course Phy 201 6/30, 8PM 017. `query 17
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Given Solution: `aJane 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique rating: OK ********************************************* Question: `qprin phy and gen phy 6.39: 950 N/m spring compressed .150 m, released with .30 kg ball. Upward speed, max altitude of ball YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: We can easily find the PE stored in the spring using the equation (1/2)kx^2 This gives us PE of 10.7 Joules. We also need to calculate the gravitational pull on the spring which would be 9.8(.150m)(.30) or .44 Joules therefore the total PE is 11.4 Joules We can then use this for the KE of the ball. 11.4=.5(.30kg)(v^2) So velocity would be about 8.75m/s. The max altitude can be found using the original 11.4 Joules/.3 Joules* 9.8m/s/s and finding the height change of 3.9m. Confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aWe 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I was very sure I did this right, but the numbers are off. I think I solved it for .150 and you used 150. Let me know if this is my only error. I think I did ok.
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Given Solution: `aFORMAL 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique rating: OK "