course Phy 231 g}ၳQ}assignment #016
......!!!!!!!!...................................
14:21:41 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?
......!!!!!!!!...................................
RESPONSE --> Gravitational force has a magnitude which is equal to m*g or weight (w) of the object. Because the object is being raised through a vertical distance, the product of w * vertical displace ('dy) = work ('dW). Therefore: 'dW = m * g * 'dy 'dW corresponds to the gravitational potential energy of the object. When the object is released down the ramp, the potential energy (PE) will be converted into kenetic energy (KE). Setting the potential energy decrease equal to the KE increase gives: 1/2 m * v^2 = m * g * 'dy Solving for v we get: v = sqrt[ 2g * 'dy ], therefore we can conclude that v is proportional to sqrt[ 'dy ] and because 'dx is proportionally related to v of the object, we can conclude that 'dx is proportional to sqrt[ 'dy ]. confidence assessment: 2
.................................................
......!!!!!!!!...................................
14:22:08 ** 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). **
......!!!!!!!!...................................
RESPONSE --> OK! self critique assessment: 3
.................................................
......!!!!!!!!...................................
14:36:39 In the preceding situation why do we expect that the kinetic energy of the ball will be proportional to `dy?
......!!!!!!!!...................................
RESPONSE --> The kenetic energy of the ball is proportional to 'dy (height that ball is raised) because as the mass is raised a greater distance ('dy), it's PE will increase. As the object rolls down the ramp, that PE will be converted to KE. Therefore, if the 'dy increases, the KE will also increase, and if the 'dy decreases, the KE will also decrease. confidence assessment: 2
.................................................
......!!!!!!!!...................................
14:38:14 ** 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. **
......!!!!!!!!...................................
RESPONSE --> OK So you tied it into the equation KE = 1/2 m v^2. Since we know the KE will be proportional to v^2, we therefore should realize that it will also be proportional to the sqrt of 'dy self critique assessment: 2
.................................................
......!!!!!!!!...................................
14:41:29 Why do we expect that the KE of the ball will in fact be less than the PE change of the ball?
......!!!!!!!!...................................
RESPONSE --> the KE will be less than the PE change of the ball because of frictional forces and a slight loss of KE due to rotational forces (which I must admit I do not yet fully understand). confidence assessment: 3
.................................................
......!!!!!!!!...................................
14:42:55 ** 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). **
......!!!!!!!!...................................
RESPONSE --> OK So even if some of the rotational KE is ""lost"", that KE is recoverable. self critique assessment: 2
.................................................
......!!!!!!!!...................................
14:43:14 prin phy and gen phy 6.18 work to stop 1250 kg auto from 105 km/hr
......!!!!!!!!...................................
RESPONSE --> skip - not my class confidence assessment: 3
.................................................
......!!!!!!!!...................................
14:43:21 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.
......!!!!!!!!...................................
RESPONSE --> skip - not my class self critique assessment: 3
.................................................
......!!!!!!!!...................................
14:43:27 prin and gen phy 6.26. spring const 440 N/m; stretch required to store 25 J of PE.
......!!!!!!!!...................................
RESPONSE --> skip - not my class confidence assessment: 3
.................................................
......!!!!!!!!...................................
14:43:33 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.
......!!!!!!!!...................................
RESPONSE --> skip - not my class self critique assessment: 3
.................................................
......!!!!!!!!...................................
14:43:41 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?
......!!!!!!!!...................................
RESPONSE --> skip - not my class confidence assessment: 3
.................................................
......!!!!!!!!...................................
14:43:46 ** 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.. **
......!!!!!!!!...................................
RESPONSE --> skip - not my class self critique assessment: 3
.................................................
......!!!!!!!!...................................
14:55:00 query 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?
......!!!!!!!!...................................
RESPONSE --> m = 0.0250 kg 'ds = 0.75 m v0 = 0 cm/s what is vf when 'ds = 0 vf = sqrt[ v0^2 + 2 a 'ds ], where a = F/m vf = sqrt[ v0^2 + 2 (F / m) 'ds ] vf = sqrt[ 2 * (70 N / 0.0250 kg) * 0.75 m ] vf = sqrt[ 5600 m/s^2 * 0.75 m ] = [4200 m^2/s^2] vf = 64.81 m/s confidence assessment: 2
.................................................
......!!!!!!!!...................................
15:01:33 ** 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) = 280 m/s, approx. **
......!!!!!!!!...................................
RESPONSE --> Interesting. So I needed to consider the work done from 0 to 200 N and then from 200 N to 70 N. The area under the graph of force vs displacement would have been about 90 Joules (or 100 Joules as you discussed, based on inconsistency between straight line graph and the curved line). Using the KE = 1/2 m v^2 formula, we can solve the equation by solving for v: v = sqrt( 2 KE / m ) v = sqrt( 2 100 Joules / 0.0250 kg ) v = sqrt( 8000 m^2/s^2) however, when I calculate the square root of 8000, I get 89.44 m/s, as opposed to your 280 m/s.
.................................................
......!!!!!!!!...................................
18:33:30 Univ. 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?
......!!!!!!!!...................................
RESPONSE --> m = 70 g = 0.07 kg g = 9.8 m/s^2 w = 9.8 m/s^2 * 0.07 kg = 0.686 m/s^2 kg (N) 'dKE = Fnet * 'ds = 'dWnet assuming bird flies 1 m high, it would be doing: 0.686 N * 1 m = 0.686 Joules If 10 flaps/sec, 0.0686 Joules/sec or 0.0686 watts (W) m = 70 kg Pmax = 1.4 kW (for a few seconds) Pave = 500 W (steady) To fly, we need to be producing min. 10 W/kg. 500 W / 70 kg = 7.14 W/kg, which is less than the required amount needed for flight. However, at 1.4 kW, we get 1400 W / 70 kg = 20 W/kg, so it appears that for a few short seconds, we might be able to attain flight - but we would not be able to sustain it. confidence assessment: 2
.................................................
......!!!!!!!!...................................
18:34:06 ** 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. **
......!!!!!!!!...................................
RESPONSE --> OK self critique assessment: 3
.................................................