course Phy 231 In the last problem of this assignment do you agree that the difference in value was largely due to a magnification of difference in rounding, or was an error in my process? ³Ñ¤™GˆÃ׃¾½ŽèPΡÀ»Ã„¨‚è¸OÞassignment #017
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14:45:54 ANSWERS/COMMENTARY FOR QUERY 17
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RESPONSE --> Ok
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14:46:05 prin phy and gen phy 6.33: jane at 5.3 m/s; how high can she swing up that vine?
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RESPONSE --> Ok confidence assessment: 3
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14:46:12 Jane is going to convert her KE to gravitational PE. We assume that nonconservative forces are negligible, so that `dKE + `dPE = 0 and `dPE = -`dKE. Jane's KE is .5 M v^2, where M is her mass. Assuming she swings on the vine until she comes to rest at her maximum height, the change in her KE is therefore `dKE = KEf - KE0 = 0 - .5 M v0^2 = - .5 M v0^2, where v0 is her initial velocity. Her change in gravitational PE is M g `dy, where `dy is the change in her vertical position. So we have `dKE = - `dPE, or -5 M v0^2 = - ( M g `dy), which we solve for `dy (multiply both sides by -1, divide both sides by M g) to obtain `dy = v0^2 / (2 g) = (5.3 m/s)^2 / (2 * 9.8 m/s^2) = 1.43 m.
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RESPONSE --> Ok self critique assessment: 3
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14:46:18 prin phy and gen phy 6.39: 950 N/m spring compressed .150 m, released with .30 kg ball. Upward speed, max altitude of ball
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RESPONSE --> Ok confidence assessment: 3
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14:46:19 prin phy and gen phy 6.39: 950 N/m spring compressed .150 m, released with .30 kg ball. Upward speed, max altitude of ball
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RESPONSE --> confidence assessment:
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14:46:28 We will assume here that the gravitational PE of the system is zero at the point where the spring is compressed. In this situation we must consider changes in both elastic and gravitational PE, and in KE. The PE stored in the spring will be .5 k x^2 = .5 ( 950 N/m ) ( .150 m)^2 = 107 J. When released, conservation of energy (with only elastic and gravitational forces acting there are no nonconservative forces at work here) we have `dPE + `dKE = 0, so that `dKE = -`dPE. Since the ball is moving in the vertical direction, between the release of the spring and the return of the spring to its equilibrium position, the ball t has a change in gravitational PE as well as elastic PE. The change in elastic PE is -107 J, and the change in gravitational PE is m g `dy = .30 kg * 9.8 m/s^2 * .150 m = +4.4 J. The net change in PE is therefore -107 J + 4.4 J = -103 J. Thus between release and the equilibrium position of the spring, `dPE = -103 J The KE change of the ball must therefore be `dKE = - `dPE = - (-103 J) = +103 J. The ball gains in the form of KE the 103 J of PE lost by the system. The initial KE of the ball is 0, so its final KE is 103 J. We therefore have .5 m vv^2 = KEf so that vf=sqrt(2 KEf / m) = sqrt(2 * 103 J / .30 kg) = 26 m/s. To find the max altitude to which the ball rises, we return to the state of the compressed spring, with its 107 J of elastic PE. Between release from rest and max altitude, which also occurs when the ball is at rest, there is no change in velocity and so no change in KE. No nonconservative forces act, so we have `dPE + `dKE = 0, with `dKE = 0. This means that `dPE = 0. There is no change in PE. The initial PE is 107 J and the final PE must also therefore be 107 J. There is, however, a change in the form of the PE. It converts from elastic PE to gravitational PE. Therefore at maximum altitude the gravitational PE must be 107 J. Since PEgrav = m g y, and since the compressed position of the spring was taken to be the 0 point of gravitational PE, we therefore have y = PEgrav / (m g) = 107 J / (.30 kg * 9.8 m/s^2) = 36.3 meters. The ball will rise to an altitude of 36.3 meters above the compressed position of the spring.
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RESPONSE --> Ok self critique assessment: 3
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14:46:34 gen phy problem A high jumper needs to be moving fast enough at the jump to lift her center of mass 2.1 m and cross the bar at a speed of .7 m/s. What minimum velocity does she require?
