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?

23.41 N

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

force tends to be low and long diustance masses are high

the components of a vector are the parts of the vectors that are to each other.

b perpendicular