course Phy 231 12/16 7:30am 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: `a** Since the initial vertical velocity is zero the time of fall for a given setup will always be the same. Therefore the horizontal range is proportional to the horizontal velocity of the projectile. The horizontal velocity is attained as a result of vertical displacement `dy, with gravitational PE being converted to KE. PE loss is proportional to `dy, so the KE of the ball as it leaves the ramp will be proportional to `dy. Since KE = .5 m v^2, v is proportional to sqrt( KE ), therefore to sqrt(y). ** This argument can be conceptually summarized in the following series of statements: The loss of gravitational PE is proportional to the vertical distance of fall. Gravitational PE is converted to KE. So KE is proportional to the vertical distance of fall. Therefore 1/2 m v^2 is proportional to the vertical distance of fall. So the squared velocity is proportional to the vertical distance of fall. Therefore the velocity is proportional to the square root of the distance of fall. Saying the same thing but more rigorously: Mass m and acceleration of gravity g are considered constant, gravitational forces are the only forces present. Initial velocity is zero. Change in vertical position is `dy, which is negative. Distance of fall is | `dy | , which is positive. Change in gravitational PE is m g `dy, a negative quantity which is proportional to | `dy |. `dKE = -`dPE So `dKE is a positive quantity, which is proportional to | `dy |. Since KE_0 = 0, `dKE = KE_f = 1/2 m vf^2. Thus vf^2 is proportional to KE, which is proportional to | `dy |. vf is therefore proportional to sqrt( |`dy |). That is, the velocity attained when dropped from rest is proportional to the square root of the distance of fall. STUDENT QUESTION: This part confuses me The horizontal velocity is attained as a result of vertical displacement `dy, with gravitational PE being converted to KE. PE loss is proportional to `dy, so the KE of the ball as it leaves the ramp will be proportional to `dy. INSTRUCTOR COMMENT: PE loss is -m g `dy. Since m and g are constant for this situation, PE loss is therefore proportional to `dy. This means, for example, that if `dy is doubled then PE loss is doubled; if `dy is halved then PE loss is halved. The KE gain is equal to the PE loss, so KE gain is also proportional to `dy. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Your explanation was much more thorough. THe initial solution you gave was very good in understanding the concept better. I'm a little lost in the in-depth version of the solution, but will continue to review for clarity. ------------------------------------------------ Self-critique rating:2 ********************************************* Question: `qIn the preceding situation why do we expect that the kinetic energy of the ball will be proportional to `dy? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Because as the ball sits at the top of the ramp, it has PE that will be coverted to KE and the ball rolls down the ramp. A confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** This question should have specified just the KE in the vertical direction. The kinetic energy of the ball in the vertical direction will be proportional to `dy. The reason: The vertical velocity attained by the ball is vf = `sqrt(v0^2 + 2 a `ds). Since the initial vertical velocity is 0, for distance of fall `dy we have vf = `sqrt( 2 a `dy ), showing that the vertical velocity is proportional to the square root of the distance fallen. Since KE is .5 m v^2, the KE will be proportional to the square of the velocity, hence to the square of the square root of `dy. Thus KE is proportional to `dy. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I like how your solutions have contained analysis using equations of motion to illustrate the answer. I will try to incorporate this concept into my answers in the future. ------------------------------------------------ Self-critique rating: 1 ********************************************* Question: `qWhy do we expect that the KE of the ball will in fact be less than the PE change of the ball? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: THe KE will be less than the original PE of the ball because there are additional forces (frictional, etc) that are non-conservative, which will be lost during the descent of the ball. Therefore, the KE will be less than the original PE of the ball at the top of the ramp. confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** STUDENT RESPONSE: Because actually some of the energy will be dissapated in the rotation of the ball as it drops? INSTRUCTOR COMMENT: Good, but note that rotation doesn't dissipate KE, it merely accounts for some of the KE. Rotational KE is recoverable--for example if you place a spinning ball on an incline the spin can carry the ball a ways up the incline, doing work in the process. The PE loss is converted to KE, some into rotational KE which doesn't contribute to the range of the ball and some of which simply makes the ball spin. ANOTHER STUDENT RESPONSE: And also the loss of energy due to friction and conversion to thermal energy. INSTRUCTOR COMMENT: Good. There would be a slight amount of air friction and this would dissipate energy as you describe here, as would friction with the ramp (which would indeed result in dissipation in the form of thermal energy). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `qprin phy and gen phy 6.18 work to stop 1250 kg auto from 105 km/hr YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: not my class confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe work required to stop the automobile, by the work-energy theorem, is equal and opposite to its change in kinetic energy: `dW = - `dKE. The initial KE of the automobile is .5 m v^2, and before calculating this we convert 105 km/hr to m/s: 105 km/hr = 105 km / hr * 1000 m / km * 1 hr / 3600 s = 29.1 m/s. Our initial KE is therefore KE = .5 m v^2 = .5 * 1250 kg * (29.1 m/s)^2 = 530,000 kg m^2 / s^2 = 530,000 J. The car comes to rest so its final KE is 0. The change in KE is therefore -530,000 J. It follows that the work required to stop the car is `dW = - `dKE = - (-530,000 J) = 530,000 J. