course PHY231 嗣tv٘Îassignment #016
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19:54:00 `q001. Note that this assignment contains 4 questions. . How long does it take for an object dropped from rest to fall 2 meters under the influence of gravity?
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RESPONSE --> The final velocity of an object that falls 2m from rest under the influence of gravity is 6.26 m/sec: vf^2 = v0^2 + 2 * a* 'ds and v0 = 0 vf = sqrt(2*a* 'ds) = sqrt (2 * 9.8 m /sec^2 * 2 m) vf = 6.26 m/sec Since the object started from rest, the average velocity is half the final velocity and is equal to 3.13 m/sec. The time it takes the object to fall can be found by rearranging vave = 'ds * 'dt: 'dt = 'ds / vave 'dt = 2m / 3.13 m/sec = .63 sec confidence assessment: 3
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19:57:33 The object has initial velocity 0, acceleration 9.8 meters/second^2, and displacement 2 meters. We can easily solve this problem using the equations of motion. Using the equation `ds = v0 `dt + .5 a `dt^2 we note first that since initial velocity is zero the term v0 `dt will be zero and can be dropped from the equation, leaving `ds = .5 a `dt^2. This equation is then easily solved for `dt, obtaining `dt = `sqrt(2 *`ds / a ) = `sqrt(2 * 2 meters / (9.8 m/s^2) ) = .64 second.
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RESPONSE --> It's more efficient to use the equation that contains the unknown and just solve for it directly. self critique assessment: 2
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19:58:17 `q002. While an object dropped from rest falls 2 meters under the influence of gravity, another object moves along a level surface at 12 meters/second. How far does the second object move during the time required for the first object to fall?
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RESPONSE --> An object traveling at 12 m/sec for .64 sec moves 7.7 m: 'ds = vave 'dt confidence assessment: 3
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19:58:23 As we have seen in the preceding problem, the first object requires .64 second to fall. The second object will during this time move a distance of 12 meters/second * .64 second = 8 meters, approximately.
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RESPONSE --> self critique assessment: 3
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20:00:52 `q003. An object rolls off the edge of a tabletop and falls to the floor. At the instant it leaves the edge of the table is moving at 6 meters/second, and the distance from the tabletop to the floor is 1.5 meters. Since if we neglect air resistance there is zero net force in the horizontal direction, the horizontal velocity of the object will remain unchanged. Since the gravitational force acts in the vertical direction, motion in the vertical direction will undergo the acceleration of gravity. Since at the instant the object leaves the tabletop its motion is entirely in the horizontal direction, the vertical motion will also be characterized by an initial velocity of zero. How far will the object therefore travel in the horizontal direction before it strikes the floor?
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RESPONSE --> Using the same strategy as earlier, the object falls for .553 sec. During that time, it has a horizontal displacement of 6 m /sec * .553 sec = 3.3 m confidence assessment: 3
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20:01:01 We analyze the vertical motion first. The vertical motion is characterized by initial velocity zero, acceleration 9.8 meters/second^2 and displacement 1.5 meters. Since the initial vertical velocity is zero the equation `ds = v0 `dt + .5 a `dt^2 becomes `ds = .5 a `dt^2, which is easily solved for `dt to obtain `dt = `sqrt( 2 `ds / a) = `sqrt( 2 * 1.5 m / (9.8 m/s^2) ) = .54 sec, approx., so the object falls for about .54 seconds. The horizontal motion will therefore last .54 seconds. Since the initial 6 meter/second velocity is in the horizontal direction, and since the horizontal velocity is unchanging, the object will travel `ds = 6 m/s * .54 s = 3.2 m, approximately.
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RESPONSE --> self critique assessment: 3
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20:08:55 `q004. An object whose initial velocity is in the horizontal direction descends through a distance of 4 meters before it strikes the ground. It travels 32 meters in the horizontal direction during this time. What was its initial horizontal velocity? What are its final horizontal and vertical velocities?
