Question

Two equal length springs having a stiffness of : K A = 300 N/m KB = 200 N/m Are nested together in order to form a shock absorber. If a two kilogram block is dropped from an at rest position 0.6 meters above the springs, what is the deformation of the springs when the block momentarily stops at the bottom of its path of travel ?

Answer 1

The net stiffness of the spring is, Kmet = KA+K; = 300 N/m + 200 N/m =500 N/m From law of conservation of energy we have, mgh Kimya (2.0 kg)(9.8m/s*)(0.6 m + x) = }(500 N/m) x? 19.6(0.6+ x) = 250.rº 250.x2 - 19.6% -11.76 = 0 As this is quadratic equation we have two values for x, by solving we get, - - b ± √b² - 4ac 2a -(-19.6)= /(-19.6) – 4(250)(-11.76) 2 (250) = +0.2596m, -0.1812 m As x cannot be negative, thus the deformation of the spring when the block momentarily stops is x=+0.2596 m.