course Phy 231 If the slope of a graph of number of paper clips needed to maintain equilibrium vs. ramp slope is 86 clips / unit of ramp slope , and if the slope of a graph of the acceleration of a cart vs. the number of paper clips attached by a string and suspended over a pulley is (11 cm/s2) / clip: *How many cm/s2 of acceleration should correspond to 1 unit of ramp slope?
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14:34:10 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 produces work that can be reversed. Examples are gravitational and elastic forces. Nonconservative forces produce irreversible work. Examples are frictional and heat dissipating forces.
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14:40:45 ** 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 --> The work done against a conservative force is conserved in the potential energy of the object that the work is done on. The work done on against the force is the total work done on the object less the work done against other forces. The work done against a nonconservative force is not conserved as potential energy.
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16:05:59 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 --> The change in KE of the system is the net work done on the system. The work done on the system will be W2 - W1 (since W1 is done by the system, -W1 is done on the system). Using a weight being pulled up a ramp by a rubber band as a simple example, the work W1 done by the conservative forces is the product of the distance the weight moves and the difference between the forces exerted on the weight by the rubber band and by gravity. The work W2 done by the nonconservative force is the product of the frictional force and the distance the weight moves The change in PE of the system is the work done by the system against conservative forces. The conservative forces in this system are gravitational force and the elastic force of the rubber band. The net conservative force is elastic - gravitational force. This force does work W2 Change in KE is (W2 - W1), which should be a positive number since we assumed the weight was moving. Change in PE is ( -W2), which will be a negative number.
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16:11:03 ** `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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16:33:42 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 only conservative forces are acting on the object: 'dKE + 'dPE = 0 Since there is a nonconservative force also acting on the object, that work will be lost to the system: 'dKE + 'dPE - W = 0 Solved for 'dPE: 'dPE = W - 'dKE
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16:35:15 ** 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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16:38:39 Give a specific example of such a process.
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RESPONSE --> The change in potential energy of a cart moving up an incline is increased by the amount of kinetic energy used to move the cart, less the amount of energy lost to overcome the force of friction.
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16:42:56 ** 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 --> I wasn't thinking about the nonconservative forces that also increase kinetic energy.
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17:05:28 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 --> Since work can be defined: mass * acceleration * distance It makes sense that the work done by gravity (acceleration) in this situation is proportional to the number of washers (mass) and the distance they fall. Increasing the number of washers or the distance through which they fall will increase the work done. The work done on the cart in this situation is the same as the work done on the washers. They will be displaced the same vertical distance. Increasing the work done will increase the vertical distance moved. That change in distance results from adding more suspended mass while holding slope constant or from decreasing the slope of the incline while holding the suspended mass constant.
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17:05:46 ** 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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17:16:14 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 against friction of the cart-incline-pulley-washer system is not reversible. The force that produces this work is nonconservative. The work done by gravity on the washers is reversible, as is the work done to raise the cart. The force that produces this work is conservative. When there is sufficient suspended mass to cause the cart to move up the incline, the gravitational force is greater than the frictional force. The work done to raise the cart is the product of the vertical distance the cart travels and the net force acting on the cart. The net force on the cart is the difference between the component of gravitational force that is parallel to the incline and the frictional force. The work done on the washers is the product of the gravitational force on them and the distance through which they move.
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17:21:13 ** 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 --> It is simpler to acknowledge that the vertical distance moved by each mass (suspended and cart) will be the same, and consider the relationship between the forces. They will be the same as the relationships among the work.
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17:23:23 What is our evidence that the acceleration of the cart is proportional to the net force on the cart?
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RESPONSE --> The net force on the cart will be proportional to the amount of suspended mass. If we graph the acceleration of the cart vs. the amount of suspended mass, the graph will be linear if the acceleration and net force are proportional.
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17:23:28 ** the graph of acceleration vs. number of washers should be linear **
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17:23:32 prin phy and gen phy prob 34: Car rolls off edge of cliff; how long to reach 85 km/hr?
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17:23:37 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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17:23:40 **** 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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17:23:44 ** 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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17:23:49 **** Gen phy what is the meaning of the slope of the graph and why should it have this meaning?
