a 1 kg mass is attached to a spring hanging vertically and hangs at rest in the equilibrium position. the spring constant of the spring is 1 n/cm. the mass is pulled downward 2 cm and released. what is the speed of the mass when it is 1 cm above the point from which it was released?

Answer 1

Max spring potential energy
= (1/2) k X^2 = (100/2)*(0.02)^2
= 0.2 Joules

When 0.01 m below the release point, the spring has 1/4 of its max potential energy, so 0.15 J is available for kinetic energy

(1/2)MV^2 = 0.15 J can be solved for the speed V. I get 0.548 m/s

It is permissible to neglect gravity in this problem when you regard kx as the force applied by the spring, including Mg, with x measured from the equilibrium position. g does not affect the answer or the period of vibration.

Answer 2

the answer is 1.73 but i don't know how can every one help me ??