Aspring is characterized by a spring constant of 60 n/m. how much potential energy does it store, when stretched by 1.0 cm?

3.0 x10^-3 J

Explanation:

The potential energy of a spring is given by PE = (0.5)k*x^2  

Where

K: Spring Constant = 60 N/m

x: displacement of the spring from its equilibrium position = 1cm = 0.01m

Then PE = 0.5(60)(.01)^2 = 0.003J = 3.0 x10^-3 J

the entire statement is true and correct.  

there's no diagram, nothing is illustrated, and no question is asked.

v=ir = 3x15 =45 v