course Phy 231 mvxOٸassignment #011
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19:52:21 set 3 problems 15-19. Explain the difference between a conservative and a nonconservative force, and give an example of each.
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RESPONSE --> A conservative force is one that is capable of storing the energy within the system, while nonconservative force cannot store the energy and the system will lose that energy. An example conservative force is gravititational force and it conservation of potential and kinetic energy as an object is raised and lowered. An example of a nonconservative force would be friction, which retards the other forces of a system, and will dissapate the energy it cannot store in other forms such as heat that will leave the system. confidence assessment: 3
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19:52:51 ** A conservative force conserves energy--you can get your energy back. For example: Push something up a hill, climb back down the hill and turn your back and it will probably return your energy to you--all at once--after regaining it as it rolls back down. So you do work against gravity, and gravity can return the energy as it pulls the thing back to you. 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. **
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RESPONSE --> Ok self critique assessment: 3
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20:18:01 If 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.
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RESPONSE --> Within this sysytem we have forces involving three objects: a rubber band, a weight, and an incline; therefore, the rubberband if assumed to be ideal will act only with conservative forces; however, the weight and ramp as the weight moves up and down the ramp will involve the nonconservative force of friction. The complete example could be a weight is attached to the rubber band and then pushed down the ramp. Assuming that the rubber bands elastic limit is not met and is able to provide sufficient force the weight will be returned to a position near it original position, but not at that position, because it will lose energy to the nonconservative force of friction. During this example work is done on the weight to make it move down the ramp from rest; therefore, work is done on the system, while the system does work against the nonconservative force of friction. The KE change present is the the weight will first gain KE as it is acted upon by another force, then as it goes down the ramp it will lost PE of gravity, then as the rubber band pulls it back it will regain that PE as it lose KE to gravity and friction. The rubber band as it stretches will gain PE as it is stretched, and lose this PE in froms of KE as it moves the weight against the forces of gravity and friction. This makes sense in common terms, because when a rubber band is stretched it is storing gaining energy as force is applied to it and it returns that force as it returns to the shape and position before stretching. For the weight this makes sense because it has the potential energy to slide down the as it rests on the slope, but is held back by the friction and rubber at rest; however, once it is pushed down the slope it no longer has the potential to slide down the slope, because it is sliding the slope, and it will regain that potential because it will return to a position near where it previously was on the ramp; however, it also makes sense not to return to the same place because friction will responsible for part of the lose of kinetic and not just gravity. confidence assessment: 3
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20:20:07 ** `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 this case then `dPE = - W2. PE decreases, thereby tending to increase KE. If work -W1 is done ON the system by a nonconservative force and W2 is done ON the system by a conservative force, the NET work done ON the system is -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 nonconservative force on the system is positive, e.g., gravity acting on an object which is falling downward (force and displacement in the same direction implies positive work), the tendency will be to increase the KE of the system and W2 would be positive. If W2 is 150 J and W1 is 50 J, this means 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. If the object was rising, 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 (which would only be possible if it had at least 200 J of KE to lose--think of an object with considerable velocity sliding up a hill). **
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RESPONSE --> Ok, I did not know you wished an example with specific values since they could be chosen at random, and thought the important aspect of the question was the concepts being discussed. self critique assessment: 2
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20:38:09 If the KE of an object changes by `dKE while the total nonconservative force does work W on the object, by how much does the PE of the object change?
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RESPONSE --> If we have an object at the top of an incline and the object is placed there initially at rest but instantaneously overcomes static friction and begins sliding down here is how the energy would change. Let's say that is was a 1 kg object 1 m high; therefore, it had the potential of 1 kg * 9.8 m/s^2 * 1m = 9.8 joules. This PE will decrease as it slide down the ramp and is converted to KE and lost the the nonconservative force of friction. Lets say that the retarding force of friction is that of .1 the mass of the object and this is lost during the entire distance of 1 m; therefore, the friction will have a energy lose of .1 * 1 kg * 9.8 m/s^2 * 1 m = .98 joules. This means that the object will lose .98 joules of PE to friction and 8.82 joules to KE. The total change of PE as it reaches the bottom of the incline would be the total 9.8 joules. confidence assessment: 3
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20:39:42 ** 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 is the work done ON the object by nonconservative forces then work -W is done BY the object against nonconservative forces, and therefore `dWnoncons = -W. We therefore have `dKE + `dPE - W = 0 so that `dPE = -`dKE + W. **
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RESPONSE --> Ok, I actually gave specific values this time and they were not necessary. I descbribed this a more specific format and did not give my answer in the appropiate general form. self critique assessment: 3
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20:40:05 Give a specific example of such a process.
