37m/s^2, then since quantities that are both equal to a third quantity are equal to one another, .98N-.21N = .37m/s^2. However N and m/s^2 are complete different units, so the left- and right-hand sides of this equality are unlike terms. Unlike terms can't possibly be equal. And of course it's very obvious that .98N-.21N =. .37m/s^2 is simply an untrue statement. Untrue statements tend to lead to confusion, e.g., when you review your work and don't necessarily remember what you were thinking when you wrote the thing down, or when your statement is viewed by someone else who doesn't already know what to expect. If you said F_net = .98N-.21N=.77N so a = F_net / m = .77 N / (2.1kg) =.37m/s^2 then you would not have any false statements in your solution, your solution would be clear to anyone who understands Newton's Second Law, and would be much more likely useful to you when reviewing your work. (Note also that N / kg = (kg m/s^2) / kg = m/s^2 It's important to maintain the habit of reducing units to fundamental units and doing the algebra of the units.) STUDENT QUESTION &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I may be wrong but I thought 1% of 2.1 = 2.1 * .01 = .021 not .21
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Given Solution: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: `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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Fnet = m * a 10.78 N - 9.821N = .959 N .959N = 2.1kg * a A = .457 m/s/s confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Its acceleration will be due to the net force. This net force will include the 10.78 Newton force in the positive direction and the 9.8 Newton force in the negative direction. It will also include a frictional force of .21 Newtons in the direction opposed to motion. Since motion is in the negative direction, the frictional force will therefore be in the positive direction. The net force will thus be Fnet = 10.78 Newtons - 9.8 Newtons + .21 Newtons = +1.19 Newtons, in contrast to the +.98 Newtons obtained when friction was neglected and the +.77 Newtons obtained when the system was moving in the positive direction. , The acceleration of the system is therefore • a = 1.19 N / (2.1 kg) = .57 m/s^2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: `q006. If friction is neglected, what will be the result of adding 100 grams to a similar system which originally consists of two 10-kg masses, rather than the two 1-kg masses in the previous examples? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Mass 1: 10 kg + 100 g = 10.1 kg Mass 2: 10 kg Total mass: 20.1 kg Force on mass 1 = 10kg * 9.8m/s.s = 98 N Force on mass 2 = 10.1 *9.8m/s/s = 98.98 N Fnet = 98N + 98.98 N = 196.98 N +.98N = 20.1 kg * a A = .488 m/s/s confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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 total mass of the system is 10 kg + 10.1 kg = 20.1 kg, so that the acceleration of the system is • a = .98 Newtons / 20.1 kg = .048 m/s^2, approx.. Comparing this with the preceding situation, where the net force was the same (.98 N) but the total mass was 2.1 kg, we see that the same net force acting on the significantly greater mass results in significantly less acceleration. Note on the direction of the frictional force: It's not quite accurate to say that the frictional force is always in the direction opposite motion. I'm not really telling you the whole story here--trying to keep things simple. Friction can indeed speed things up, depending on your frame of reference. The more accurate statement is that forces exerted by kinetic friction act in the direction opposite the relative motion of the two surfaces. (Forces exerted by static friction act in the direction opposite the sum of all other forces). • For example a concrete block, free to slide around in the bed of a pickup truck which is accelerating forward, is accelerated by the frictional force between it and the truckbed. So the frictional force is in its direction of motion. If the block doesn't slide, it is static friction that accelerates it and there is no relative motion between the surfaces of the block and the truckbed. If the block does slide, the frictional force is still pushing it forward relative to the road, and relative to the road it accelerates in its direction of motion, but the frictional force isn't sufficient to accelerate it at the same rate as the truck; it therefore slides backward relative to the truckbed. Relative to the truckbed the block slides backward while the frictional force pushes it forward--the frictional force is in the direction opposite the relative motion. • If the block is sliding, it is moving toward the back of the truck while friction is pushing it toward the front. So in this case the frictional force acts in the direction opposite the relative motion of the two surfaces. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: `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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Total weight: 10kg + 10kg + .1kg = 20.1 kg Fnet = 20.1N * 9.8m/s/s = 196.98 1% of 196.98N = 196.98 N * .01 = 1.9698N 98.98 N - 98N = .98N .98 - 1.9698 = -.9898 N -.9898 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. STUDENT QUESTION I understand how we got the 197 newtons, but I do not understand why we need to find the 1 percent, Is friction always 1 percent of of the total weight of the object, or were we just to assume from the previous problem?????? I understand why it is negative and how we get 1.97 newtons INSTRUCTOR RESPONSE Friction is a percent of the force 'pressing' two objects together; the percent depends on the nature of the two surfaces (e.g., ice sliding over smooth plastic will be a small percent, while rubber sliding over asphalt is much higher). That percent is usually called the 'coefficient of friction', and is generally expressed in decimal form (e.g., a 1% coefficient of friction would be .01). