course Phy 231 6/29 10:30 016. `query 16*********************************************
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Given Solution: A quick synopsis: • The object accelerates uniformly downward, so the distance it falls is proportional to the square of the time of fall. Thus the time of fall is proportional to the square root of the distance fallen. • The object's horizontal velocity is constant, so its horizontal distance is proportional to the time of fall. • So the horizontal distance is proportional to the square root of the distance it falls. More details: The distance of vertical fall, starting with vertical velocity 0, is • `dy = v0 `dt + .5 a `dt^2 = 0 `dt + .5 a `dt^2 = .5 a `dt^2, so `dy is proportional to `dt^2. • Equivalently, therefore, `dt is proportional to sqrt(`dy). The horizontal distance is • `dx = v_horiz * `dt so `dx is proportional to `dt. • `dx is proportional to `dt, and `dt is proportional to sqrt(`dy), so `dx is proportional to sqrt(`dy). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique rating #$&* OK ********************************************* Question: `qIn the preceding situation why do we expect that the vertical kinetic energy of the ball will be proportional to `dy? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Dy is the change in vertical position. The change in kinetic energy is proportional to the velocity of the object. The velocity of the object is proportional to the change in position because the velocity increases at a steady rate from the acceleration of gravity. Therefore the change in position is proportional to the change in kinetic energy. Also KE is proportional to PE because PE decreases as dy increases. Confidence rating #$&*: 2
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Given Solution: This could be argued by analyzing the motion of the object, and using the definition of kinetic energy: 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. ** In terms of energy the argument is simpler: • 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.) • KE gain is equal to the PE loss, so KE gain is also proportional to `dy. Self-critique (if necessary): OK Self-critique rating #$&* OK ********************************************* Question: `qWhy do we expect that the KE of the ball will in fact be less than the PE change of the ball? Your solution: The conversion of energy from potential to kinetic is not completely exact. Some is lost as thermal energy or in resistance from the air, so the kinetic energy of the object will be less than the change in PE. Confidence rating #$&*: 3
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Given Solution: `a** There is some air friction, which dissipates some of the energy. PE is lost and the lost PE goes into an increase in KE, and into dissipated energy. The KE increase and dissipated energy 'share' the 'lost' PE. Air friction doesn't lower the PE; it's the change in altitude that lowers the PE (an object's gravitational potential energy gets lower as it descends). The loss of PE results in an increase in KE, though not as great an increase as if air friction wasn't present. Related to the idea that `dKE and `dPE should be equal and opposite (which is sometimes the case and sometimes not): The energy situation is governed by the work-energy theorem in the form • `dW_noncons_ON = `dKE + `dPE. In general, a nonconservative force can increase the KE and/or the PE in any way at all. It can increase one without changing the other. `dKE and `dPE are not generally equal and opposite. For example you can speed up a cart by pushing or pulling it along a level surface. The force you exert is nonconservative, and it increases the KE of the cart without changing its PE. Or you could lift an object at a constant speed; its KE wouldn't change but its gravitational PE would. However in some situations nonconservative forces are either absent or negligible. For example if you toss a steel ball a couple of meters into the air and let it fall to the ground, it doesn't attain enough speed for air resistance to become significant, so once you release it the ball pretty much behaves as if nonconservative forces were absent. Then as it rises it slows down, decreasing its KE and increasing its PE. As it then falls its PE decreases but it speeds up, increasing its KE. Changes in KE and PE turn out, in this case, to be equal and opposite. • Formally, since `dW_noncons_ON = `dKE + `dPE, it follows that if there are no nonconservative forces `dKE + `dPE = 0 and `dKE and `dPE are equal and opposite. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique rating #$&* OK ********************************************* Question: `qprin phy and gen phy 6.18 work to stop 1250 kg auto from 105 km/hr Your solution: 105km/hr(1000m/1km)(1hr/3600s) = 29.2 m/s -KE=work=(1/2)(1250kg)(29.2m/s)^2 KE = -532900 Joules Work – 532900 J Confidence rating #$&*: 3
