#$&* course Phy 241 October 15 around 5pm 016. `query 16*********************************************
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Given Solution: `a** A quick synopsis: • The object accelerates uniformly downward, so the distance it falls is proportional to the square of the time of fall. Thus the time of fall is proportional to the square root of the distance fallen. • The object's horizontal velocity is constant, so its horizontal distance is proportional to the time of fall. • So the horizontal distance is proportional to the square root of the distance it falls. More details: The distance of vertical fall, starting with vertical velocity 0, is • `dy = v0 `dt + .5 a `dt^2 = 0 `dt + .5 a `dt^2 = .5 a `dt^2, so `dy is proportional to `dt^2. • Equivalently, therefore, `dt is proportional to sqrt(`dy). The horizontal distance is • `dx = v_horiz * `dt so `dx is proportional to `dt. • `dx is proportional to `dt, and `dt is proportional to sqrt(`dy), so `dx is proportional to sqrt(`dy). STUDENT RESPONSE WITH INSTRUCTOR COMMENTS: When the object is accelerating downward the distance it falls is equal to the square root of the time it takes to fall. Right idea but that would be the square of the time, not the square root. So the time of our fall is equal to proportional to, not equal to; again you have the right idea the square root of the distance the object fell. Our horizontal velocity is constant which will make our horizontal distance equal to the time of the fall. Because of this, the horizontal distance in equal to the square root of the distance that our object falls. Good. To summarize: The vertical distance is proportional to the square of the time of fall, so the time of fall is proportional to the square root of the distance fallen. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ self-critique rating #$&*: ********************************************* 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: If we use the concept of KE then we can say: Vf=sqrt (v0^2+2a’ds) If v0=0 then: Vf=sqrt (2a’ds) Vertical velocity then equals sqrt (‘dy) KE=1/2mv^2 KE=v^2 then KE=’dy (if I did everything correctly) confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** This could be argued by analyzing the motion of the object, and using the definition of kinetic energy: The vertical velocity attained by the ball is vf = `sqrt(v0^2 + 2 a `ds). Since the initial vertical velocity is 0, for distance of fall `dy we have vf = `sqrt( 2 a `dy ), showing that the vertical velocity is proportional to the square root of the distance fallen. Since KE is .5 m v^2, the KE will be proportional to the square of the velocity, hence to the square of the square root of `dy. Thus KE is proportional to `dy. ** In terms of energy the argument is simpler: • PE loss is -m g `dy. • Since m and g are constant for this situation, PE loss is therefore proportional to `dy. (This means, for example, that if `dy is doubled then PE loss is doubled; if `dy is halved then PE loss is halved.) • KE gain is equal to the PE loss, so KE gain is also proportional to `dy. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ self-critique rating #$&*: ********************************************* 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: Well it’s obvious that air friction will get rid of some of the energy. This will then lose PE which will make the KE increase. The PE loss converts to KE. 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). ** STUDENT QUESTION Ok, so air friction, dissipating energy, lowers the PE and Increases the KE. So does this mean that they do not neccessarily have to be equal and opposite of each other? INSTRUCTOR RESPONSE Air friction doesn't lower the PE; it's the change in altitude that lowers the PE (an object's gravitational potential energy gets lower as it descends). The loss of PE results in an increase in KE, though not as great an increase as if air friction wasn't present. Related to the idea that `dKE and `dPE should be equal and opposite (which is sometimes the case and sometimes not): The energy situation is governed by the work-energy theorem in the form • `dW_noncons_ON = `dKE + `dPE. In general, a nonconservative force can increase the KE and/or the PE in any way at all. It can increase one without changing the other. `dKE and `dPE are not generally equal and opposite. For example you can speed up a cart by pushing or pulling it along a level surface. The force you exert is nonconservative, and it increases the KE of the cart without changing its PE. Or you could lift an object at a constant speed; its KE wouldn't change but its gravitational PE would. However in some situations nonconservative forces are either absent or negligible. For example if you toss a steel ball a couple of meters into the air and let it fall to the ground, it doesn't attain enough speed for air resistance to become significant, so once you release it the ball pretty much behaves as if nonconservative forces were absent. Then as it rises it slows down, decreasing its KE and increasing its PE. As it then falls its PE decreases but it speeds up, increasing its KE. Changes in KE and PE turn out, in this case, to be equal and opposite. • Formally, since `dW_noncons_ON = `dKE + `dPE, it follows that if there are no nonconservative forces `dKE + `dPE = 0 and `dKE and `dPE are equal and opposite. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): There’s more to it then what I put, but after reading the solution, it made more sense and understandable. ------------------------------------------------ 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: If graphed, the area beneath the curve would be about 90J. Since KE=1/2mv^2, we can re-arrange the equation to solve for v. V=sqrt (2KE/m) V=sqrt (2(90)/0.0250) V=84.85m/s confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The work done to pull the bow back is represented by the area beneath the force vs. displacement curve. The curve could be approximated by a piecewise straight line from about 0 to 200 N then back to 70 N. The area beneath this graph would be about 90 N m or 90 Joules. The curve itself probably encloses a bit more area than the straight line, so let's estimate 100 Joules (that's probably a little high, but it's a nice round number). If all the energy put into the pullback goes into the arrow then we have a .0250 kg mass with kinetic energy 100 Joules. Solving KE = .5 m v^2 for v we get v = sqrt(2 KE / m) = sqrt( 2 * 100 Joules / ( .025 kg) ) = sqrt(8000 m^2 / s^2) = 89 m/s, approx. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Rounding made a slight difference in solutions. ------------------------------------------------ self-critique rating #$&*: ********************************************* 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: Convert 70g to kg. (0.07kg) 0.07kg*(10watt/kg)=0.7 J/sec to FLY At 10 flaps per second, 0.07J wingbeat 0.07kg*(25watt/kg)=0.175J wingbeat Not sure how to get the human’s calculations… 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"