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): 3 ------------------------------------------------ Self-critique rating: ********************************************* Question: `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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Force = Slope * weight = .07 * 3 kg (9.8m/s/s) = 2.058 N After looking at the solution I realized how to do the problem confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. ** STUDENT NOTE: i didn't think to use the acceleration of gravity for this one, it said the object was paralell INSTRUCTOR RESPONSE: The situation talks about the weight having two components, one being parallel to the incline, and the instruction tells you how to find that parallel component when the slope of the incline is small. We know from experience that an object will pick up speed along the incline, as opposed to the direction perpendicular to the incline (to move in the perpendicular direction the object would have to leave the incline, either burrowing down into the incline or levitating up off the incline). The direction along the incline is parallel to the incline. So its acceleration is parallel to the incline, and the net force must be parallel to the incline. In the absence of other forces, only gravity has a component parallel to the incline. Therefore in this ideal case the gravitational component parallel to the incline is the net force. In reality there are other forces present (e.g., friction) but the parallel gravitational component is nevertheless present, and contributes to the net force in the direction of motion. STUDENT QUESTION If we are finding the force at 29.4 newtons is that the perpendicular, and the slope * weight is parallel, why is this considered the weight instead of force in the calculation, is this because the weight is the force that is moving the object INSTRUCTOR RESPONSE The weight of an object is the force exerted on it by gravity. Since objects near the surface of the Earth accelerates downward, if free to do so, at 9.8 m/s^2, the weight of an object is 9.8 m/s^2 multiplied by its mass. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: `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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: M = 3 kg Slope = .07 F = 2.058 N F = m * a A = F / m A = 2.058 N / 3 kg = .686 m/s/s confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The weight component perpendicular to the incline is balanced by the perpendicular force exerted by the incline. The only remaining force is the parallel component of the weight, which is therefore the net force. The acceleration will therefore be a = F / m = 2.1 Newtons / (3 kg) = .7 m/s^2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: `q010. What would be the acceleration of the cart in the previous example if friction exerted a force equal to 2% of the weight of the cart, assuming that the cart is moving down the incline? [Note that friction is in fact a percent of the perpendicular force exerted by the incline; however for small slopes the perpendicular force is very close to the total weight of the object]. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: M = 3 kg Friction = 3 kg (9.8) = 29.4 * .02 = .588 F = 3kg (9.8) * .07 = 2.058 Fnet = 2.1 - .59 = 1.47 A = F / m A = 1.47N / 3kg = .49 m/s/s confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. STUDENT QUESTION (instructor comments in bold) ok, so the weight is equal to the force that is making the onject go down the incline (because force and mass are equal here????) the force is equal to the mass times the acceleration of gravity the mass is 3 kg, the weight is 29.4 Newtons the mass or weight was 29.4newtons. 2 percent of this is .02 * 29.4 = .59 newtons which is the frictional force, and that should be -.59 since friction works in the opposite direction of the system. To find the net force of the system with add -.59 + 2.1Newtons = 1.5 newtons So the mass of the systme = 3kg a= f/m = 1.5/3kg = .5m/s^2 I am a little confused, I am not sure about the weight (being the gravity 9.8m/s^2 * the mass 3kg= 29.4newtons) is related to the force of the system slope .07 * weight 29.4????? That goes back to the statement in the preceding problem, about how the gravitational force splits into two components, one being parallel to the incline and equal to slope * weight (provided slope is small). We will see soon, in terms of vectors, why this is so, and also how to deal with the situation where the slope isn't small. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating: ********************************************* Question: `q011. Given the conditions of the previous question, what would be the acceleration of the cart if it was moving up the incline? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 2.058 N + .588N = 2.65 N A = f / m A = 2.65N / 3 kg = .882 m/s/s confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. STUDENT QUESTION ?????????????????????????????????if the cart is moving downhill and this is considered in the positive direction parallel to the weight of the object which is also considered to be positve, why if it is going in the opposite direction would that not be considered negative since the systme is moving against gravity and the force of it own weight ( which made it go down the incline in the previous problem) ?????????????????????????? INSTRUCTOR RESPONSE The direction of gravity does not determine the positive direction; the positive direction is simply declared in the solution, and you would be free to use either direction as positive. Once the positive direction is declared, all forces, displacements, velocities and accelerations will be positive or negative depending on whether they are in the positive direction, or opposite to it. The positive direction as chosen in the given solution is down the incline. The displacement is up the incline, as is the velocity, so both displacement and velocity are negative. The frictional force and the component of the gravitational force along the incline are both positive, according to the choice of positive direction. So the object has a negative velocity and a positive acceleration, meaning that it is moving in the negative direction but slowing. It takes some thinking to get used to this idea; the idea is far from trivial. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: `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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The cart will have is moving in the negative direction and will have a positive acceleration, which will slow the cart down eventually bringing it to a stop.