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RESPONSE --> The time it takes to the object to fall 4 meters is sqrt(s 'ds /a) = .9 sec The object travels 32 meters in .9 sec with a constant velocity of 35.4 m/sec. The initial vertical velocity is zero, and the final vertical velocity is sqrt ( 2 a 'ds) = 17.64 m/sec confidence assessment: 2
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20:13:55 We analyze the vertical motion first. The vertical motion is characterized by initial velocity zero, acceleration 9.8 meters/second^2 and displacement 4 meters. Since the initial vertical velocity is zero the equation `ds = v0 `dt + .5 a `dt^2 becomes `ds = .5 a `dt^2, which is easily solved for `dt to obtain `dt = `sqrt( 2 `ds / a) = `sqrt( 2 * 4 m / (9.8 m/s^2) ) = .9 sec, approx., so the object falls for about .9 seconds. The horizontal displacement during this .9 second fall is 32 meters, so the average horizontal velocity is 32 meters/(.9 second) = 35 meters/second, approximately. The final vertical velocity is easily calculated. The vertical velocity changes at a rate of 9.8 meters/second^2 for approximately .9 seconds, so the change in vertical velocity is `dv = 9.8 m/s^2 * .9 sec = 8.8 m/s. Since the initial vertical velocity was zero, final vertical velocity must be 8.8 meters/second in the downward direction. The final horizontal velocity is 35 meters/second, since the horizontal velocity remains unchanging until impact.
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RESPONSE --> I understand how to get the final velocity simply using the acceleration and 'dt. My less efficient calculation also included an error in the value of 'ds self critique assessment: 2
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ŏ֥ݧ assignment #016 016. `query 16 Physics I 06-25-2007
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20:53:10 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 --> The potential energy of the object the moment before it drops is equal to the kinetic energy it gains during the fall: m * g * 'dy = m/2 * v^2 , and both sides can be divided by mass for g * 'dy = 1/2 * v^2 The equation can be rearranged to: v^2 = 2 * g * 'dy. This shows v^2 is proportional to 'dy. The graph of 'dx^2 vs. 'dy is linear, suggesting a proportional relationship between 'dx^2 and 'dy. If dx^2 and v^2 are both proportional to 'dy, they are proportional to each other. confidence assessment: 2
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21:32:06 ** 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 --> Horizontal range ('dx) depends on horizontal velocity. Horizontal velocity can vary for the same setup if the initial position of the ball changes. This is the result of different initial potential energy of the ball, which is due to the 'dy. The work-energy theorem showed the proportional relationship between velocity and the square root of 'dy. self critique assessment: 2
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21:36:18 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 --> The kinetic energy of the ball is equal to the potential energy lost during its fall. This is equal to the product of 'dy and the net force acting on the ball. confidence assessment: 3
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21:39:48 ** 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 --> self critique assessment: 3
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21:42:06 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 --> The KE of the ball will be less than the PE change of the ball because the ball does work against friction. Some of the KE will be in the form of rotational motion, so translational velocity is not the total velocity. confidence assessment: 3
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21:42:32 ** 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 --> self critique assessment: 3
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21:42:38 prin phy and gen phy 6.18 work to stop 1250 kg auto from 105 km/hr
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RESPONSE --> confidence assessment:
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21:42:42 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 --> self critique assessment:
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21:42:45 prin and gen phy 6.26. spring const 440 N/m; stretch required to store 25 J of PE.
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RESPONSE --> confidence assessment:
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21:42:48 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 --> self critique assessment:
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21:42:51 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 --> confidence assessment:
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21:42:55 ** 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 --> self critique assessment:
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21:54:48 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 --> The speed of the arrow as it leaves the bow is 83.5 m/sec: v = sqrt(2*KE / m) Neglecting thermal energy loss and friction, the kinetic energy of the arrow will be equal to the amount of work done by pulling the arrow back against the bowstring. This is equal to the area under the force vs displacement curve. The area under the curve can be approximated by a triangle from zero to 200 N and by a trapezoid from 200N to 70 N. The triangle and trapezoid meet at x = .4m. The area under the curve is approximately 87.25 J. Substituting this value for KE and .025 kg for the mass gives a velocity of 83.5 m/sec confidence assessment: 3
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22:01:54 ** 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 --> 200 / .0025 = 80000 sqrt (80000) = 282.8 self critique assessment: 3
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22:15:55 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 --> The average power per kg of bird's mass is .5 *(25 + 10)Watts = 17.5 J / sec Each second, the bird must exert 17.5 J /kg. For a .07 kg bird, this corresponds to 1.225 Joules. (17.5 J/kg * .07 kg) Each second, this bird produces 10 wingbeats. This correlates to .1225 Joules per wingbeat. A 70-kg human has a steady power output of 500 Watts and a short burst power output of 1.4 kW. At 17.5 W/kg, the power output required for flight is 1.2 kW. This is much more than the steady output of an athlete. Using the lower range of power requirement (10 Watts/kg), a 70-kg would require 700 Watts of power. This is close to the steady output of an athlete and might be possible. confidence assessment: 3
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22:16:13 ** 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: 3
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