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RESPONSE --> Ok confidence assessment: 3
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14:47:04 FORMAL SOLUTION: Formally we have `dPE + `dKE = 0. `dPE = M * g * `dy = M * 20.6 m^2 / sec^2, where M is the mass of the jumper and `dy is the 2.1 m change in altitude. `dKE = .5 M vf^2 - .5 M v0^2, where vf is the .7 m/s final velocity and v0 is the unknown initial velocity. So we have M g `dy + .5 M vf^2 - .5 M v0^2 = 0. Dividing through by M we have g `dy + .5 vf^2 - .5 v0^2 = 0. Solving for v0 we obtain v0 = sqrt( 2 g `dy + vf^2) = sqrt( 2* 9.8 m/s^2 * 2.1 m + (.7 m/s)^2 ) = sqrt( 41.2 m^2/s^2 + .49 m^2 / s^2) = sqrt( 41.7 m^2 / s^2) = 6.5 m/s, approx.. LESS FORMAL, MORE INTUITIVE, EQUIVALENT SOLUTION: The high jumper must have enough KE at the beginning to increase his PE through the 2.1 m height and to still have the KE of his .7 m/s speed. The PE change is M * g * 2.1 m = M * 20.6 m^2 / sec^2, where M is the mass of the jumper The KE at the top is .5 M v^2 = .5 M (.7 m/s)^2 = M * .245 m^2 / s^2, where M is the mass of the jumper. Since the 20.6 M m^2 / s^2 increase in PE must come at the expense of the initial KE, and since after the PE increase there is still M * .245 m^2 / s^2 in KE, the initial KE must have been 20.6 M m^2 / s^2 + .245 M m^s / s^2 =20.8 M m^s / s^2, approx. If initial KE is 20.8 M m^s / s^2, then .5 M v0^2 = 20.8 M m^s / s^2. We divide both sices of this equation by the jumper's mass M to get .5 v0^2 = 20.8 m^2 / s^2, so that v0^2 = 41.6 m^2 / s^2 and v0 = `sqrt(41.6 m^2 / s^2) = 6.5 m/s, appprox.
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RESPONSE --> Ok self critique assessment: 3
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15:05:34 query Univ. 7.42 (7.38 in 10th edition). 2 kg block, 400 N/m spring, .220 m compression. Along surface then up 37 deg incline all frictionless. How fast on level, how far up incline?
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RESPONSE --> We have the spring having a potential energy of PEel = 1/2 k * x^2; therefore, we find PEel = 1/2 (400N/m) * (.220 m)^2 = 9.68 J. This allows to to find the vf since the block has an initial velocity of zero this means the KE value, which will be that of the converted PE, will represent the vf of the block; therefore, 9.68 J = 1/2 (2 kg) * vf^2; 9.68 J = vf^2; 3.11 m/s at release. To find the distance traveled we will use the formula 'ds = (vf^2 - v0^2)/2 a, where we have a value of a = g * sin 37 = 5.90 m/s^2. Now we find that 'ds = (3.11 m/s)^2/ 2 (5.9 m/s^2) = 9.68 J/ ( 11.8 m/s^2) = 0.82 m confidence assessment: 3
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15:11:11 ** The spring exerts a force of 400 N / m * .220 m = 84 N at the .220 m compression. The average force exerted by the spring between equilibrium and this point is therefore (0 N + 84 N) / 2 = 42 N, so the work done in the compression is `dW = Fave * `ds = 42 N * .220 m = 5.0 Joules, approx. If all this energy is transferred to the block, starting from rest, the block's KE will therefore be 5.0 Joules. Solving KE = .5 m v^2 for v we obtain v = sqrt(2 KE / m) = sqrt(2 * 5.0 Joules / (2 kg) ) = 2.2 m/s, approx.. No energy is lost to friction so the block will maintain this speed along the level surface. As it begins to climb the incline it will gain gravitational PE at the expense of KE until the PE is 5.0 J and the KE is zero, at which point it will begin to slide back down the incline. After traveling through displacement `ds along the incline the height of the mass will be `ds sin(37 deg) = .6 `ds, approx., and its gravitational PE will be PE = m g h = m g * .6 `ds = .6 m g `ds. Setting this expression equal to KE we obtain the equation .6 m g `ds = KE, which we solve for `ds to obtain `ds = KE / (.6 m g) = 5.0 Joules / (.6 * 2 kg * 9.8 m/s^2) = .43 meters, approx. **
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RESPONSE --> I agree completely with the equations used here; however, in the calculation of the energy and the following velocity I have different values. When I inputed in the exact format the values given for you answer I still found the answer that I gave, which is that the block would have 9.8 J not 5 J. Looking at this logically I don't see how .22 m * 42 N = 5 J, because .22 is roughly 1/5, and 1/5 of 40 is 8. Also my value of average force was 44 N not 42, because 400 N * .22 = 88 N not 84. I'm not sure why are values differ, but as I said even when I input the values in the same format as given here I still find the values I gave earlier and not these answers. self critique assessment: 3
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16:52:07 query univ phy 7.50 62 kg skier, from rest, 65 m high. Frict does -10.5 kJ. What is the skier's speed at the bottom of the slope? After moving horizontally over 82 m patch, air res 160 N, coeff frict .2, how fast is she going? Penetrating 2.5 m into the snowdrift, to a stop, what is the ave force exerted on her by the snowdrift?