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Given Solution: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I don’t not understand the positive and negative directions?
#$&* course phy201 3/161:06 011. `query 11
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Given Solution: `a** A conservative force conserves energy--you can get your energy back. For example: Push something massive up a hill, then climb back down the hill. The object, by virtue of its position, has the potential to return most of your energy to you, after regaining it as it rolls back down. You will have done work against gravity as you move along a path up the hill, and gravity can return the energy as it follows its path back down the hill. In this sense gravity conserves energy, and we call it a conservative force. However, there is some friction involved--you do extra work against friction, which doesn't come back to you. And some of the energy returned by gravity also gets lost to friction as the object rolls back down the hill. This energy isn't conserved--it's nonconservative. ** Another more rigorous definition of a conservative force is that a force is conservative if the work done to get from one point to another independent of the path taken between those two points. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qIf a system does work W1 against a nonconservative force while conservative forces do work W2 on the system, what are the change in the KE and PE of the system? Explain your reasoning from a commonsense point of view, and include a simple example involving a rubber band, a weight, an incline and friction. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I don’t know how to approach this problem confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** `dKE is equal to the NET work done ON the system. The KE of a system changes by an amount equal to the net work done on a system. If work W1 is done BY the system against a nonconservative force then work -W1 is done ON the system by that force. `dPE is the work done BY the system AGAINST conservative forces, and so is the negative of the work done ON the system BY nonconservative forces. • In the present case W2 stands for the work done on the system by conservative forces, so `dPE = - W2. PE decreases, thereby tending to increase KE. So work -W1 is done ON the system by nonconservative forces and work W2 is done ON the system by a conservative force. • The NET work done ON the system is therefore `dW_net_on = -W1 + W2. The KE of the system therefore changes by `dKE = -W1 + W2. If the nonconservative force is friction and the conservative force is gravity, then since the system must do positive work against friction, W1 must be positive and hence the -W1 contribution to `dKE tends to decrease the KE. e.g., if the system does 50 J of work against friction, then there is 50 J less KE increase than if there was no friction. If the work done by the conservative force on the system is positive, e.g., gravity acting on an object which is falling downward, then since force and displacement in the same direction implies positive work, gravity does positive work and the tendency will be to increase the KE of the system and W2 would be positive. A couple of numerical examples: If W2 is 150 J and W1 is 50 J, then in terms of the above example of a falling object, this would mean that gravity tends to increase the KE by 150 J but friction dissipates 50 J of that energy, so the change in KE will be only 100 J. This is consistent with `dW_net_ON = -W1 + W2 = -50 J + 150 J = 100 J. The previous example was of a falling object. If the object was rising (e.g., a ball having been thrown upward but not yet at its highest point), displacement and gravitational force would be in opposite directions, and the work done by gravity would be negative. In this case W2 might be, say, -150 J. Then `dKE would be -150 J - 50 J = -200 J. The object would lose 200 J of KE. This would of course only be possible if it had at least 200 J of KE to lose. For