Elastic potential energy: An athlete stretches a spring an extra 40.0 cm beyond its initial length. How much energy has he transferred to the spring, if the spring constant is 52.9 N/cm?

Question

Quizplus · Accepted Answer

The energy stored in a spring is given by the formula $E = \frac{1}{2} k x^2$, where $E$ is the elastic potential energy, $k$ is the spring constant, and $x$ is the displacement from the equilibrium position. Here, $k = 52.9 \, \text{N/cm}$ and $x = 40.0 \, \text{cm}$. First, convert $k$ to N/m for consistency with $x$ in meters: $52.9 \, \text{N/cm} = 5290 \, \text{N/m}$. Then, convert $x$ to meters: $40.0 \, \text{cm} = 0.4 \, \text{m}$. Plugging these values into the formula gives $E = \frac{1}{2} \times 5290 \times (0.4)^2 = 423 \, \text{J}$.

Elastic Potential Energy: an Athlete Stretches a Spring an Extra