If you compress a horizontal spring with spring constant $k$ from its equilibrium position by a distance $x$, how much work was done by the spring on your hand? $k x^{2}$ $-k x^{2}$ $\frac{1}{2} k x^{2}$ $-\frac{1}{2} k x^{2}$

Step 1 ：The problem is asking for the work done by the spring on your hand when you compress a horizontal spring with spring constant \(k\) from its equilibrium position by a distance \(x\).

Step 2 ：The work done by a spring is given by the formula \(W = \frac{1}{2} k x^{2}\), where \(k\) is the spring constant and \(x\) is the distance the spring is compressed or extended from its equilibrium position.

Step 3 ：This is because the force exerted by a spring is proportional to the distance it is stretched or compressed, and work is the integral of force over distance.

Step 4 ：Substituting the given values into the formula, we get \(W = \frac{1}{2} k x^{2}\).

Step 5 ：\(\boxed{W = \frac{1}{2} k x^{2}}\) is the final answer.