example, in order to lose 200 J of KE, the ball thrown upward would have to be moving upward fast enough that it has 200 J of KE. STUDENT COMMENT I find this really confusing. Could this be laid out in another way? INSTRUCTOR RESPONSE If you find this confusing at this point, you will have a lot of company. This is a challenge for most students, and these ideas will occupy us for a number of assignments. There is light at the end of the tunnel: It takes awhile, but once you understand these ideas, the basic ideas become pretty simple and even obvious, and once understood they are usually (but not always) fairly easy to apply This could be laid out differently, but would probably be equally confusing to any given student. Different students will require clarification of different aspects of the situation. If you tell me what you do and do not understand about the given solution, then I can clarify in a way that will make sense to you. I also expect that in the process of answering subsequent questions, these ideas will become increasingly clear to you. In any case feel free to insert your own interpretations, questions, etc. into a copy of this document (mark insertions with &&&& so I can locate them), and submit a copy. STUDENT QUESTION If the system goes against the force will this always make it negative? INSTRUCTOR COMMENT If a force and the displacement are in opposite directions, then the work done by that force is negative. If the system moves in a direction opposite the force exerted BY the system, the work done BY the system is negative. Note, however, that if this is the case then any equal and opposite force exerted ON the system will be in the direction of motion, so the force will do positive work ON the system. A separate document related to this problem is located in the document • work_on_vs_by_dKE_dPE_etc_questions_answers.htm STUDENT COMMENT This is a little confusing and I have read over the link that you gave. It will take some time to get use to the concepts. So, almost all of the factors are equal and opposite of each other? INSTRUCTOR COMMENT In terms only of forces acting ON an object or system, we have the following: 1. The object or system can be subjected to any number of forces acting ON the system. The net force F_net_ON is the sum, the net effect, of all those forces. 2. On any given interval the work done by the net force is equal to the change in the KE of the object or system. 3. This is summarized in the work-kinetic energy theorem `dW_net_ON = `dKE 4. Each force acting on the object or system can be classified as some combination of conservative and nonconservative forces, so 5. the net force can be expressed as the sum of a net conservative and a net nonconservative force: F_net_ON = F_net_cons_ON + F_net_noncons_ON. 6. Thus `dW_net_ON = `dW_net_cons_ON + `dW_net_noncons_ON. 7. Change in PE can be defined to be equal and opposite the work done ON the system by conservative forces: `dW_net_cons_ON = - `dPE 8. Since `dW_net_ON = `dW_net_cons_ON + `dW_net_noncons_ON, the work-kinetic energy theorem becomes `dW_net_cons_ON + `dW_net_noncons_ON = `dKE. 9. Since `dW_net_noncons_ON = -`dPE this can be written -`dPE + `dW_net_noncons_ON = `dKE. 10. This can be rearranged to `dW_net_noncons_ON = `dKE + `dPE. In the above we have explained the relationships among six quantities: `dKE F_net_ON `dW_net_ON `dW_net_ON_cons `dW_net_ON_noncons `dPE The main relationships are `dW_net_ON = `dKE and `dW_net_ON_noncons = `dKE + `dPE. If we replace the word ON by the word BY (indicating forces exerted and work done BY rather than ON the system), the force and therefore the work reverse sign. In particular this gives us `dKE + `dPE + `dW_net_BY_noncons = 0, a form which is useful in understanding some problems. For more practice, you may apply these principles to the suggested exercises at the link query_11_suggested_exercise.htm &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Very confused by the explanation, I just don’t understand what is being explained.
