Question

I will give brainlist

A spring of spring constant k=500 N/m is elongated from its initial length by a distance x=2 cm, then the force applied by the spring is:

A. F= 3N.

B. F= 10 N.

C. F=500 N.

D. F= 0N.

Answer 1

Answer:

1.96 kg * m2/s2

Explanation:

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 energy from the equation PE = 12kx2, which equals 1.96 Joules, or kg * m2/s2 if you want SI units.