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RESPONSE --> I gave a specific example of this process in my previous answer. confidence assessment: 3
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20:40:55 ** For example if 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 then 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. This 100 J goes into the PE of the object. **
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RESPONSE --> Ok self critique assessment: 3
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20:44:18 Class notes #10. Why does it make sense that the work done by gravity on a set of identical hanging washers should be proportional to the product of the number of washers and the distance through which they fall? Why is this consistent with the idea that the work done on a given cart on an incline is proportional to the vertical distance through which the cart is raised?
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RESPONSE --> Work is defined as the product of mass times distance and is directly proportional to both values; therefore, it makes sense work is directly proportional to the increased mass of the washers and the distance from the ground. This is consitant with a cart on an incline because another part of the definition of work is only the force applied parrallel to the direction traveld is applicable to the work be calculated; therefore, only the vertical change in distance is applicable for a cart on a ramp. confidence assessment: 3
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20:45:00 ** 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. **
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RESPONSE --> Ok self critique assessment: 3
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20:54:24 How 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?
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RESPONSE --> The work done by gravity is conservative, and is acting to convert the PE of the system to KE, while friction is nonconservative acting to change the PE into another form of energy lost by the system. The greatest work is that of the gravity on the washers, because it is acting with the greatest force on the washers which are pulling the cart and the cart has mass equal or less than the washers and has equal or less force acting upon it, and the friction has to have less force than either or there would be no motion of the system. The relationship of the three is force of the washers is equal to or greater than the force of the cart and friction combined. confidence assessment: 3
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20:55:37 ** 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. The work friction + work against gravity to raise cart = work by gravity on the hanging weights. **
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RESPONSE --> Ok self critique assessment: 3
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20:57:11 What is our evidence that the acceleration of the cart is proportional to the net force on the cart?
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RESPONSE --> It is proportional because of the relationship defined by a = F/ m, which states that as Force goes up so does the acceleration in directy proportion, because on a given mass more force allows for greater change in velocity. confidence assessment: 3
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21:01:43 ** the graph of acceleration vs. number of washers should be linear **
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RESPONSE --> The acceleration should be linear, because there would be no acceleration if all forces involved were equal and opposite; therefore, regardless of mass if you have 0 / mass of any value you have 0 for acceleration. self critique assessment: 3
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21:01:54 prin phy and gen phy prob 34: Car rolls off edge of cliff; how long to reach 85 km/hr?
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RESPONSE --> Ok confidence assessment: 3
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21:02:06 We 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 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..
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RESPONSE --> Ok self critique assessment: 3
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21:02:14 **** 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?
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RESPONSE --> Ok confidence assessment: 3
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21:02:23 ** 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 distance traveled, 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. **
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RESPONSE --> Ok self critique assessment: 3
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21:02:31 **** Gen phy what is the meaning of the slope of the graph and why should it have this meaning?
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RESPONSE --> Ok confidence assessment: 3
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21:02:38 ** 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. **
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RESPONSE --> Ok self critique assessment: 3
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21:02:46 Gen phy what is the meaning of the area under the curve, and why should it have this meaning?
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RESPONSE --> Ok confidence assessment: 3
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21:02:55 ** 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 we can annalyze it as one for estimation puposes. **
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RESPONSE --> Ok self critique assessment: 3
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21:03:00 Gen phy what is the area of a rectangle on the graph and what does it represent?
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RESPONSE --> Ok confidence assessment: 3
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21:03:06 ** The area of a rectange on the graph represents a distance. **
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RESPONSE --> Ok self critique assessment: 3
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21:35:42 univ phy problem 2.90 from 10th edition (University Physics students should solve this problem now). Superman stands on the top of a skyscraper 180 m high. A student with a stopwatch, determined to test the acceleration of gravity for himself, steps off the top of the building but Superman can't start after him for 5 seconds. If Superman then propels himself downward with some init vel v0 and after that falls freely, what is the minimum value of v0 so that he catches the student before that person strikes the ground?
