course Phy 201 ????E??`??????i?assignment #016016. `query 16
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11:37:40 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 --> 'dx=sqrt('dy) When the ball starts to fall, its vo=0. For this reaon, the tiem it takes fro the object to fall when all other variables remains the same will be constant. confidence assessment: 2
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11:38:46 ** 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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RESPONSE --> I left out a point. I understand that the horizontal velocity can be obtained after finding the vertical displacement (represented by y). PE loss is proportional to displacement, and from that we find the KE as the ball leaves the ramp is proportional to displacement. I Understand there is the equation that we can use just justify this. self critique assessment: 2
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11:39:18 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 --> I am honestly not sure... confidence assessment: 0
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11:40:00 ** 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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RESPONSE --> The KE of the ball in the vertical direction is proportional to 'ds as depicted by the equation in which vo=0 is eliminated and vf is left equal to the square root of the distance, 'dy. self critique assessment: 2
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11:40:17 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 --> My guess is that some of the energy is dissipated and is not regained. confidence assessment: 2
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11:40:43 ** 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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RESPONSE --> The PE loss is converted to KE. I Understand this, but after reading this I am unsure of the exact answer to the question. self critique assessment: 2
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11:48:36 prin phy and gen phy 6.18 work to stop 1250 kg auto from 105 km/hr
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RESPONSE --> 'dW='dKE 105km/h=29.1m/s 29.1m/s=Vo KE=.5mv^2 =.5(1250kg)(29.1)^2 =530,000J Since the object of this process is to make the auto come to a rest, we can assume that the final velocity will be 0. KEf=.5mv^2 =.5(1250kg)(0) =0m/s KEf-KEo=0-530,000=-530,000 J 'dw=-'dKE 'dw=-(-530,000J) 'dw=530,000J confidence assessment: 2
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11:48:44 The 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.
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RESPONSE --> I understand. self critique assessment: 2
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11:49:54 prin and gen phy 6.26. spring const 440 N/m; stretch required to store 25 J of PE.
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RESPONSE --> This one looks hard. I really don't even know where to start. confidence assessment: 0
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12:00:51 The 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.
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RESPONSE --> To find the force at any given position you use the formula -kx, and at equilibrium the spring will always have force 0. To find the force between the point and the equilibrium, one should merely average the two, yielding the equation -1/2kx. After finding the force between the position and the equilibrium, the process becomes more familiar. 'dW=F*'ds In this case, the force is the equation that is derived after averagining the equilibrium force and the position force multiplied by the position x. The resulting work equation is 'dW=-1/2kx^2 Generally, springs only exert the conservative elastic force. From this, the PE change in the spring can be calculated by finding the opposite of the Work formula, ultimately yielding the PE change formula as the positive form of the negative work done formula. PE=1/2k^2 really means how much potential energy the spring ""stores."" Now, more specifically in terms of this problem, we can fill in the variables using the equations that I have explained above. I have read over the actual number part and understand it. The actual setup was what I was having difficulty with. self critique assessment: 2
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12:04:02 gen 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?
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RESPONSE --> Fnet=110N m=88g=.088kg 'ds=78 cm=.78m 'dW=110N*.78m=about 86 J .5mv^2=KE We don't know velocity, so I can use this equation to find velocity. V=sqrt(2*KE/m)=sqrt(2*86J/.088kg)=about 44 m/s confidence assessment: 3
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12:04:09 ** 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.. **
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RESPONSE --> I understand. self critique assessment: 3
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12:04:12 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 --> confidence assessment:
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12:04:16 ** 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 --> self critique assessment:
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12:04:26 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 --> confidence assessment: \
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12:04:30 ** 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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RESPONSE --> self critique assessment:
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