course Phy 121 016. `query 16*********************************************
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Given Solution: `a** 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&* OK ********************************************* Question: `qIn the preceding situation why do we expect that the vertical kinetic energy of the ball will be proportional to `dy? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: PE= -mgy. M and g are constant leaving only y. KE= -PE so KE is also related to y. confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** 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 #$&* OK ********************************************* Question: `qWhy do we expect that the KE of the ball will in fact be less than the PE change of the ball? Your solution: Air resistance mainly. confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&* OK ********************************************* Question: `qprin phy and gen phy 6.18 work to stop 1250 kg auto from 105 km/hr Your solution: ½(m)v^2=1/2(1250)(29.1^2)=529,256 J confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 #$&* OK ********************************************* Question: `qprin and gen phy 6.26. spring const 440 N/m; stretch required to store 25 J of PE. Your solution: PE=1/2(k)x^2 25J=1/2(440)x^2 x=.337 confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&* OK "