course PHY 121 1/4 016. `query 16*********************************************
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Given Solution: ** 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). ** This argument can be conceptually summarized in the following series of statements: • The loss of gravitational PE is proportional to the vertical distance of fall. • Gravitational PE is converted to KE. • So KE is proportional to the vertical distance of fall. • Therefore 1/2 m v^2 is proportional to the vertical distance of fall. • So the squared velocity is proportional to the vertical distance of fall. • Therefore the velocity is proportional to the square root of the distance of fall. Saying the same thing but more rigorously: • Mass m and acceleration of gravity g are considered constant, gravitational forces are the only forces present. • Initial velocity is zero. • Change in vertical position is `dy, which is negative. Distance of fall is | `dy | , which is positive. • Change in gravitational PE is m g `dy, a negative quantity which is proportional to | `dy |. • `dKE = -`dPE • So `dKE is a positive quantity, which is proportional to | `dy |. • Since KE_0 = 0, `dKE = KE_f = 1/2 m vf^2. • Thus vf^2 is proportional to KE, which is proportional to | `dy |. • vf is therefore proportional to sqrt( |`dy |). • That is, the velocity attained when dropped from rest is proportional to the square root of the distance of fall. STUDENT QUESTION: This part confuses me 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. INSTRUCTOR COMMENT: PE loss is -m g `dy. Since m and g are constant for this situation, PE loss is therefore proportional to `dy. This means, for example, that if `dy is doubled then PE loss is doubled; if `dy is halved then PE loss is halved. The KE gain is equal to the PE loss, so KE gain is also proportional to `dy. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I think I have the general concept but should probably review a few more times to make sure. ------------------------------------------------ Self-critique rating: ---- ********************************************* Question: `qIn the preceding situation why do we expect that the kinetic energy of the ball will be proportional to `dy? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The `dy or vertical displacement is relative to the KE because: KE = .5(m)(v^2). This shows that the KE is related to the velocity of the system. Another equation, vf^2 = vi^2+ 2*a*`ds, with `ds or distance traveled representative also of `dy or vertical displacement, shows that the vertical displacement is proportional to the velocity. So if the KE is proportional to the velocity, and velocity proportional to the displacement, then KE is inadvertently proportional to displacement. confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** 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. * &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: ---- ********************************************* Question: `qWhy do we expect that the KE of the ball will in fact be less than the PE change of the ball? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: In a perfect world the KE and PE change would be equal, but there are other factors like friction that influence the KE of the ball as it travels. Confidence rating: 3
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Given Solution: ** 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). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: ---- ********************************************* Question: `qprin phy and gen phy 6.18 work to stop 1250 kg auto from 105 km/hr YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: m = 1250kg. v = 105km/hr. 105km/hr* (1000m/km)*(1hr/60m)*(1m/60s) = 29.2m/s. KE = .5(m)(v^2) = .5(1250kg)(29.2m/s) = 530,000J. Since this is how much KE moves the auto, we need just as much to stop the auto. Confidence rating: 2
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Given Solution: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I could have explained it a little better. ------------------------------------------------ Self-critique rating: ---- ********************************************* Question: `qprin and gen phy 6.26. spring const 440 N/m; stretch required to store 25 J of PE. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: To determine the position at which the spring could store a PE of 25J, we have to use the equation PE = .5k(x^2). 25J = .5k(x^2), with k being the 440N/m and x being the position we need. 25J = .5(440N/m) x^2 25J = 220 x^2 25J/220 = x^2, taking square root of both sides = x = 0.34 confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. • Brief summary of elastic PE, leaving out a few technicalities: • 1/2 k x is the average force, x is the displacement so the work is 1/2 k x * x = 1/2 k x^2 • F = -k x • Work to stretch = ave stretching force * distance of stretch • ave force is average of initial and final force (since force is linear) • applying these two ideas the work to stretch from equilibrium to position x is 1/2 k x * x, representing ave force * distance the force is conservative, so this is the elastic PE at position x STUDENT QUESTION: What does the kx stand for? INSTRUCTOR RESPONSE: The premise is that when the end of the spring is displaced from its equilibrium position by displacement x, it will exert a force F = - k x back toward the equilibrium point. Since the force is directed back toward the equilibrium point, it tends to 'restore' the end of the spring to its equilibrium position. Thus F in this case is called the 'restoring force'. The force is F = - k x, with F being proportional to x, i.e,. the first power of the displacement. A graph of F vs. x would therefore be a straight line, and the restoring force is therefore said to be linear. A function is linear if its graph is a straight line. So we say that F = - k x represents a linear restoring force. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I forgot to add the +/- to the position, but that makes sense since the square root of something could be either negative or positive, because neg^2 = positive. I have been disregarding that in all the other problems where I took the square root of both sides. ------------------------------------------------ Self-critique rating: "