#$&* course Phy 231 7/30/11 If your solution to stated problem does not match the given solution, you should self-critique per instructions at
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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): OK ------------------------------------------------ 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: PEs = 0.5 * k *x^2 PEs = 0.5 * 950 * 0.150^2 PEs = 10.69 J PEg = mgy 10.69 J = 0.3 kg * 9.8 m/sec^2 * y y = 3.64 m KE = 0.5 * m * v^2 10.69 J = 0.5 * 0.3 kg * v^2 v = 8.44 m/sec 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: KE0 + PE0 = KE + PE 0.5 * m * v0^2 + mgy0 = 0.5 * m * vf^2 + mgy 0.5 * v0^2 + gy0 = 0.5 * vf^2 + gy 0.5 * v0^2 + = 0.5 * 0.7^2 m/sec + 9.8 m/sec^2 * 2.1 m v0^2 = (0.245 m/sec + 20.58 m^2/ sec^2)*2 v0^2 = 41.65 v0 = 6.45 m/sec 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): OK ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: `qquery 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: PEs = 0.5 * k * x^2 PEs = 0.5 * 400 N/m * 0.22^2 m PEs = 9.68 J PEg = mgy 9.68 J = 2 kg * 9.8 m/sec^2 * sin(37) * y y = 0.82 m KE = 0.5 * m * v^2 9.68 J = 0.5 * 2 kg * v^2 v = 3.11 m/sec confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The spring exerts a force of 400 N / m * .220 m = 88 N at the .220 m compression. The average force exerted by the spring between equilibrium and this point is therefore (0 N + 88 N) / 2 = 44 N, so the work done in the compression is `dW = Fave * `ds = 44 N * .220 m = 9.7 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 * 9.7 Joules / (2 kg) ) = 3.2 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 9.7 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) = 9.7 Joules / (.6 * 2 kg * 9.8 m/s^2) = .82 meters, approx. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: `qquery 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: PE = mgy PE = 62 kg * 9.8 m/sec^2 * 65 m PE = 39494 J KE = 39494 - 10500 KE = 28994 J KE = 0.5 * m * v^2 28994 = 0.5 * 62 kg * V^2 v= 30.58 m/sec Then 160 N +(0.2 * 62 kg * 9.8 m/sec^2) = 281.52 N W = F * x W = 281.52 N * 82 m w = 23,084.60 J KE = 28,994 J - 23,084.60 kJ KE = 5,909.40 J KE = 0.5 * m * v^2 5909.40 J = 0.5 * 62 kg * v^2 v = 13.81 m/sec dw = F_Ave * ds 5909.40 J = F_Ave * 2.5 m F_Ave = 2,363.76 N confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** 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.** STUDENT QUESTION If the PE is = 20.6, then why is the initial KE= 20.6 so that we are adding the .245 to the initial KE of 20.6 to get 20.8, I thought that the KE was equal and opposite to the PE why would we not subtract here? INSTRUCTOR RESPONSE `dKE + `dPE = 0, provided there are no nonconservative forces acting on the system. In such a case, PE goes up as the mass rises, so KE goes down. Another way of looking at it: All or part of the KE converts to PE. The mass can only go as high as the initial KE permits. Once the initial KE is 'used up', no increase in PE is possible (recall the assumption that no nonconservative forces act during this phase of motion). At maximum height the mass is still moving in the horizontal direction, so not all of the KE converts to PE. In this case PE increases by 20.6 M m^2/s^2, .245 M m^2/s^2 of KE is still present at the highest point, so about 20.8 M m^2 / s^2 of KE must have been present initially. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: 3 " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!