Respuesta :

egenriether
As AL2006 correctly pointed out the formula is 1/2 kx^2.  I was thinking of force and work is the integral of force over the distance applied.  So now
[tex]W= \frac{1}{2}k x^{2} [/tex]
and
[tex] \frac{2*6.0}{0.04^{2} } = k =7500 \frac{J}{m^{2}}[/tex]

Answer:

Spring constant or the force constant, k = 7500 N/m.

Explanation:

Given that,

Energy stored in the spring, E = 6 J

Stretching in the spring, x = 4 cm = 0.04 m

To find,

The spring constant (force constant) of this spring.

Solution,

We know that the work done by the spring when it is stretched is given by :

[tex]W=\dfrac{1}{2}kx^2[/tex]

k is the spring constant

[tex]k=\dfrac{2W}{x^2}[/tex]

[tex]k=\dfrac{2\times 6}{(0.04)^2}[/tex]

k = 7500 N/m

So, the spring constant (force constant) of this spring is 7500 N/m.