________________________________________________________________________ PROBLEM (12.88) A car of weight W = 15 kips is stopped by means of spring bumper which it will hit at a speed of v = 25 mph (Fig. P12.87). The largest allowable horizontal force is P = 180 kips. Calculate: (a) The maximum deflection of the spring. (b) The required spring constant k. SOLUTION 25(5280 ×12) = 440 ips 3600 W 15, 000 m= = = 38.82 lb in. s 2 g 32.2(12) v= (a) Energy absorbed by the spring is 1 1 Pδ max = mv 2 2 2 or mv 2 38.82(440) 2 δ max = = = 41.8 in. P 180 ×103 (b) Spring rate is P 180(103 ) = = 4.31 kip in. k= 41.8 δ max Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.