course phy 121 The book's answers and your answers are in disagreement for text problem #39 ôý‚Íìì˜ð¼•—¢eçõò›ÅŽÅöNÙ£¸assignment #017
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08:36:32 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 --> Enter, as appropriate, an answer to the question, a critique of your answer in response to a given answer, your insights regarding the situation at this point, notes to yourself, or just an OK. Always critique your solutions by describing any insights you had or errors you makde, and by explaining how you can make use of the insight or how you now know how to avoid certain errors. Also pose for the instructor any question or questions that you have related to the problem or series of problems. This is one of the questions I sent to you last night. It seems with other problems that involve horizontal and vertical motion we have had to figure things out seperately, but this solution acts as if there is no difference. So, I am unsure why this solution ""works."" I am also having difficulty with the algebra involved. When you divide both sides by mg why does .5mv0^2/mg = v0^2/2g? In my mind the m's don't cancel out because one is a half m and the other is a whole m. I am also unclear how you get two g's. self critique assessment: 1
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09:09:29 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 --> This is another one I sent to you last night. I couldn't come up with the right answer at all. This is what I did: KE1 + PE1 = KE2 + PE2 I finally decided that you have to account for the effect of gravity on the ball so I changed the equation to KE1 + PEspring 1 + PEgravity1 = KE2 + PEspring2 +PEgravity2 1/2mv0^2 + 1/2kx1^2 + mgy1 = 1/2mvf^2 + 1/2kx2^2 + mgy2 I decided that up was positive v0 = 0 m = .30kg k = 950N/m g= -9.8m/s^2 (because the gravity is in the opposite direction of movement) x1 = -.150m (b/c the spring is compressed) x2 = 0 (point of spring at equilibrium and point of release of ball) y1 = 0 (there will be no gravity PE at the spring's downward compression, so this has to be zero) y2 = .150m (at the point of release of the ball the effect of gravity has to be figured as .150m above the zero mark) 1/2 (.30kg)(0^2m/s) + 1/2 (950N/m)(-.150^2m) + .30kg(-9.8m/s^2)(0m) = 1/2 (.30kg) vf^2 + 1/2 (950N/m)(0^2m) + .30kg(-9.8m/s)(.150m) 0 + 10.6875 + 0 = .15vf^2 + 0 - 0.441 11.1285 = .15vf^2 74.19 = vf^2 vf =8.6 m/s I realized what I did wrong last night ( I redid this one problem over and over last night for more than an hour!!!). I forgot to square the x1. I used -.150m instead of 0.0225m My answer is .3 off from the book, but it is close enough for me. Max altitude: vf^2 = v0^2 + 2a*ds vf = 0 (when the ball stops in midair and turns around) v0 = 8.6m/s ( the vel when it leaves the spring) a = -9.8m/s^2 0 = 8.6^2 + 2 (-9.8m/s^2) * ds -73.96 = -19.6 * ds -54.36 = ds Not sure why I get a negative distance, it needs to be positive. 54.36 The question asks for from the compression point with is .150m below the point where the ball leaves the spring so add that to 54.36 54.51m Oops, that's not what the book says. confidence assessment: 1
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09:12:36 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 --> 26 m/s is closer to what I was getting last night. The book says the answer is 8.3m/s The book says 3.64m for the distance self critique assessment: 1
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09:12:58 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 --> NA confidence assessment: 3
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