course phy 121 You might note an edge of frustration in some my responses, I apologize if I appear rude. Just when I start thinking I am understanding I get confused all over again and that is very frustrating. °¢í¯bô†ôõÙãîŒ÷Øàã{yû’¯µassignment #016 016. `query 16 Physics I 08-30-2008
......!!!!!!!!...................................
10:24:48 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 --> I haven't got a clue. I even went back and reread class notes #15 and I still don't understand. I initially don't see why we seem to just guess from a graph that dx^2 = dy. I see that it does from the new graph. As the ball rolls down the ramp PE changes to KE. PE = m*g*dy and KE = .5mv^2 so m*g*dy = .5mv^2. When you solve for v it equals the sqrt of 2g*dy. Therefore v is proportional to sqrt dy. I assume because 2 and g are constant. Then if v is proportional to sqrt dy then it is also proportional to dx...And honestly I am just spitting back to you the notes from my reading. I don't know what I am saying, or (once again) if any of it is on the right track. confidence assessment: 0
.................................................
......!!!!!!!!...................................
10:31:42 ** 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 --> PE loss is determined by how far down the ball travels. If it vertically travels farther then it is going to lose more PE and gain more KE. A shorter vertical distance will mean less PE loss and less KE gain. So both KE and PE are proportional to the dy (the vertical distance down the ramp). Don't know why v is therefore proportional to sqrt y. self critique assessment: 1
.................................................
......!!!!!!!!...................................
10:35:58 In the preceding situation why do we expect that the kinetic energy of the ball will be proportional to `dy?
......!!!!!!!!...................................
RESPONSE --> That's the part I understand. PE decreases as it travels through distance dy. As PE decreases KE increases. So as the object travels through distance dy KE will increase by the same amount that PE decreases and PE is dependent on the dy. confidence assessment: 2
.................................................
......!!!!!!!!...................................
10:47:55 ** 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 --> No, I don't get any of this. I can't keep it straight in my head. I even have a running list of formulas and notes and I still can't keep it straight! Every equation can be turned around and looked at a different way and I can't ever figure out which one we're starting with. This one starts with vf^2 = v0^2 + 2 a*ds. Change the ds to dy and get rid of the v0^2 because it is zero, rearrange things and we have vf = sqrt (2a*dy) Take KE = .5m*vf^2 substitute sqrt (2a*dy) in for vf and you get KE = .5 m * (sqrt (2a*dy)^2 or KE = .5m * 2a*dy m and a are both constant so... KE is proportional to dy. This reminds me of a silly little nonsense ""saying"" we had when I was younger about why fireengines are red. Fire engines are red and books are too (read). Two and two are four, four times three is twelve, there are twelve inches in a ruler, Queen Elizabeth is a ruler, she is also a ship, ships sail in the sea, fish swim in the sea, fish have fins, the Fins fought the Russians, the Russians are ""red"", fire engines are always rushin' , therefore fireengines are red. It can all be connected, but does it really make sense? self critique assessment: 1
.................................................
......!!!!!!!!...................................
10:49:23 Why do we expect that the KE of the ball will in fact be less than the PE change of the ball?
......!!!!!!!!...................................
RESPONSE --> Because of friction? Friction will slow the ball down, and it will lose energy as heat? confidence assessment:
.................................................
......!!!!!!!!...................................
10:50:53 ** 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 --> I had written friction and the loss of energy to heat, but when I went to click OK or whatever it is to accept my answer I accidentally clicked cancel. self critique assessment: 2
.................................................
......!!!!!!!!...................................
10:58:34 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.
......!!!!!!!!...................................
RESPONSE --> Enter, as appropriate, an answer to the question, a critique of your answer in response to a given answer, your insights regarding the situation at this point, notes to yourself, or just an OK. Always critique your solutions by describing any insights you had or errors you makde, and by explaining how you can make use of the insight or how you now know how to avoid certain errors. Also pose for the instructor any question or questions that you have related to the problem or series of problems. Accidentally clicked next quetion twice. I did the problem prior to looking at the answer. I followed the same steps, but I forgot to change km/h to m/s. I then intuitively knew to change the negative Joules to positive, although the dW = -dKE and other such equations continue to confuse me. self critique assessment: 2
.................................................
......!!!!!!!!...................................
10:59:55 prin and gen phy 6.26. spring const 440 N/m; stretch required to store 25 J of PE.
......!!!!!!!!...................................
RESPONSE --> k = 440N/m PE = 25J ds = ? PE elastic = .5*k*x^2 25J = .5 (440N/m) x^2 x^2 = .114m x = .33m confidence assessment: 2
.................................................
......!!!!!!!!...................................
11:00:57 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 --> I didn't account for the fact that the spring could be compressed -.33m or stretched +.33m, but I do understand that. self critique assessment: 3
.................................................
......!!!!!!!!...................................
11:01:07 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 --> NA confidence assessment: 3
.................................................
......!!!!!!!!...................................
11:01:17 ** 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 --> self critique assessment: 3
.................................................
......!!!!!!!!...................................
11:01:34 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 --> NA confidence assessment: 3
.................................................
......!!!!!!!!...................................
11:01:40 ** 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 --> self critique assessment: 3
.................................................