#$&* course PHY 201 17. query 17 submitted 6 Mar 11 around 9:40 PM. 17. query 17
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Given Solution: Outline of solution: Jane has KE. She goes higher by increasing her gravitational PE. Her KE is 1/2 m v_0^2, where m is her mass and v0 is her velocity (in this case, 6.3 m/s^2). If she can manage to convert all her KE to gravitational PE, her KE will decrease to 0 (a decrease of 1/2 m v0^2) and her gravitational PE will therefore increase by amount 1/2 m v_0^2. The increase in her gravitational PE is m g `dy, where m is again her mass and `dy is the increase in her altitude. Thus we have PE increase = KE loss In symbols this is written m g `dy = 1/2 m v0^2. The symbol m stands for Jane's mass, and we can also divide both sides by m to get g `dy = 1/2 v0^2. Since we know g = 9.8 m/s^2 and v0 = 6.3 m/s, we can easily find `dy. `dy = v0^2 / (2 g) which is easily evaluated to obtain `dy = 1.43 m. MORE DETAILED SOLUTION: 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 - 1/2 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. STUDENT QUESTION: I’m confused as to where the 2 g came from INSTRUCTOR RESPONSE: You are referring to the 2 g in the last line. We have in the second-to-last line - 1/2 M v0^2 = - ( M g `dy). Dividing both sides by - M g, and reversing the right- and left-hand sides, we obtain `dy = - 1/2 M v0^2 / (M g) = 1/2 v0^2 / g = v0^2 / (2 g). STUDENT QUESTION do we get dy'=v0^2/2g will this always be the case? INSTRUCTOR RESPONSE Most basic idea: On the simplest level, this is a conversion of PE to KE. This is the first thing you should understand. The initial KE will change to PE, so the change in PE is equal to the initial KE. In this case the change in PE is m g `dy. For other situations and other conservative forces the expression for `dPE will be very different. The simplest equation for this problem is therefore init KE = increase in PE so that 1/2 m v0^2 = m g `dy More general way of thinking about this problem: More generally we want to think in terms of KE change and PE change. We avoid confusion by not worrying about whether each change is a loss or a gain. Whenever conservative forces are absent, or being regarded as negligible, we can set the expression for KE change, plus the expression for PE change, equal to zero. • In the present example, KE change is (final KE - initial KE) = (0 - 1/2 m v^2) = -1/2 m v^2, while PE change is m g `dy. • We get the equation -1/2 m v0^2 + m g `d y = 0. • This equation is easily rearranged to get our original equation 1/2 m v0^2 = m g `dy. The very last step in setting up the problem should be to write out the expressions for KE and PE changes. • The expression for PE change, for example, depends completely on the nature of the conservative force. For gravitational PE near the surface of the Earth, that expression is m g `dy. For gravitational PE where distance from the surface changes significantly the expression would be G M m / r1 - G M m / r2. For a spring it would be 1/2 k x2^2 = 1/2 k x1^2. • The expression for KE change is 1/2 m vf^2 - 1/2 m v0^2; this is always the expression as long as mass doesn't change. In this particular case the equation will read • 1/2 m vf^2 - 1/2 m v0^2 + m g `dy = 0 If we let vf = 0, the previous equations follow. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): 2, I had to refer to the given solution and more detailed solution for a better understanding. I still not that clear. ------------------------------------------------ Self-critique rating: 3 ********************************************* 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: Given: x = .150 m k = 950 N/m m = .30 kg The PE stored = ½ k x^2 PE = ½ (950 N/m) (.150 m)^2 PE = 10.7 J and the change in elastic PE is -10.7 J. The change in gravitational PE is m * g *dy = .30 kg * 9.8 m/s^2 * .150 m = +.44 J. So the total change in PE is therefore -10.7 J + 4.4 J = -10.3 J. dy = dPE / (m * g) dy = 10.7 J / (.30 kg * 9.8 m/s^2) dy = 10.7 J / (2.9 N) dy = 10.7 N * m / (2.9 N) dy = 3.7 m confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a We being with a few preliminary observations: • 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. • We also observe that no frictional or other nonconservative forces are mentioned, so we assume that nonconservative forces do no work on the system. • It follows that `dPE + `dKE = 0, so the change in KE is equal and opposite to the change in PE. The PE stored in the spring will be .5 k x^2 = .5 ( 950 N/m ) ( .150 m)^2 = 10.7 J. 