Question

A .2kg object attached to a vertical spring causes it to stretch 5.0cm. Another .2kg object is added to the spring. Describe what happens to the following:
Energy of the spring
k of the spring
stretch of the spring

Answer 1

weight = .2 g = F
F = k x
.2 g = k * 0.050 Meters
k = .2 g / 0.050 = about 4 g Newtons/meter
if g = 9.81, k = 39.24 Newtons/meter
potential energy in spring =(1/2) k x^2 = (1/2)(39.24) (0.05)^2 Joules
if you double the force
x, the stretch distance, also doubles
the energy is proportional to x^2 so it is 4 times the original
k of course does not change