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Below is a problem that I have been trying to figure out for a very long time and am getting no where. I've been fine doing several other Hooke's Law problems, but this one is really getting me. Please help explain the steps to work through this problem and an answer. (I do care to understand how to do it). Thanks in advance.

Here is the problem:

Work of 3 Joules is done in stretching a spring from its natural length to 18 cm beyond its natural length. What is the force (in Newtons) that holds the spring stretched at the same distance (18 cm)?

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Assuming that the spring is ideally following Hooke's law, i.e. $$\mathbf F=k\mathbf x$$ the amount of the work done in displacement of the spring would be: $$W=\int\mathbf F.d\mathbf x=\int k\,xdx=\frac 12 kx^2\\ \implies\frac 12 k\times 0.18^2=3$$ The force that holds the spring stretched at the same distance is $$F=0.18k=\frac{2\times 3}{0.18}=\frac{100}3$$

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    Thank you. You were incredibly helpful. I now feel that I understand the concept. – 2017-01-25
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    @anonym Anytime ;) – 2017-01-25
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Work (in joules) is given by integrating $$F=kx$$ where k is the spring constant and f is the force. Thus, work is $$ W =\int_{x_0}^{x_1} kx$$ You can just rearrange this equation to solve for $k$, and then go back to Hooke's law and obtain the force.