Question

A lightweight spring of 50 N/m is suspended vertically on a ceiling and in an initial equilibrium state. An object with a mass of 0,25 kg is attached to the end of the spring so that the system is in a final equilibrium state. a. Draw the free body diagram of the object in the final equilibrium state. b. What is the length of the spring in the final equilibrium state relative to its length in the initial equilibrium state? Subsequently, the object is pulled down as far as 0,1 m and released so that the system undergoes a simple harmonic oscillation about the final equilibrium point. c. What is the speed of the object as it passes through the final equilibrium point when the object is moving upward? 3.

Answer 1

initial equilibrium state. Io.25kg final equilibrium state. we know, f = kn, Twy spring constant] here, k = 50 N/m f = 0.25x9.8 * FK = 0.25X9.8W el = 0.049 m The length of the spring in the final equilibrium state belative to its initial equilibrium state is 0.049 m. Herre, A=o.im : Total energy a 2 KAN - potabial energy at the end of the spring. to be the velocity of the object. bet

KAN = £mort Inn 1n20.049 m on 2 KAM = 1/2 x 0.256 + 1 x 50 x(0.049) an, 50% (0.gr = 0.2560 + 50% (0.049 on, 0.25o = 50x60.15-50(0.049)~ or 0.25o = 0.3 7995 or on 1.5198 U 1.2328 : The speed of the object is 1.2328 m/s as it passes through the final equilibrium point when the object is moving upward.