Question

If the work done to stretch an ideal spring by 4.0 cm is 6.0J, what is the spring constant (force constant) of this spring?

Answer 1

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
$W= \frac{1}{2}k x^{2}$
and
$\frac{2*6.0}{0.04^{2} } = k =7500 \frac{J}{m^{2}}$

Answer 2

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 :

$W=\dfrac{1}{2}kx^2$

k is the spring constant

$k=\dfrac{2W}{x^2}$

$k=\dfrac{2\times 6}{(0.04)^2}$

k = 7500 N/m

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