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17:23:53 ** 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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17:23:59 Gen phy what is the meaning of the area under the curve, and why should it have this meaning?
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17:24:03 ** 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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17:24:07 Gen phy what is the area of a rectangle on the graph and what does it represent?
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17:24:12 ** The area of a rectange on the graph represents a distance. **
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17:43:28 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 --> Considering the top of the building to be position 0, the student will hit the ground when he has traveled 180 m. 'ds = v0 'dt + a/2 'dt^2 Solved for the distance the student's has traveled: 'ds = 4.9 t^2 The student will reach the ground at t = sqrt( 36.734) = 6.061sec If Superman waits 5 seconds before leaving position 0, he has 1.061 seconds to reach the student as the student reaches the ground. Solving the fundamental equation for v0: v0 = ('ds / 'dt) - (a 'dt / 2) Substituting t = 1.061 sec gives v0 = 164.5 m/sec. Graphing the positions of Superman and the student as their displacement, on the same graph; Student position = 4.9t^2 Superman position = 164.5 (t - 5) = 4.9(t - 5)^2 The y axis represents the displacement in meters, the x axis represents the time in seconds. The curves intersect at 6.061 sec, which corresponds to 179.98 m If the building is less than 122.5 m, the student will reach the ground in 5 seconds, so Superman will not be able to reach him before he reaches the ground.
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17:48:03 univ phy what is Superman's initial velocity, and what does the graph look like (be specific)?
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RESPONSE --> Superman's initial velocity is 164.5 m/sec. The graph shows the student's position as a curve that increases at an increasing rate, beginning at the origin. Superman's position crosses the x-axis at t = 5 and appears to increase at a constant rate.
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17:48:41 ** 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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17:48:52 sketch 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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17:54:48 ** 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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j {{~ Student Name: assignment #012
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19:17:16 `q001. Note that there are 4 problems in this set. Two 3 kg masses are suspended over a pulley and a 1 kg mass is added to the mass on one side. Friction exerts a force equal to 2% of the total weight of the system. If the system is given an initial velocity of 5 m/s in the direction of the lighter side, how long will a take the system to come to rest and how far will it travel?
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RESPONSE --> The force on the heavier side = 4 kg * 9.8 m/sec^2 = 39.2N The force on the lighter side = 3 kg * 9.8 m/sec^2 = 29.4 N The weight of the system = 7 kg * 9.8 m/sec^2 = 68.6 N The frictional force = 68.6 N * .02 = 1.37 N If motion is in the direction of the lighter side, it is in the negative direction, so velocity is -5 m/sec. When the system comes to a rest, the change in velocity will be +5 m/sec. The net force on the system without friction is 39.2 N - 29.4 N = 9.8 N The frictional force opposes motion in the positive direction, which is toward the heavier side, so the net force on the system is 8.4 N. The acceleration will be 8.4N / 7kg = 1.2 m/sec^2 5 m/sec / 1.2 m/sec^2 = 4.2 sec The distance in traveled in 4.2 sec will be vAve * 4.2. Since the object ended at rest, the average velocity is 1/2 the initial velocity. 2.5 * 4.2 =10 meters traveled by the system
