Question

LU Help System Announcements PLINTER VERSION HACK NEXT JRCES Chapter D2, Problem D2/030 Incorrect The cart impacts the safety barrier with speed vo- 3.85 m/s and is brought to a stop by the nest of nonlinear springs which provide a deceleration 8 --kıx - Kyx, where x is the amount of spring deflection from the undeformed position and k, and ky are positive constants. If the maximum spring deflection is 480 mm and the velocity at half-maximum deflection is 3.53 m/s, determine the values for the constants ky and am Undeformed position Study СУУРУУ WWW АЛУУСУ Answers: ki - 134.85 key = 1255.93 m22 Click if you would like to Show Work for this questioni Open Show Work LINK TO TEXT 255 PM here to search

Answer 1

The impact with barries expression de 3.85 givers is a = -K, X - K2 33 substitute var = a thens v dv - -K, X -K2 x - var =(-6, sak ₂ x dne 0 integrate above expression within the limites. The cart has a velocity of 3.85 m/s when the springs are in undeforured state that means x=0 and ha&a velocity of v at spring 'de fossngation of se. lvar- (1-1, x = K22²) de - 1-K, ex*: (29) Cut .95) -- [K, Ca, P) +R, CES - ;] C - 3.84*) -- [K, (T) +K2C At maximum spring deformation, the Velocity of cast will be uso. (-3.) = -(K, CI + K2(e)] -7.41125 =-{k, (2) + K₂ ( 7 The maximum spring deflection is X = 480mm = 0.480m - 3.85

-7.4025 =-{k,( 0.480") +Kz (0.4804,7 7.41125 = 0.1152 K, + 0.01327104 K2 ke = 7.41125 -0.015 0.01327104 substitute v= 3.53 m/s and half maximum deflection a= 480mm = 240mm = 0.240m in ( - 3.85") -- [K, (%) + Kz (ry] (3.53* 3.85*) =-{k, (0.240") +K2 (0.2017 -1-1808 - 10.0288 K, #8-2944x16%)K2] 1. 1808 = 0.0288K, +(8.29448104)KZ 1.19 08 = 0.0298 K + (8.2944X104) (7.4125–0.89 ki 1 0.01327104 ] 1.1808 = 0.0288 K, +0.4632 -Go2x103) ki ki = 33. 222 5-2

from ③ 7041125 - 0.1152 (33.222) K2 = 2 0.01327104 1 K₂ = 270.0674 m2 52 |