First find K of the spring:
Mg = Kx
here, x = elongation of the spring
= 13.40 - 12.00
= 1.40 cm
= 0.014 m
K = Mg/x
= 3.15*9.8/0.014 = ??
Having got K, find X from:
KX^2/2 = 10.0J
where X is the elongation when stored PE is 10.0 Joules.
The total length of spring then would be:
X + 0.12 m
an ideal spring of negligible mass is 12.00 cm long when noting is attached to it. When you hang a 3.15kg weight from it, you measures its length to be 13.40 cm.
if you wanted to store 10.0 J of potential energy in this spring, what would be its total length?
Assume that it continues to obey hooke's Low.
1 answer
1 answer