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A block is fastened to a horizontal spring. The block is pulled to a distance $$x = 10$$ cm from its equilibrium position (at $$x = 0$$) on a frictionless surface from rest. The energy of the block at $$x = 5$$ cm is $$0.25$$ J. The spring constant of the spring is ______ N m$$^{-1}$$.
Correct Answer: 50
$$E_{\text{total}} = \frac{1}{2}kA^2$$
Since mechanical energy is conserved, the energy given at $$x = 5\text{ cm}$$ is simply the total energy of the system
$$0.25 = \frac{1}{2} \cdot k \cdot (0.1)^2$$
$$k = \frac{0.50}{0.01}$$
$$k = 50\text{ N m}^{-1}$$
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