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RESPONSE --> The problem in the book has a small modification from the values presented here. Instead of a 62 kg skier the book has a 60 kg skier, which is value I used during the problem. The skier has a potential energy determined by the vertical 'ds, gravity, and mass; therefore, PE = 60 kg * 9.8 m/s^2 * 65 m = 35280 J. This will be converted to KE as the skier moves down the hill, and will lose 10.5 kJ to fricion; therfore, KE at the bottom will be 35.28 kJ - 10.5 kJ = 24.78 kJ. This gives us a vf determined by the change in KE, and since the skier has no initial energy this will be found by the KE at the bottom of 24,780 J = 1/2 (60 kg) * vf^2. This gives us 826 m^2/s^2 = vf^2, which gives us vf = 28.74 m/s. Now we find how fast she is going at the bottom of the slope by determing the net force acting against the her as she moves across the patch. Since we already have air resistance in Newtons we only need to determine the Ffriction, which is Ffric = .2 * 60 kg * 9.8 m/s^2 = 117.6, and we take this and the air resistance force as being against her; therefore, we find the Fnet = -117.6 N - 160 N = -277.6 N. With this we find that the skier has acceleration of -277.6 N/ 60 kg = -4.63 m/s^2. This allows us to determine the speed of skier at the end of the patch to be vf = 'sqrt [ ( 28.74 m/s)^2 + 2 (-4.63 m/s^2) ( 82 m)] = 'sqrt (826 m^2/s^2 - 759.32 m^2/s^2) = 8.17 m/s. We now find the force acting upon the skier necessary to stop her in 2.5 m by using the energy of the final velocity 8.17 m/s; therefore, KE = 1/2 (60 kg) (8.17)^2 = 30 kg * (66.68 m^2/s^2) = 2000 J approximately. With this much energy we know that the KE/ 'ds = F; therefore, 2,000 J/ 2.5 m = 800 N = Fnet. confidence assessment: 3
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17:05:38 ** The gravitational PE of the skier decreases by 60 kg * 9.8 m/s^2 * 65 m = 38 kJ, approx. (this means 38 kiloJoules, or 38,000 Joules). The PE loss partially dissipated against friction, with the rest converted to KE, resulting in KE = 38 kJ / 10.5 kJ = 27.5 kJ. Formally we have `dKE + `dPE + `dWnoncons = 0, where `dWnoncons is the work done by the skier against friction. Since friction does -10.5 kJ of work on the skier, the skier does 10.5 kJ of work against friction and we have `dKE = -`dPE - `dWnoncons = - (-38 kJ) - 10.5 kJ = 27.5 kJ. The speed of the skier at this point will be v = sqrt( 2 KE / m) = sqrt( 2 * 27,500 J / (65 kg) ) = 30 m/s, approx. Over the 82 m patch the force exerted against friction will be .2 * 60 kg * 9.8 m/s^2 = 118 N, approx., so the force exerted against nonconservative forces will be 118 N + 160 N = 280 N approx.. The work done will therefore be `dWnoncons = 280 N * 82 m = 23 kJ, approx., and the skier's KE will be KE = 27.5 kJ - 23 kJ = 4.5 kJ, approx. This implies a speed of v = sqrt( 2 KE / m) = 12 m/s, approx. To stop from this speed in 2.5 m requires that the remaining 4.5 kJ of KE be dissipated in the 2.5 m distance. Thus we have `dW = Fave * `ds, so that Fave = `dW / `ds = 4500 J / (2.5 m) = 1800 N. This is a significant force, about 3 times the weight of the skier, but distributed over a large area of her body will cause a good jolt, but will not be likely to cause injury.**
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RESPONSE --> Ok, but I believe that you intended for their to be a ' - ' symbol in the final equation of the first paragraph here, becuase 38 kg / 10.5 kJ is not equal to 27.5 kJ, but 38 kJ - 10.5 kJ is. During this problem there were some differences due to rounding, and this caused a magnification of differences starting with the determination vf after the patch, which then have us greatly different force values when she hit the bank of snow. self critique assessment: 3
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