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19:31:03 We know the initial velocity of the system and, for the period until it comes to rest, we know that its final velocity will be 0 m/s. If we can find the acceleration, we will have three of the five necessary variables with which to analyze the motion and we can therefore determine the time interval `dt and displacement `ds corresponding to this period. We begin by analyzing the forces acting on the system. Before we do so we declare the positive direction of motion for this system to be the direction in which the system moves as the greater of the two hanging masses descends, i.e., the direction of the net force on the system Gravity exerts forces of 4 kg * 9.8 m/s^2 = 39.2 Newtons on the 4 kg mass and 3 kg * 9.8 m/s^2 = 29.4 Newtons on the 3 kg mass. Taking the positive direction to be the direction in which the system moves when the 4 kg mass descends, as stated earlier, then these forces would be +39.2 Newtons and -29.4 Newtons. The total mass of the system is 7 kg, so its total weight is 7 kg * 9.8 m/s^2 = 68.4 Newtons and the frictional force is therefore frictional force = .02 * 68.4 Newtons = 1.4 Newtons, approx.. If the system is moving in the negative direction, then the frictional force is opposed to this direction and therefore positive so the net force on the system is +39.2 Newtons - 29.4 Newtons + 1.4 Newtons = +11.2 Newtons. This results in an acceleration of +11.2 N / (7 kg) = 1.6 m/s^2. We now see that v0 = -5 m/s, vf = 0 and a = 1.6 m/s^2. From this we can easily reason out the desired conclusions. The change in velocity is +5 m/s and the average velocity is -2.5 m/s. At the rate of 1.6 m/s^2, the time interval to change the velocity by +5 m/s is `dt = +5 m/s / (1.6 m/s^2) = 3.1 sec, approx.. At an average velocity of -2.5 m/s, in 3.1 sec the system will be displaced `ds = -2.5 m/s * 3.1 s = -7.8 meters. These conclusions could also have been reached using equations: since vf = v0 + a `dt, `dt - (vf - v0) / a = (0 m/s - (-5 m/s) ) / (1.6 m/s^2) = 3.1 sec (appxox). Since `ds = .5 (v0 + vf) * `dt, `ds = .5 (-5 m/s + 0 m/s) * 3.1 s = -7.8 meters.
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RESPONSE --> Friction always opposes motion, regardless of direction.
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19:47:22 `q002. If the system in the previous example was again given an initial velocity of 5 m/s in the direction of the 3 kg mass, and was allowed to move for 10 seconds without the application of any external force, then what would be its final displacement relative to its initial position?
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RESPONSE --> Using the corrected values of -7.8 m traveled in 3.125 sec, and continuing for 6.875 sec for a total of 10 seconds: The net force on the system is now 39.2 N - 29.4 N - 1.2 N = 8.4 N The acceleration is 8.4 N / 7 kg = 1.2 m/sec^2 in 6.875 sec, the change in velocity is 8.25 m/sec. The average velocity is 4.125 m/sec The displacement is 4.25 m/sec * 6.875 sec = 28.36 m. The displacement relative to its initial position is 28.36 m - 7.8 m = 20.56 m
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19:48:08 Since the acceleration of the system is different if it is moving in the positive direction than when it is moving in the negative direction, if we have both positive and negative velocities this problem must the separated into two parts. As seen in the previous example, in the first 3.2 seconds the system displaces -7.8 meters. This leaves 6.8 seconds after that instant during which the system may accelerate from rest in the positive direction. We therefore analyze the motion from the instant the system comes to rest until the remaining .8 seconds has elapsed. The frictional force during this time will be negative, as it must oppose the direction of motion. The net force on the system will therefore be + 39.2 Newtons -29.4 Newtons - 1.4 Newtons = 8.4 Newtons, and the acceleration will be 8.4 Newtons / (7 kg) = 1.2 m/s^2, approx.. The initial velocity during this phase is 0, the time interval is 6.8 sec and the acceleration is 1.2 m/s^2. We therefore conclude that the velocity will change by 1.2 m/s^2 * 6.8 sec = 8.2 m/s, approx, ending up at 8.2 m/s since this phase started at 0 m/s. This gives an average velocity of 4.1 m/s; during 6.8 sec the object therefore displaces `ds = 4.1 m/s * 6.8 sec = 28 meters approx.. These results could have also been easily obtained from equations. For the entire 10 seconds, the displacements were -7.8 meters and +28 meters, for a net displacement of approximately +20 meters. That is, the system is at this instant about 20 meters in the direction of the 4 kg mass from its initial position.
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20:11:01 `q003. An automobile has a mass of 1400 kg, and as it rolls the force exerted by friction is .01 times the 'normal' force between its tires and the road. The automobile starts down a 5% incline (i.e., an incline with slope .05) at 5 m/s. How fast will it be moving when it reaches the bottom of the incline, which is 100 meters away (neglect air friction and other forces which are not part of the problem statement)?
