course Phy 231 4/2 6 016. `query 16*********************************************
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Given Solution: `a** A quick synopsis: The object accelerates uniformly downward, so the distance it falls is proportional to the square of the time of fall. Thus the time of fall is proportional to the square root of the distance fallen. The object's horizontal velocity is constant, so its horizontal distance is proportional to the time of fall. So the horizontal distance is proportional to the square root of the distance it falls. More details: The distance of vertical fall, starting with vertical velocity 0, is `dy = v0 `dt + .5 a `dt^2 = 0 `dt + .5 a `dt^2 = .5 a `dt^2, so `dy is proportional to `dt^2. Equivalently, therefore, `dt is proportional to sqrt(`dy). The horizontal distance is `dx = v_horiz * `dt so `dx is proportional to `dt. `dx is proportional to `dt, and `dt is proportional to sqrt(`dy), so `dx is proportional to sqrt(`dy). STUDENT RESPONSE WITH INSTRUCTOR COMMENTS: When the object is accelerating downward the distance it falls is equal to the square root of the time it takes to fall. Right idea but that would be the square of the time, not the square root. So the time of our fall is equal to proportional to, not equal to; again you have the right idea the square root of the distance the object fell. Our horizontal velocity is constant which will make our horizontal distance equal to the time of the fall. Because of this, the horizontal distance in equal to the square root of the distance that our object falls. Good. To summarize: The vertical distance is proportional to the square of the time of fall, so the time of fall is proportional to the square root of the distance fallen. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique rating #$&*3 ********************************************* Question: `qIn the preceding situation why do we expect that the vertical kinetic energy of the ball will be proportional to `dy? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: KE is proportional to PE PE is proportional to dy so, it is expected that KE is proportional to dy confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** This could be argued by analyzing the motion of the object, and using the definition of kinetic energy: 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. ** In terms of energy the argument is simpler: 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.) KE gain is equal to the PE loss, so KE gain is also proportional to `dy. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique rating #$&*3 ********************************************* Question: `qWhy do we expect that the KE of the ball will in fact be less than the PE change of the ball? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: KE of the ball I changed by friction, non-conservative PE of the ball depends on gravity, conservative KE will be less because energy is lost. confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** There is some air friction, which dissipates some of the energy. PE is lost and the lost PE goes into an increase in KE, and into dissipated energy. The KE increase and dissipated energy 'share' the 'lost' PE. STUDENT RESPONSE: Because actually some of the energy will be dissipated in the rotation of the ball as it drops? INSTRUCTOR COMMENT: Good try. However there is KE in the rotation, so rotation accounts for some of the KE but doesn't dissipate 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. Dissipated energy is not recoverable. 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):ok ------------------------------------------------ Self-critique rating #$&*3 ********************************************* Question: `qprin phy and gen phy 6.18 work to stop 1250 kg auto from 105 km/hr YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: No enrolled in either class confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 #$&* ********************************************* Question: `qprin and gen phy 6.26. spring const 440 N/m; stretch required to store 25 J of PE. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Not enrolled in either class confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 #$&* ********************************************* Question: `qgen phy text problem 6.19 88 g arrow 78 cm ave force 110 N, speed? What did you get for the speed of the arrow? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Not enrolled in this class confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 #$&* ********************************************* Question: `qquery univ phy 6.84 (6.74 10th edition) bow full draw .75 m, force from 0 to 200 N to 70 N approx., mostly concave down. What will be the speed of the .0250 kg arrow as it leaves the bow? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: M = 0.025 kg dt = 0.75 m F = 90 N W = F * dt w = 90 N * 0.75 m w = 67.5 J w = k = ½ * m*v^2 67.5 J = ½ * 0.025 kg * v^2 v = 73.5 m/s confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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):ok ------------------------------------------------ Self-critique rating #$&*3 ********************************************* Question: `qUniv. 6.90 (6.78 10th edition) requires 10-25 watts / kg to fly; 70 g hummingbird 10 flaps/sec, work/wingbeat. Human mass 70 kg, 1.4 kW short period, sustain 500 watts. Fly by flapping? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 70 g = 0.07 kg If we use 10 watts/kg to fly, it would be: 0.07 kg * 10 watts/kg = 0.7 watts If its a70 kg human 70 kg * 10 watts/kg = 700 watts which would be too much for a flight overall. confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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):ok ------------------------------------------------ Self-critique rating #$&*3