a mass of 2.1 kg attached to a verticalspring stretches the spring 6.2 cm from its original equilibriumposition, what is the spring constant?

Answer 1

Weight in Newtons = m g = 2.1 * 9.81 = F
stretch in meters = 0.062 = x
F = k x
2.1 * 9.81 = k * 0.062