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Given Solution: `a** We have `dKE + `dPE + `dWbyNoncons = 0: The total of KE change of the system, PE change of the system and work done by the system against nonconservative forces is zero. Regarding the object at the system, if W_nc is the work done ON the object by nonconservative forces then work -W_nc is done BY the object against nonconservative forces, and therefore `dWnoncons_on = -W_nc. We therefore have `dKE + `dPE - W_nc = 0 so that `dPE = -`dKE + W_nc. ** Equivalently, the work-energy theorem can be stated • `dW_ON_nc = `dKE + `dPE In this example the work done on the system by nonconservative forces is labeled W_nc, without the subscript ON and without the `d in front. However it means the same thing, so the above becomes W_nc = `dKE + `dPE and we solve for `dPE to get `dPE = -`dKE + W_nc STUDENT COMMENT I’m still confused on how to understand when the energy is done on the object and when the energy is done against the object. INSTRUCTOR RESPONSE In an application, that can be the difficult question. However in this case it is stated that W_nc is the work done by nonconservative forces ON the object. STUDENT COMMENT: I had the same logic as the given solution, however I got ‘dPE = -‘dKE - W_nc as the answer. I some how got an extra negative. Maybe Work can only be positive….?? INSTRUCTOR RESPONSE: In this problem W_nc was specified as the work done on the object by nonconservative forces. You have to be careful about whether W_nc is ON the system or BY the system. You used the equation `dKE + `dPE + W_nc = 0; however that equation applies to the work done BY the system against nonconservative forces. Written more specifically the equation you used would be ‘dKE + ‘dPE + W_nc_BY = 0 so `dPE = - `dKE - `W_nc_BY. W_nc_BY = - W_nc_ON so `dPE = - `dKE + W_nc_ON. STUDENT RESPONSE WITH INSTRUCTOR'S COMMENTS (instructor comments in bold): ok, so dke + dPe - W_nc = 0 W_nc is total nonconservative forces doing work on the object, Right up to here this increases kinetic energy and decreases potential energy. there is no assumption about the sign of any of these quantities; any quantity could be positive or negative, as long as `dKE + `dPE - `dW_nc_on = 0 If `dW_nc_ON is positive then `dPE + `dKE is positive, but this could occur with positive `dKE and `dPE, or with a negative `dPE with lesser magnitude than a positive `dKE, or with a negative `dKE with lesser magnitude than a positive `dPE. All you would know is that `dKE + `dPE would be positive. If `dW_nc_ON is negative then `dPE + `dKE is negative, but this doesn't tell you anything about the sign of either of the two quantities. All we can say is that `dPE = `dW_nc_on - `dKE. Since it is decreasing the potential energy it is negative. dKE is the kinetic energy which is positive since the potential energy is increased. If `dW_nc_on = 0, for example, an increase in either KE or PE implies a decrease in the other. KE would increase due to a decrease in PE (e.g., if you drop an object), while an increase in PE would be associated with a decrease in KE (e.g., an object thrown upward gains PE as it loses KE). So an increase in KE tends to decrease PE, though `dW_nc_on can be such that KE and PE both increase. In this problem we solve for PE. So that dPE= - dKE + W_nc. As potential energy increasess kinetic energy decreases and the non conservative work is positive because it is going with the direction of force more so than against it. An increase in PE could be the result of loss of KE and/or positive work done by nonconservative forces. PE could also increase along with KE as long as `dW_nc_on is positive and large enough (e.g., a rocket increases both PE and KE due to nonconservative forces (the nonconservative forces result from ejecting fuel at high speed, i.e., from the rocket engines). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Still no clue??? ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qGive a specific example of such a process. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Need to look at solution, I don’t understand all of these symbols and the relationships between them. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** For example suppose I lift an object weighing 50 N and in the process the total nonconservative force (my force and friction) does +300 J of work on the object while its KE changes by +200 J. • The 300 J of work done by my force and friction is used to increase the KE by 200 J, leaving 100 J to be accounted for. More formally, `dW_noncons_ON = +300 J and `dKE = +200 J. Since `dW_noncons_ON = `dKE + `dPE, So +300 J = +200 J + `dPE, and it follows that `dPE = +100 J. • This 100 J goes into the PE of the object. ** STUDENT QUESTION (instructor responses inserted in bold) &&&&&&I read your example first, and it makes no sense to me. Your force and the friction does 300J of work on the object.... your force should be positive and friction should be negative right....so did you just add these two numbers together and get a positve number??? Right. No numbers were assumed for the work I do and the work done by friction. These two forces make up the nonconservative force on the system, and we just assumed a single total. If you wish you can assume, say, that I do 350 J of work and friction does -50 J. However the breakdown of the individual nonconservative forces isn't the point here. All we really need is the total work done on the object by nonconservative forces. How did you know that is KE changed by 200J.....did you just make that up or did you mathmatically figure that out??? The phrase starts with 'for example, then reads 'suppose I lift an object weighing 50 N