Assignment__2 - Assignment #2 4/21/2010 PROBLEM 2.2 PROBLEM...

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Assignment #2 4/21/2010 PROBLEM 2.2 PROBLEM STATEMENT: The official weight of a baseball is 5 oz (0.3125 lbf). What is the kinetic energy of a 90 mph fastball in ft-lbf, kJ, and Btu? DIAGRAM DEFINING SYSTEM AND PROCESS: GIVEN: W = 0.3125 lbf, v = 90 mph ASSUME: Standard gravitational acceleration, g=32.17 ft/s 2 FIND: Kinetic energy (in ft - lbf, kJ, and Btu) GOVERNING RELATIONS: Kinetic energy, 2 c m KE = 2g v . Weight c mg W g = QUANTITATIVE SOLUTION: Mass of the ball: 2 c 2 Wg 0.3125lbf 32.17lbm ft / s lbf m 0.3125lbm g 32.17ft / s × = = = g g The kinetic energy is: a. ( 29 ( 29 ( 29 2 2 2 2 2 c c 2 s m 1 miles 5280 ft 1 hr KE lbf = = 0.3125 lbm 90 2g 2g hr miles 3600 s lbm ft = 2722.5 ft 32.174lbm ft / s lbf s = 84.62 ft lbf   ×     × ÷ v g g g g b. ( 29 ( 29 1 1 KE kJ = KE ft lbf = 84.62 737.56 737.56 = 0.115 kJ × × g c. ( 29 ( 29 1 1 KE Btu = KE kJ = 0.115 1.055 1.0551 = 0.109 Btu × × DISCUSSION OF RESULTS: Since it was assumed that the gravitational acceleration was standard (32.17 ft/s 2 ), the mass in lbm has the same numerical value as the weight in lbf. If g were not standard, this would not be the case, since the value of g c (a units conversion constant) is always 32.17 in USCS units, irrespective of the gravitational force. v = 90 mph W = 0.3125 lbf
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4/21/2010 PROBLEM 2.4 PROBLEM STATEMENT: Just before reentry from a 250 mile high orbit, the space shuttle weighs approximately 250,000 lbf and has a velocity of 17,300 mph. At this altitude the gravitational acceleration is 29 ft/s 2 . What is the total kinetic plus potential energy of the shuttle in ft-lbf and in Btu? GIVEN: h = 250 mile, W = 250000 lbf, v = 17300 mph, g = 29 ft/s 2 FIND: Total kinetic and potential energy (ft - lbf and Btu) GOVERNING RELATION: 2 total c c m mgh E = KE+PE = + 2g g v , c mg W g = QUANTITATIVE SOLUTION: The kinetic energy of the space shuttle is computed as follows: ( 29 ( 29 2 2 c c c 2 2 2 2 2 12 Wg m 1 KE = = 2g 2g g 250000 lbf 1 miles 5280 ft 1 hr = 17300 2 hr miles 3600 s 29 ft/s KE = 2.775 10 ft lbf     × ×         × v v g The potential energy is: ( 29 c c c 11 Wg h mgh PE = = g g 5280 ft = 250000 lbf 250 mile mile = 3.3 10 ft lbf × × × g Thus, the total energy of the shuttle is ( 29 ( 29 12 11 total 12 E = KE+PE = 2.775 10 + 3.3 10 = 3.11 10 ft lbf × × × g which is also equivalent to ( 29 12 total 9 1 Btu E = 3.11 10 ft lbf 778.16 ft lbf = 3.99 10 Btu × × × g g DISCUSSION OF RESULTS: This is clearly a situation in which kinetic and potential energy are not negligible! When converted to internal energy during reentry, the air temperature and shuttle
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This note was uploaded on 04/20/2010 for the course M E 320 taught by Professor Deinert during the Spring '08 term at University of Texas at Austin.

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Assignment__2 - Assignment #2 4/21/2010 PROBLEM 2.2 PROBLEM...

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