Question

14.10. Two springs we neated together and have the che sths shown as.kg ball is dropped from restat a height of 5 m onto the springs, determine the maximum deformation of each spring needed to momentary stop the motion Solution We will establish the datum at the initial position of the ball and assume both springs deform such that A deforms and B deforms (84 -0.05 m). Thus, T: + V = T: +3 60034 - 69.0581 - 19.12 = 0 Choosing the positive root A = 0.245 m Therefore, 88 = 0.245 m-0.05 m = 0.195 m Ans.
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Answer 1

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thtxn (pym) o4mcB00N/m Hexe KA 800N/m KB Y0ON/m Tham; total balanca; forn leght to point of stopping Potential ene m ht a) Eresgy gained bSpaiAn 2 Spei ng Kg (x8. ® 16) Gain speingoma iuating ARt -0 VBXAXO05 245-25+ 19.054= 600 - 20XA +D. 600 X 9. 05A - 2475 0 The cots ase 69#6sx oots an iL.comig!! Solutien But if eqaion wvuld ka aa 600 %-69.0 5 ZA -9 12 20 -0-1900 m o 245m Then Yoot Than positine yoot o215 m oA0245m 245m-D 0Sm-0195m Than S0 Scanned with CamScanner