and in the process the total nonconservative force (my force and friction) does +300 J of work on the object while its KE changes by +200 J' So all these quantities are simply assumed, for the sake of a numerical example. How does your 300J increase the KE by 200J??? I understand that if 'dKE is only 200J then the other 100J is 'dPE....I'm just not sure about the rest. There is no specified connection between the 300 J of work I do and the 200 J of KE increase. We just assume these quantities. Once we have these quantities (in this case, simply by assumption; in other problems we will often find these quantities from other information), they dictate the PE increase. There are a number of ways to think about this intuitively. For example: • If I do 300 J of work on an object, then if my force is the only force acting on it, the its KE will increase by 300 J. • If I do more than 300 J of work, but friction reduces the net force on the object to 300 J, then KE will increase by 300 J. • If all the nonconservative forces together (e.g., my force plus frictional force) do 300 J of work on the system, and if no other forces act, then the KE will increase by 300 J. • If all nonconservative forces together do 300 J of work and the KE increases by only 200 J, then the nonconservative forces can't be the whole story, because the work done by the net force is equal to the change in KE. The conclusion is that other forces must also be acting, and since they aren't nonconservative (we've assumed that all nonconservative forces together are accounted for in that 300 J), those forces must be conservative. And they must do -100 J of work on the system, so that the net force does 300 J - 100 J = 200 J of work. • Of course we can also resort to equations. Since `dW_NC_on = `dPE + `dKE, it follows that `dPE = `dW_NC_on - `dKE = 300 J - 200 J = 100 J. &&&&& STUDENT COMMENT The problem didn’t seem to fit the equation we got earlier. This stuff is very confusing. I am going to read over it again.
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Given Solution: `a** Informally: • The more clips, the more gravitational force, and the more the clips descend the more work is done by that force. • The amount of work depends on how many clips, and on how far they descend. • The number of clips required is proportional to the slope (as long as the slope is small). More formally, the force exerted by gravity is the same on each clip, so the total gravitational force on the hanging clips is proportional to the number of clips. The work done is the product of the force and the displacement in the direction of the force, so the work done is proportional to product of the number of washers and the vertical displacement. To pull the cart up a slope at constant velocity the number of washers required is proportional to the slope (for small slopes), and the vertical distance through which the cart is raised by a given distance of descent is proportional to the slope, to the work done is proportional to the vertical distance thru which the cart is raised. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qHow does the work done against friction of the cart-incline-pulley-washer system compare with the work done by gravity on the washers and the work done to raise the cart? Which is greatest? What is the relationship among the three? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Dw (noncom) = dKE + dPE The greatest force is the force exerted by gravity on the washers or the cart would not move. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The force exerted by gravity on the hanging weights tends to move the system up the incline. The force exerted by gravity on the cart has a component perpendicular to the incline and a component down the incline, and the force exerted by friction is opposed to the motion of the system. In order for the cart to move with constant velocity up the incline the net force must be zero (constant velocity implies zero accel implies zero net force) so the force exerted by gravity in the positive direction must be equal and opposite to the sum of the other two forces. So the force exerted by gravity on the hanging weights is greater than either of the opposing forces. So the force exerted by friction is less than that exerted by gravity on the washers, and since these forces act through the same distance the work done against friction is less than the work done by gravity on the washers. The work done against gravity to raise the cart is also less than the work done by gravity on the washers. • Work done against friction + work against gravity to raise cart = work by gravity on the hanging weights. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qWhat is our evidence that the acceleration of the cart is proportional to the net force on the cart? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The acceleration is equal to the net sum of forces divided by the mass of the object. As the acceleration increases or decreases, the sum of forces increases or decreases at a proportional rate. