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 Physics Help!
 Planespotting From United States of America, joined Apr 2004, 3547 posts, RR: 5Posted Fri Oct 8 2004 17:00:45 UTC (10 years 5 months 2 weeks 6 days 19 hours ago) and read 1851 times:

 I went to the Southwest Airlines Pigskin Plane Pull on Thursday. They get two teams of 16, one team of U-Texas Fans and one team of Oklahoma U Fans, and see who can pull a 737-700 50 yards the fastest. The OU team won, getting a time of 40 seconds. A Boeing 737-700 weighs roughly 83,000 lbs, which equates to about 37,648 kg. 50 yards is 45.72 meters. I am trying to figure out the basics of the winning pull. it has been awhile since Physics in high school and im running into some problems. Here is what i have so far: T=40 seconds D=45.72 Meters Vo=0 Vave= D/T, Vave = 45.72m/40sec, Vave = 1.143 m/s PE=MA, PE = 37,648.2kg(9.8m/s^2), PE = 368,952.36 kg m/s^2 (or newtons) at rest. I am trying to figure out how to determine Velocity Final, acceleration, and then moving on to figure out how much work was exerted on the 737 over the 50 yards, and the kinetic energy of the 737 at the end of the pull. the formula i have found and tried to use for the acceleration and Vf is: V^2=Vo^2+2ax I come up with an acceleration of .0143m/s^2. But the Vf has to have been faster then the average speed of 1.143m/s. And then the acceleration that i have is wrong. arghh. I just cannot seem to figure out how to solve this problem. Thank you in advance.
 FredT From United Kingdom, joined Feb 2002, 2185 posts, RR: 26 Reply 1, posted Fri Oct 8 2004 17:08:00 UTC (10 years 5 months 2 weeks 6 days 19 hours ago) and read 1809 times:

 v = t * a s = t^2*a / 2 a = 2*s/t^2 v_final = a*t W = s*F F = m*a W = s*m*a This is neglecting the (significant!) friction![Edited 2004-10-08 17:11:49]
 I thought I was doing good trying to avoid those airport hotels... and look at me now.
 Planespotting From United States of America, joined Apr 2004, 3547 posts, RR: 5 Reply 2, posted Fri Oct 8 2004 17:21:10 UTC (10 years 5 months 2 weeks 6 days 19 hours ago) and read 1785 times:

 yes of course. so now im getting a = .05715 m/s^2 and Vf= 2.286 m/s
 Vikkyvik From United States of America, joined Jul 2003, 10750 posts, RR: 26 Reply 3, posted Sat Oct 9 2004 03:29:54 UTC (10 years 5 months 2 weeks 6 days 9 hours ago) and read 1554 times:

 I'm not sure how much you want to simulate what really occurred in the plane pull, but remember that with these equations, you're calculating for a constant acceleration, which I'm sure was not the case at the actual event. Anyway, good luck! ~Vik
 Do all philosophers have an "s" in them?
 Backfire From Germany, joined Oct 2006, 0 posts, RR: 0 Reply 4, posted Mon Oct 11 2004 09:19:24 UTC (10 years 5 months 2 weeks 4 days 3 hours ago) and read 1392 times:

 There's a simple reason why you're having difficulty: You can't solve the problem with the limited information you have - it is just not possible. There is no information on the acceleration, nor any on the final velocity. You only have distance and time - but there's a zillion ways in which to move an object 45m in 40 seconds. For example: You could spend 39 seconds getting it halfway, and then accelerate to cover the other half in just one second, in which case your final velocity will be very high. Or you could maintain a gradual constant acceleration throughout the pull. The most likely scenario, in my view, is that there would be an initial burst of acceleration to get the aircraft moving, but that the velocity would then remain fairly constant throughout the pull - in which case you can calculate a rough idea of average speed. Not very useful, though. In short, you either have to make some assumptions on the acceleration profile or come up with some more figures for variables such as the final velocity. Without these, it is impossible to come up with the answers you're looking for.
 DeltaGuy From , joined Dec 1969, posts, RR: Reply 5, posted Tue Oct 12 2004 07:34:29 UTC (10 years 5 months 2 weeks 3 days 5 hours ago) and read 1296 times:

 That friction can't be ignored for long either...just the force to break those tires away from the ground must have taken alot of energy..that conversion from PE to KE. Great thread though!   DeltaGuy
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