course phy121 €??y????????????assignment #017017. `query 17
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13:55:55 ANSWERS/COMMENTARY FOR QUERY 17
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RESPONSE -->
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13:57:26 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 --> `dy = v0^2 / (2*g) = 5.3m/s^2 / (2 * 9.8m/s/s) = 1.43m confidence assessment: 2
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13:59:22 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 --> self critique assessment: 3
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14:25:06 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 --> PE of spring = .5 k x^2 = .5 * 980N/m * .150m^2 = .5 * 980N/m = 490N/m .150m^2 = .0225m^2 490N/m * .0225m^2 = 11.03J since the PE of the spring will decrease to 0, increasing the KE of the spring to 11.03J, the spring will do 11.03J of work on the ball. At the point where the spring finished working on the ball, the ball has a KE of 11.03J. KE = .5 m v^2 so v = sqrt[(2KE)/m] = 8.58m/s `dy = v0^2 / (2g) = 8.58m/s / (2 * 9.8m/s/s) = .438m confidence assessment: 1
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14:29:19 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 --> There are some discrepencies here: The spring rate was 980N/m. .5 * 980 * .15^2 = ~11J. According to my findings, y should be y = PEgrav / (m g) = 11.04J / (.3kg * 9.8m/s/s) = 3.76m I got .438m before. self critique assessment: 0
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