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The net force is the sum of the gravitational force on the weights, and the frictional force (one force being positive, the other negative). The acceleration is the net force divided by total mass (mass of cart plus hanger plus washers). Washers are progressively transferred from the cart to the hanger, which keeps the mass of the system constant while increasing the net force. So the acceleration increases with the number of washers on the hanger. The gravitational force on the weights is therefore proportional to the number of washers on the hanger. • With each added washer we get the same additional force, so we get the same additional acceleration. • So the graph is linear. However the acceleration is not proportional to the number of weights. • The net force on the system is equal to the gravitational force on the weights, plus the frictional force (which is of opposite sign, so while we are in fact adding quantities of opposite signs, it 'feels' like we're subtracting frictional force from gravitational force). • The gravitational force on the weights is proportional to the number of washers, but when we add in the effect of the frictional force, our force is no longer proportional to the number of weights. Still linear, but not proportional to ... . STUDENT COMMENT I did not go into this great of detail. I simply used the one equation. Is this necessary to go in this thoroughly or was this just for our knowledge? INSTRUCTOR RESPONSE The question is asking about 'our evidence'. The fact that the law has been thoroughly tested by centuries of engineering and physics is what makes it a law, but the question here is whether the evidence obtained in the experiment (in the Class Notes) indicates that the acceleration of this particular cart is proportional to the net force acting on it. Of course our experiments had better agree with the law, within their limits of precision. Our results will be more a test of our experimental design, and our execution of that design, than a test of the established law. The law itself turns out to be valid only within certain restrictions, becoming invalid at relativistic velocities and at the quantum level of matter (realms far beyond our everyday experience and perception of the world). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qprin phy and gen phy prob 34: Car rolls off edge of cliff; how long to reach 85 km/hr? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: V0 = 0 m/s A = 9.8 m/s/s Vf = 85 km/hr * (1000/3600) = 23.61 m/s A = dv / dt 9.8 m/s/s = 23.61 m/s / dt Dt = 2.41s confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aWe know that the acceleration of gravity is 9.8 m/s^2, and this is the rate at which the velocity of the car changes. The units of 85 km/hr are not compatible with the units m/s^2, so we convert this velocity to m/s, obtaining velocity • 85 km/hr ( 1000 m/km) ( 1 hr / 3600 sec) = 23.6 m/s. Common sense tells us that with velocity changing at 9.8 m/s every second, it will take between 2 and 3 seconds to reach 23.6 m/s. More precisely, the car's initial vertical velocity is zero, so using the downward direction as positive, its change in velocity is `dv = 23.6 m/s. • Its acceleration is a = `dv / `dt, so • `dt = `dv / a = 23.6 m/s / (9.8 m/s^2) = 2.4 sec, approx.. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q**** prin phy and gen phy problem 2.52 car 0-50 m/s in 50 s by graph How far did the car travel while in 4 th gear and how did you get the result? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Dv = 45 m/s - 36.5 m/s = 8.5 m/s Dt = 27.5 s - 16s = 11.5s Ds = (v0 + vf) / 2 * dt Ds = 40.75 * 11.5 = 468.63 meters confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** In 4th gear the car's velocity goes from about 36.5 m/s to 45 m/s, between clock times 16 s and 27.5 s. Its average velocity on that interval will therefore be vAve = (36.5 m/s + 45 m/s) / 2 = 40.75 m/s and the time interval is 'dt = (27.5s - 16s) = 11.5 s. We therefore have 'ds = vAve * `dt = 40.75 m/s * 11.5 s = 468.63 m. The area under the curve is the displacement of the car, since vAve is represented by the average height of the graph and `dt by its width. It follows that the area is vAve*'dt, which is the displacement `ds. The slope of the graph is the acceleration of the car. This is because slope is rise/run, in this case that is 'dv/'dt, which is the ave rate of change of velocity or acceleration. We already know `dt, and we have `dv = 45 m/s - 36.5 m/s = 8.5 m/s. The acceleration is therefore a = `dv / `dt = (8.5 m/s) / (11.5 s) = .77 m/s^2, approx. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `q **** Gen phy what is the meaning of the slope of the graph and why should it have this meaning? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The slope is the rate of change of velocity versus clock time confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The graph is of velocity vs. clock time, so the rise will be change in velocity and the run will be change in clock time. So the slope = rise/run represents change in vel / change in clock time, which is acceleration. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qGen phy what is the meaning of the area under the curve, and why does it have this meaning? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The area of the curve is the displacement of distance covered by the car confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The area under the curve is the distance traveled. This is so because 'ds = vAve*'dt. 'dt is equal to the width of the section under the curve and vAve is equal to the average height of the curve. The area of a trapezoid is width times average height. Although this is not a trapezoid it's close enough that we for the purpose of estimation can analyze it as such. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qGen phy what is the area of a rectangle on the graph and what does it represent? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The area of the rectangle represents the total distance covered confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The area of a rectangle on the graph represents a distance. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3 " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!