course PHY 121 υTDދʬw{assignment #016
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14:35:36 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 --> Average velocity * the change in time = the change in x.
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14:36:41 ** 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 --> KE = .5m v^2 is proportional to sqrt(KE), then to the sqrt(y)
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14:39:45 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 change in y (work) is final KE - initial KE
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14:40:11 ** 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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14:41:25 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 therma E dissipates
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14:42:11 ** 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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14:56:42 prin phy and gen phy 6.18 work to stop 1250 kg auto from 105 km/hr
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RESPONSE --> Vo = 105km/hr m = 1250kg Vf = 0 vAve = 52.5km/hr dv = 105km/hr a= 9.8m/s^2 105km/hr = 105000m/3600s = 29.17m/s 29.17m/s / 9.8m/s^2 = 97.2s
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15:00:09 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 am way off base. Vo = .5m v^2 KE = .5m v^2 = .5 * 1250kg * (29.1m/s)^2 = 530,000kg m^2/s^2 = 530,000J -530,000kg for change in KE dw = -dke = (-530,000J) = 530,000J How do I know to use these steps. I thought I was on the right path when I worked the problem.
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15:01:16 prin and gen phy 6.26. spring const 440 N/m; stretch required to store 25 J of PE.
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15:04:39 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 --> (0 + (-kx))/2 = -1/2kx = average force exerted dw = F * ds = -1/2 k s*x = -1/2 k x^2 dPE = -dw = -(-1/2kx^2) = 1/2k x^2
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15:04:39 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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15:06:17 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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15:08:45 ** 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 --> 110N * .78m = 86J .088Kg with 86J of KE Why did you convert to.088kg? .5m v^2 = KE for v v = sqrt(2*KE/m) = sqrt (2*86J/.088kg) =44m/s
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ßVx||䥁ՖϢ Student Name: assignment #016
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14:02:29 `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 --> Vo = 0 dt = ? ds = 2m g = 9.8cm/s^2 vf^2 = vo^2 + 2a ds vf^2 = 0 + 2(9.8m/s^2) 2m vf^2 = 19.6m/s^2 * 2m vf = sqrt39.2m/s^2 vf = 6.26m/s 6.26m/s / 2 = 3.13m/s = Vave dt = 2m / 3.13m/s = .64seconds
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14:03:02 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 --> I have the same answer but used different formulas.
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14:07:28 `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 --> dt = .64s vAve = 12m/2 = 6m ds = vAve * dt ds = 6m * .64s = 3.84m/s
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14:07:39 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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14:15:14 `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 --> dsy = 1.5m Voy = 0 Vox = 6m/s 1/2 Vox = 3m/s 3m/s / 1.5m = 2s = dt
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14:17:48 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 --> I am not sure what I did that I obtained 2m. I know I used the formula 1/2 Vox/ ds of 1.5m then I multiplied that answer by the 6m/s.
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14:22:56 `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 --> ds = 4m Vox = 32m Voy = 0 dx =32m I am not sure what to do next?
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14:31:00 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 --> Voy = 0 a = 9.8cm/s^2 ds = 4m ds = vo dt + .5a dt^2 4m = 0 + .5(9.8cm/s^2) dt^2 4m = 4.9cm/s^2 dt^2 4m dt^2 = 4.9cm/s^2
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