course Phy 121 s|wsGî}Kwassignment #016 016. `query 16 Physics I 02-26-2008 }ϧoϦn assignment #015 wy Physics I Class Notes 02-26-2008
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00:27:24 If the object was released from rest and and allowed to fall freely through a downward distance `dy equal to the vertical distance traveled on the ramp, its gravitational potential energy will convert to kinetic energy. In this case, setting the potential energy decrease equal to the kinetic energy increase (i.e., `dKE = -`dPE) gives.5 m v^2 = m g `dy. We solve to obtain v = `sqrt(2 g `dy). This demonstrates that v is proportional to`sqrt(`dy).
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RESPONSE --> If the object was released from rest and and allowed to fall freely through a downward distance `dy equal to the vertical distance traveled on the ramp, its gravitational potential energy will convert to kinetic energy. In this case, setting the potential energy decrease equal to the kinetic energy increase (i.e., `dKE = -`dPE) gives.5 m v^2 = m g `dy. We solve to obtain v = `sqrt(2 g `dy). This demonstrates that v is proportional to`sqrt(`dy).
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00:27:35 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 --> If the object was released from rest and and allowed to fall freely through a downward distance `dy equal to the vertical distance traveled on the ramp, its gravitational potential energy will convert to kinetic energy. In this case, setting the potential energy decrease equal to the kinetic energy increase (i.e., `dKE = -`dPE) gives.5 m v^2 = m g `dy. We solve to obtain v = `sqrt(2 g `dy). This demonstrates that v is proportional to`sqrt(`dy). confidence assessment: 3
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00:28:11 ** 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 --> self critique assessment: 3
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00:29:13 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 KE is an object in motion doing work. 'dy is equal to the distance traveled on the ramp. confidence assessment: 1
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00:29:47 ** 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 --> OK 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 self critique assessment: 3
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00:31:01 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 --> Because as the object falls KE is decreased while PE is gained. confidence assessment: 1
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00:31:51 ** 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 --> OK self critique assessment: 2
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00:52:01 prin phy and gen phy 6.18 work to stop 1250 kg auto from 105 km/hr
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RESPONSE --> First we need to convert 105km/hr into m/s: 105km/hr * 1000m/km * 1hr/3600s = 29.17m/s Now we use the equation: Wnet = -Fd = .5(mv2)^2 - .5(mv1)^2 -Fd = 0 - .5(1250kg)(29.17m/s)^2 -Fd = -625kg(850.9m^2/s^2) Fd = -531,812.5 Joules / -1 = 531,812.5 J confidence assessment: 3
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00:52:34 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 --> Got It self critique assessment: 3
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01:12:15 prin and gen phy 6.26. spring const 440 N/m; stretch required to store 25 J of PE.
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RESPONSE --> Using the formula Elastic PE = .5(k)(x)^2 I substitue and get the following: 25 Joules = .5(440N/m)(x)^2 25 J = 220 Newton/m * x^2 25 J / 220 Newton/m = x^2 .114m^2 = x^2 'sqrt.114m^2 = x = .338 Meters confidence assessment: 2
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01:12:39 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 --> Got it self critique assessment: 3
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