Question

A spring having a spring constant of 180 N/m is stretched 0.51 m from its equilibrium position. How much elastic potential energy does the spring possess?

Answer 1

I'm assuming that you are asking what is the elastic
potential energy stored in the spring at the position
stretched by 16.5 cm.
Since you know the spring constant k, 144 N/m and
the spring stretch from the equilibrium position x, is
16.5 cm, or 0.165, you find the spring's potential
1
energy from the equation PE
-ka?, which equals
2
1.96 Joules, or kg * m?/s? if you want SI units.

Answer 2

23.41 N

Step-by-step explanation:

Formula to find elastic potential energy is, P.E=0.5*k*delta x^2

Where, P.E. = Elastic potential energy

k is the spring constant

delta x is the length of the stretched spring.

So, P.E.=0.5*180*0.51^2 = 23.41 N