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 has a change in gravitational PE as well as elastic PE. • The change in elastic PE is -10.7 J, and the change in gravitational PE is m g `dy = .30 kg * 9.8 m/s^2 * .150 m = +.44 J. • The total change in PE is therefore -10.7 J + 4.4 J = -10.3 J. Summarizing what we know so far: • Between release and the equilibrium position of the spring, `dPE = -10.3 J During this interval, the KE change of the ball must therefore be `dKE = - `dPE = - (-10.3 J) = +10.3 J. Intuitively, the ball gains in the form of KE the 10.3 J of PE lost by the system. The initial KE of the ball is 0, so its final KE during its interval of contact with the spring is 10.3 J. We therefore have • .5 m v^2 = KEf so that • vf=sqrt(2 KEf / m) = sqrt(2 * 10.3 J / .30 kg) = 8.4 m/s. To find the max altitude to which the ball rises, we consider the interval between release of the spring and maximum height. • At the beginning of this interval the ball is at rest so it has zero KE, and the spring has 10.7 J of elastic PE. • At the end of this interval, when the ball reaches its maximum height, the ball is again at rest so it again has zero KE. The spring also has zero PE, so all the PE change is due to the gravitational force encountered while the ball rises. • Thus on this interval we have `dPE + `dKE = 0, with `dKE = 0. This means that `dPE = 0. There is no change in PE. Since the spring loses its 10.7 J of elastic PE, the gravitational PE must increase by 10.7 J. • The change in gravitational PE is equal and opposite to the work done on the ball by gravity as the ball rises. The force of gravity on the ball is m g, and this force acts in the direction opposite the ball's motion. Gravity therefore does negative work on the ball, and its gravitational PE increases. If `dy is the ball's upward vertical displacement, then the PE change in m g `dy. • Setting m g `dy = `dPE we get `dy = `dPE / (m g) = 10.7 J / ( .30 kg * 9.8 m/s^2) = 10.7 J / (2.9 N) = 10.7 N * m / (2.9 N) = 3.7 meters. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: `qgen 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Given: dy = 2.1 m M = mass of the jumper I thought this question was really hard. I will have to read this question and solution several times before I can get a better understanding. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. STUDENT QUESTION I used the equation 'dy=v0^2 / (2g). Isn't that easier? INSTRUCTOR RESPONSE Good, but that equation only applies under certain conditions. Your solution didn't account for the final KE, which doesn't make a lot of difference but does make enough to decide the winner of a competitive match. In general you don't want to carry an equation like 'dy=v0^2/(2g) around with you. If you carry that one around, there are about a hundred others that apply to different situations, and you'll overload very quickly. Among other things, that equation doesn't account for both initial and final KE. It applies only when the PE change is gravitational, only near the surface of the Earth, and only when the final KE is zero. Way too many special conditions to keep in mind, way too much to remember. You want to start your reasoning from `dKE + `dPE + `dW_noncons_ON = 0. We assume that nonconservative forces are negligible, so that `dW_noncons_ON is itself zero, giving us `dPE + `dKE = 0. For this situation `dPE = m g `dy, `dKE = KE_f - KE_0 = 1/2 m vf^2 - 1/2 m v0^2, and the equation becomes m g `dy + 1/2 m v0^2 - 1/2 m vf^2 = 0. In a nutshell, there are only three things you need in order to analyze similar situations: • `dKE + `dPE + `dW_noncons = 0 • KE = 1/2 m v^2 • `dPE = m g `dy (in the vicinity of the Earth's surface) &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): 1 ------------------------------------------------ Self-critique rating: 1 " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: