Question

F(lb) For protection, the barrel barrier is placed in front of the bridge pier. If the relation between the force and deflection of the barrier is F = (90(10%)x12) lb, where x is in ft, determine the car's maximum penetration in the barrier. The car has a weight of 4000 lb and it is traveling with a speed of 75 ft/s just before it hits the barrier. F = 90(10)*x12 (R)

Answer 1

Solution. For this problem we need a basic idea regouding work - energy theorem According to the theorem,' įm og ² - Impe? U= Work happens during the change in velocity cie change in kinetic energy. Of= final velocitra Vi= initial velocita Also, - S Fodx Now in our case; the barrel barrier is placed in front of bridge pier. so that this will prevent the car. For that reason the car will penetrate into the barrier for some distance then the car will stop re of=o. Now we know if the force is working in the opposite direction of displacement then the work is negative. (the force Fis opposing the car to move forward) - SE.de so U= U- -| go (103)V2 dx Uz - 900w) () U = 90(109) (3x+2)

Given; that Weight = 4000 lb so mass = 4000 32.2 greavita (g) so, 32.2 32.2 since, Wang Imler? byl 4000 ) x (76)² meeg? = x (4000)xo²= henne -30 (103) (2 x ²), or ļx (2000) was? es -90 (10)(1394). - 1 (0)223" )x75= goxto gox 103 ( 1 2 3 4 4 x 4000x75x 3 xbox/03 or, 4000 32.2 2 3127 or, x 2 32.2 213 u= 4000X752 x 3 2x32.2 x 2x90x103 09, 2= 3.2366 ory na 3.24ft since the car stops after displaced by a distence so that this will be the maximum penetreation of the car in the barrier.cie. dmax = 3.24ft (Answer)