Question

Consider a vertical spring with spring constant 23.45 N/m hanging from the ceiling. A small object with mass kg is added to the spring and the spring stretches to its equilibrium position. The object is then pulled down and released. The speed of the object a distance 8.815 cm from the equilibrium point is 2.580 times 10^-1 m/s. How far was the object pulled down? (in m) 1.401 times 10^-2 1.863 times 10^-2 2.477 times 10^-2 3.295 times 10^-2 4.382 times 10^-2 5.828 times 10^-2 7.752 times 10^-2 1.031 times 10^-1

Answer 1

using law of conservation of energy

energy at the 8.815 cm = energy at the maximum distance from equilibrium

0.5*k*x^2 + 0.5*m*v^2 = 0.5*k*A^2

(0.5*23.45(8.815*10^-2)^2) + (0.5*1.007*2.58^2*10^-2) = 0.5*23.45*A^2

A = 0.10309 m = 1.03*10^-1 m

so correct option is H) 1.03*10^-1 m