course PHY201 ?????????q???\?assignment #017
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14:49:19 ANSWERS/COMMENTARY FOR QUERY 17
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15:06:05 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 --> I thought about this one for a while, and I think it is a pretty simple one. I beleive that Jane's forward momentum will be working against gravity, and no matter how long than the vine is she will reach the same height. If the vine is shorter than the height that should would attain in the jump, then she might not reach her full height since she would start moving backwards once she reached the point where the vine was completely parallel to the ground. They did specify a ""tall tree"" though, so this is doubtful. a = -9.8 m/s/s v0 = 5.3 m/s vf = 0 0 = 5.3 m/s ^2 + 2 * -9.8 m/s/s * `ds 0 = 28.09 m^2/s^2 - 19.6 m/s/s * `ds `ds = -28.09 m^2/s^2 / 19.6 m/s/s `ds = -1.433 m She can swing to a height of 1.433 m confidence assessment: 2
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17:28:45 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 --> Same result, different method. I need to try and start thinking in terms of KE and PE. I don't remember seeing the m * g * `dy = grav PE formula before though, but I guess it is common sense. self critique assessment: 2
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16:18:16 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 --> k = 950 N/m `ds = .150 KE = 950 N/m * .150 m = 142.5 N 142.5 N = .30kg * a a = 142.5 N / .30 kg a = 475 m/s/s v = sqrt(2KE/m) v = sqrt(2 * 142.5 N / .30kg) = sqrt(950 m^2/s^2) = 30 m/s KE = m * g * `dy 142.5 N = .3 kg * 9.8 m/s/s * `dy `dy = 142.5 N / 2.94 kg m/s/s `dy = 48.47 m Solve for velocity at max height of spring: vf^2 = v0^2 + 2a`ds vf^2 = 0 + 2 * 475 m/s/s * .15 m vf^2 = 142.5 m^2/s^2 vf = 11.94 m/s Solve for height of flight: v0 = 11.94 m/s vf = 0 m/s a = -9.8 m/s/s 0 = 11.94 m/s + -9.8m/s/s * `dt `dt = -11.94 m/s / 9.8 m/s/s `dt = -1.22 s `ds = 11.94 m/s / 2 * 1.22 = -7.2834 Height of flight: 7.28 m Speed at launch:11.94 m/s I tried to do this both ways, using energy and equations of motion. I was pretty unsuccessful with energy, and even with equations of motion my signs and units weren't working out. confidence assessment: 1
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16:24:27 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 --> I just need to work with this more. I can see that I am at roughly the same level of discomfort with these energy problems that I was at with the uniform motion problems, and I am comfortable with those now. This is just going to take some time. self critique assessment: 2
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16:29:17 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 --> Something happened and I managed to click through the next few frames at once. Even looking at the answer to the question, I don't quite understand it. I just need to study this more. confidence assessment: 1
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16:29:34 ** 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 --> self critique assessment: 3
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16:29:46 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 --> confidence assessment: 3
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16:29:51 ** 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 --> self critique assessment: 3
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