contestada

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?

Respuesta :

Answer:

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

cashg57
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.

Otras preguntas