course Phy 231
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17:36:37 Class notes #15 When a projectile rolls off a ramp with its velocity in the horizontal direction, why do we expect that its horizontal range `dx will be proportional to the square root of its vertical displacement `dy rolling down the ramp?
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RESPONSE --> Its kinetic energy will equal the work done to lift the object against gravity by the vertical displacement 'dy: .5mv^2 = mg'dy v = 'sqrt(2g'dy) -> 2g is constant, so v and therefore 'dx will be proportional to the square root of 'dy.
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17:36:56 ** 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). **
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17:37:34 In the preceding situation why do we expect that the kinetic energy of the ball will be proportional to `dy?
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RESPONSE --> Because the final kinetic energy should equal the work required to lift the ball against gravity a distance 'dy.
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17:37:44 ** 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. **
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17:38:13 Why do we expect that the KE of the ball will in fact be less than the PE change of the ball?
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RESPONSE --> Some energy is lost to nonconservative forces.
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17:38:28 ** 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). **
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17:38:30 gen phy text problem 6.22 80 g arrow 80cm ave force 95 N, speed? What did you get for the speed of the arrow?
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17:38:32 ** 95 N acting through 80 cm = .8 m does work `dW = 95 N * .8 m = 76 Joules. If all this energy goes into the KE of the arrow then we have a mass of .08 kg with 76 Joules of KE. We can solve .5 m v^2 = KE for v, obtaining | v | = sqrt( 2 * KE / m) = sqrt(2 * 76 Joules / (.08 kg) ) = sqrt( 1900 kg m^2 / s^2 * 1 / kg) = sqrt(1900 m^2 / s^2) = 44 m/s, approx.. **
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17:40:33 query 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?
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RESPONSE --> At full draw Fx = 60 N. We want to find the Work done and therefore the kE gain and thus velocity of the arrow. The work is equal to the area under the Fx vs. l curve, which can be approximated by a line from (0,0) to (40,200) then back to (75,60). The area under this approximation is about 100 J, so kE = .5 * .0250 kg * v^2 v = 'sqrt(100 J/(.5*.0250 kg)) = 89.4 m/s
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17:46:14 ** 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) = 280 m/s, approx. **
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RESPONSE --> This is not the result I got - 'sqrt(8000) is 89.4, I believe.
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17:49:34 Univ. 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?
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RESPONSE --> a) Work per wingbeat: minimum (10 W/kg)(.070 kg) = .70 W .70 W = dW/.1 s dW = .07 J maximum (25 W/kg)(.070 kg) = 1.75 W dW = 1.75 W/.1 s = .175 J b) minimum work for a human: (10 W/kg)(70 kg) = 700 W So an athlete can maintain this power output for a very short period of time, but not sustain it.
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17:49:50 ** 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. **
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