Question

Dana pulls a spring with a spring constant k=300 N/M, stretching it from its rest length of 0.40 m to 0.50 m. What is the elastic potential energy stored in the spring?

Answer 1

Answer:

$1.5\:\mathrm{J}$

Explanation:

The elastic potential energy of a spring is given by the following:

$PE_s=\frac{1}{2}k\Delta x^2$, where $k$ is the spring constant and $\Delta x$ is displacement.

The displacement of the string is $0.50-0.40=0.10\: \mathrm{m}$.

Plugging in our given values, we get:

$PE_s=\frac{1}{2}\cdot300\cdot0.10^2=\fbox{ 1.5\:\mathrm{J} }$.