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `qprin and gen phy 6.26. spring const 440 N/m; stretch required to store 25 J of PE. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: not my class confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe force exerted by a spring at equilibrium is 0, and the force at position x is - k x, so the average force exerted between equilibrium and position x is (0 + (-kx) ) / 2 = -1/2 k x. The work done by the spring as it is stretched from equilibrium to position x, a displacment of x, is therefore `dW = F * `ds = -1/2 k x * x = -1/2 k x^2. The only force exerted by the spring is the conservative elastic force, so the PE change of the spring is therefore `dPE = -`dW = - (-1/2 kx^2) = 1/2 k x^2. That is, the spring stores PE = 1/2 k x^2. In this situation k = 440 N / m and the desired PE is 25 J. Solving PE = 1/2 k x^2 for x (multiply both sides by 2 and divide both sides by k, then take the square root of both sides) we obtain x = +-sqrt(2 PE / k) = +-sqrt( 2 * 25 J / (440 N/m) ) = +- sqrt( 50 kg m^2 / s^2 / ( (440 kg m/s^2) / m) )= +- sqrt(50 / 440) sqrt(kg m^2 / s^2 * (s^2 / kg) ) = +- .34 sqrt(m^2) = +-.34 m. The spring will store 25 J of energy at either the +.34 m or the -.34 m position. Brief summary of elastic PE, leaving out a few technicalities: 1/2 k x is the average force, x is the displacement so the work is 1/2 k x * x = 1/2 k x^2 F = -k x Work to stretch = ave stretching force * distance of stretch ave force is average of initial and final force (since force is linear) applying these two ideas the work to stretch from equilibrium to position x is 1/2 k x * x, representing ave force * distance the force is conservative, so this is the elastic PE at position x STUDENT QUESTION: What does the kx stand for? INSTRUCTOR RESPONSE: The premise is that when the end of the spring is displaced from its equilibrium position by displacement x, it will exert a force F = - k x back toward the equilibrium point. Since the force is directed back toward the equilibrium point, it tends to 'restore' the end of the spring to its equilibrium position. Thus F in this case is called the 'restoring force'. The force is F = - k x, with F being proportional to x, i.e,. the first power of the displacement. A graph of F vs. x would therefore be a straight line, and the restoring force is therefore said to be linear. A function is linear if its graph is a straight line. So we say that F = - k x represents a linear restoring force. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `qgen phy text problem 6.19 88 g arrow 78 cm ave force 110 N, speed? What did you get for the speed of the arrow? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: not my class confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** 110 N acting through 78 cm = .78 m does work `dW = 110 N * .78 m = 86 Joules appxo.. If all this energy goes into the KE of the arrow then we have a mass of .088 kg with 86 Joules of KE. We can solve .5 m v^2 = KE for v, obtaining | v | = sqrt( 2 * KE / m) = sqrt(2 * 86 Joules / (.088 kg) ) = sqrt( 2000 kg m^2 / s^2 * 1 / kg) = sqrt(2000 m^2 / s^2) = 44 m/s, approx.. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `qquery univ phy 6.84 (6.74 10th edition) bow full draw .75 m, force from 0 to 200 N to 70 N approx., mostly concave down. What will be the speed of the .0250 kg arrow as it leaves the bow? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: vert: a = 9.8 m/s^2 v0 = 0 m/s 'ds = 0.75 m 'ds = 'dt v0 + 0.5 a 'dt^2 0.75 m = 0 + 0.5 (9.8 m/s^2) ('dt^2) 0.75 m / 4.9 m/s^2 = 'dt^2 sqrt[ 'dt^2 ] = sqrt[ 0.153 s^2 ] 'dt = 0.391 s horiz: 'ds = 0.75 m m = 0.0250 kg v0 = 0 m/s Fnet = 70 N vf = ? 'dt = 0.391 s (vf + v0) / 2 = 'ds / 'dt vf + 0 m/s = 2 * ( 0.75 m / 0.391 s ) vf = 3.836 m/s confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The work done to pull the bow back is represented by the area beneath the force vs. displacement curve. The curve could be approximated by a piecewise straight line from about 0 to 200 N then back to 70 N. The area beneath this graph would be about 90 N m or 90 Joules. The curve itself probably encloses a bit more area than the straight line, so let's estimate 100 Joules (that's probably a little high, but it's a nice round number). If all the energy put into the pullback goes into the arrow then we have a .0250 kg mass with kinetic energy 100 Joules. Solving KE = .5 m v^2 for v we get v = sqrt(2 KE / m) = sqrt( 2 * 100 Joules / ( .025 kg) ) = sqrt(8000 m^2 / s^2) = 89 m/s, approx. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I knew I hadn't calculated correctly because I totally neglected the Force given in the equation; this information likely would not have been given if it wasn't needed. the equation you used makes total sense to me - I should have known this, as I recall glancing at it not too long ago and utilized it in an assignment I was reviewing yesterday. KE = 1/2 m v^2 It was interesting though that it would take 200 N to pull the bow back half the way, but only 70 N to pull it back to it's full draw length. I would have thought the farther back you draw, the more Force is needed, but not here. Anyway, a very interesting question.
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Given Solution: `a** A 70 gram = .070 kg hummingbird would require between .070 kg * 10 watts / kg = .7 watts = .7 Joules / second in order to fly. At 10 flaps / second that would be .07 Joules per wingbeat. A similar calculation for the 25 watt level shows that .175 Joules would be required per wingbeat. A 70 kg human being would similarly require 700 watts at 10 watts / kg, which would be feasible for short periods (possibly for several minutes) but not for a sustained flight. At the 25 watt/kg level flight would not be feasible. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I wasn't sure what to call a w/kg - is there a notation for this I should be using? I used Power / mass, but I'm sure there's a better way to do it.