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RESPONSE --> Automobile has a weight of 1400 * 9.8 m/sec^2 = 13720 N. The force exerted by friction is 13720 * .01 = 137.2 N The force exerted parallel to the incline by gravity (since the slope is small) = 13720 N * .05 = 686 N. The net force = 686 N - 137.2 N = 548.8 N The acceleration = 548.8 N / 140 kg = .39 m/sec^2 Using the fundamental equation vf^2 = v0^2 + 2a 'ds, knowing that the final velocity will be positive: vf^2 = 25 m^2 / sec^2 + .784 m/sec^2 * 100 m solved for vf = 10.16 m/sec
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20:11:16 We are given the initial velocity and displacement of the automobile, so we need only find the acceleration and we can analyzes problem as a standard uniform acceleration problem. The automobile experiences a net force equal to the component of its weight which is parallel to the incline plus the force of friction. If we regard the direction down the incline as positive, the parallel component of the weight will be positive and the frictional force, which must be in the direction opposite that of the velocity, will be negative. The weight of the automobile is 1400 kg * 9.8 m/s^2 = 13720 Newtons, so the component of the weight parallel to the incline is parallel weight component = 13720 Newtons * .05 = 686 Newtons. The normal force between the road and the tires is very nearly equal to the weight of the car because of the small slope, so the magnitude of the frictional force is approximately frictional force = 13720 Newtons * .01 = 137 Newtons, approx.. The frictional force is therefore -137 Newtons and the net force on the automobile is Fnet = 686 Newtons - 137 Newtons = 550 Newtons (approx.). It follows that the acceleration of the automobile must be a = Fnet / m = 550 Newtons / 1400 kg = .4 m/s^2 (approx.). We now have a uniform acceleration problem with initial velocity v0 = 5 meters/second, displacement `ds = 100 meters and acceleration a = .4 m/s^2. We can easily find the final velocity using the equation vf^2 = v0^2 + 2 a `ds, which gives us vf = +- `sqrt(v0^2 + 2 a `ds) = +- `sqrt( (5 m/s)^2 + 2 * .4 m/s^2 * 100 m) = +- `sqrt( 25 m^2 / s^2 + 80 m^2 / s^2) = +-`sqrt(105 m^2 / s^2) = +- 10.2 m/s. It is obvious that the final velocity in this problem is the positive solution +10.2 m/s, since the initial velocity and acceleration are both positive.
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20:23:55 `q004. If the automobile in the previous example started at the bottom of the incline with velocity up the incline of 11.2 m/s, how far up the hill would it be able to coast?
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RESPONSE --> The net force will be opposing the direction of motion, as will the frictional force, so net force is negative with a magnitude of 686 + 137.2 = 823.2 N The final velocity is zero, so average velocity is -5.6 m/sec Acceleration is -823.2 / 1400 = -.588 m / sec^2 Time until velocity = zero: -11.2 cm/sec / -.588 cm/sec^2 = 19.0 sec 'ds = 5.6 m/sec * 19.0 sec = 106.7 sec
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20:24:06 We again have a uniform acceleration situation where the initial velocity is known and we have the information to determine the acceleration. The other quantity we can deduce is the final velocity, which at the furthest point up the hill will be zero. Since the automobile is coasting up the incline, we will take the upward direction as positive. The frictional force will still be 137 Newtons and will again be directed opposite the velocity, so will therefore be negative. The parallel component of the weight will be the same as before, 686 Newtons, but being directed down the incline will be in the direction opposite to that if the velocity and will therefore also be negative. The net force on the automobile therefore be net force = -686 Newtons - 137 Newtons = -820 Newtons (approx.). Its acceleration will be a = Fnet / m = -820 Newtons / 1400 kg = -.6 m/s^2 (approx.). We see now that v0 = 11.2 m/s, vf = 0 and a = -.6 m/s^2. Either by direct reasoning or by using an equation we easily find that `dt = (-11.2 m/s) / (-.6 m/s^2) = 19 sec (approx) and `ds = (11.2 m/s + 0 m/s) / 2 * 19 sec = 5.6 m/s * 19 sec = 106 meters (approx).
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