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RESPONSE --> If the student jumped off a building 180 m high he would have a total fall time of 0 = 180 m + .5( -9.8 m/s^2) 'dt^2; -360 m = -9.8 m/s^2 * 'dt^2; 36.73 s^2 = 'dt^2; 6.06 s = 'dt. This means that in order to reach the student in time superman must have initial velocity great enough to catch him in 6.06 s - 5 s = 1.06 s or less. For this to happen he must have an initial velocity determined by vAve = 180 m / 1.06 s = 169.8 m/s; therefore, the initial velocity would be equal to that of the average velocity minus half of the acceleration he would eperience during 1.06 s. This gives 169.8 m/s - .5( 9.8 m/s^2) (1.06 s) = 169.8 m/s - 5.194 m/s = 164.6 m/s = v0. confidence assessment: 3
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21:46:40 univ phy what is Superman's initial velocity, and what does the graph look like (be specific)?
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RESPONSE --> 164.6 m/s as initial velocity. As for a graph I am unsure as to which kind of graph you would like for me to describe, because I have the 11th edition of the text book, and the problems do not properly align; therefore, of the choices of position, velocity, or acceleration v.s time, or velocity or acceleratoin vs. position I am unsure as to which graph to describe here. confidence assessment: 3
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21:48:47 ``a** In time interval `dt after leaving the building the falling student has fallen through displacement `ds = v0 `dt + .5 a `dt^2, where v0 = 0 and, choosing the downward direction to be positive, we have a = -9.8 m/s^2. If `ds = -180 m then we have `ds = .5 a `dt^2 and `dt = sqrt(2 * `ds / a) = sqrt(2 * -180 m / (-9.8 m/s^2)) = 6 sec, approx.. Superman starts 5 seconds later, and has 1 second to reach the person. Superman must therefore accelerate at -9.8 m/s^2 thru `ds = -180 m in 1 second, starting at velocity v0. Given `ds, `dt and a we find v0 by solving `ds = v0 `dt + .5 a `dt^2 for v0, obtaining v0 = (`ds - .5 a `dt^2) / `dt = (-180 m - .5 * -9.8 m/s^2 * (1 sec)^2 ) / (1 sec) = -175 m/s, approx. Note that Superman's velocity has only about 1 second to change, so changes by only about -9.8 m/s^2, or about -10 m/s^2. **
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RESPONSE --> Ok confidence assessment: 3
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21:59:23 ``qsketch a graph of Superman's position vs. clock time, and on the same graph show the student's position vs. clock time, with clock time starting when the person begins falling
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RESPONSE --> Now I can describe the graph since I know what I am comparing. With a graph of position vs. clock time we have the student represented by one curve that would begin with a value of 0s, 180 m and curve downwards towards the point 0 m, 6 s. This curve would also open downwards and have a slope of determined by the acceleration of gravity being -9.8 m/s^2. Supermans curve would begin at 5 s, 180 m. His curve would sharply curve downwards to meet the point of 180 m, 6s, and open upwards as his position would change at a slower rate as he approached the student free falling at the same rate. This would give his curve a sharp drop represented by an acceleration that is mathematically correct for the graph but false since we know he had instantaneous acceleration to a higher initial velocity, at least that is what we are assuming. This curve would not accurately derive acceleration, because while he was in free fall for the entire second he had an initial velocity greater than that of the student. confidence assessment: 3
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22:00:23 ** If we start our clock at t = 0 at the instant the student leaves the top of the building then at clock time t the student's `dt will be just equal to t and his position will be x = x0 + v0 t + .5 a t^2 = .5 a t^2, with x0 = 180 m and a = -9.8 m/s^2. A graph of x vs. t will be a parabola with vertex at (0,180), intercepting the t axis at about t = 6 sec. For Superman the time of fall will be `dt = t - 5 sec and his position will be x = x0 + v0 (t-5sec) + .5 a (t-5sec)^2, another parabola with an unspecified vertex. A graph of altitude vs. t shows the student's position as a parabola with vertex (0, 180), concave downward to intercept the t axis at (6,0). Superman's graph starts at (5,180) and forms a nearly straight line, intercepting the t axis also at (6,0). Superman's graph is in fact slightly concave downward, starting with slope -175 and ending with slope -185, approx. **
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RESPONSE --> Ok self critique assessment: 3
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