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Given Solution: `aThe 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): OK Self-critique rating #$&* OK ********************************************* Question: `qprin and gen phy 6.26. spring const 440 N/m; stretch required to store 25 J of PE. Your solution: dPE = -dKE dPE = -(-(1/2)(k)(x)^2) 25J = (1/2)(440N/m)(x^2) .114 m^2 = x^2 X = +- .337 meters Confidence rating #$&*: 3
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Given Solution: `aThe 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 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. The force exerted by a spring is F = - k x, where x is how far it is stretched (positive x) or compressed (negative x): If a spring is stretched beyond its equilibrium length it 'pulls back' with a tension that increases the more it is stretched. If x is how far it is stretched past its equilibrium length, then a positive stretch x results in a tension force in the negative direction. For an ideal spring, a graph of force vs. stretch is a straight line through the origin. The equation of this line is F = - k x. For any ideal spring k is a constant number which, when multiplied by the stretch, then by -1 (because the direction of the tension is opposite that of the stretch), gives us the tension force. A negative value of x corresponds to compressing the spring rather than stretching it. If a spring is compressed it 'pushes back', exerting a force in the positive direction. For an ideal spring the equation F = - k x continues to apply, since a negative value of x will appropriately result in a positive value of F. • The constant number k is called the spring constant. At position x the potential energy of the spring is 1/2 k x^2: When stretched/compressed x units from the equilibrium length, for the reasons indicated in the given solution, the average force required is 1/2 k x and the displacement from equilibrium is x so the work done is 1/2 k x * x = 1/2 k x^2. The force exerted by an ideal spring is conservative, so between equilibrium and position x the PE of the spring increases by amount 1/2 k x^2. Taking the PE of the spring to be 0 when x = 0, it follows that at position x its PE is 1/2 k x^2: • PE_spring = 1/2 k x^2. In the current example the spring constant is k = 440 Newtons / meter, and we want to find the stretch needed to give us PE of 25 Joules. We therefore solve the equation PE = 1/2 k x^2 for x, obtaining x = +- sqrt( 2 PE / k), and substitute our given values of k and PE to get the result. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique rating #$&* OK ********************************************* Question: `qgen phy text problem 6.19 88 g arrow 78 cm ave force 110 N, speed? What did you get for the speed of the arrow? Your solution: .088kg arrow, .78m, 110N Work = .78m * 110N = 85.8J 85.8J = (1/2)(.088kg)(v^2) V^2 = 1950 m^2/s^2 V = +-44.2 m/s Speed = 44.2m/s confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** 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.. ** Self-critique (if necessary): OK Self-critique rating #$&* OK ********************************************* Question: `qquery univ phy 6.84 (6.74 10th edition) bow full draw .75 m, force from 0 to 200 N to 70 N approx., mostly concave down. What will be the speed of the .0250 kg arrow as it leaves the bow? Your solution: Work = 95J = ~area under the force graph 95J = (1/2)(.025kg)(v^2) V^2 = 7600 m^2/s^2 V = +-87.1m/s Confidence rating #$&*: 3
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Given Solution: `a** 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) = 89 m/s, approx. ** Self-critique (if necessary): OK Self-critique rating #$&* OK ********************************************* Question: `qUniv. 6.90 (6.78 10th edition) requires 10-25 watts / kg to fly; 70 g hummingbird 10 flaps/sec, work/wingbeat. Human mass 70 kg, 1.4 kW short period, sustain 500 watts. Fly by flapping? Your solution: .07kg *10 watts/kg = .7 watts = .7 Joules / second .7J/s / 10beats/s = .07 J/ beat .07kg*25watts/kg = 1.75 J/s 1.75 J/s / 10beats/s = .175 J/beat Human: 70 kg * 10watts/kg = 700 J/s 70kg*25watts/kg = 1750 J/s Flight might be possible for the lower limit, but not sustained. For the upper limit, no flight would be possible. confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** 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. ** Self-critique (if necessary): OK Self-critique rating #$&* OK "