Question

a compressed spring has 16.2 j of elastic potential energy when it is compressed 0.30 m. what is the spring constant of the spring?
90 n/m
108 n/m
180 n/m
360 n/m

Explanation:

Step1: Recall elastic - potential - energy formula

The formula for elastic potential energy is $U = \frac{1}{2}kx^{2}$, where $U$ is the elastic potential energy, $k$ is the spring constant, and $x$ is the displacement from the equilibrium position.

Step2: Rearrange the formula for $k$

We can rewrite the formula $U=\frac{1}{2}kx^{2}$ to solve for $k$. Multiply both sides by 2 to get $2U = kx^{2}$, then $k=\frac{2U}{x^{2}}$.

Step3: Substitute the given values

We are given that $U = 16.2\ J$ and $x = 0.30\ m$. Substitute these values into the formula for $k$: $k=\frac{2\times16.2}{(0.30)^{2}}$.
First, calculate the numerator: $2\times16.2 = 32.4$.
Then, calculate the denominator: $(0.30)^{2}=0.09$.
Finally, divide: $k=\frac{32.4}{0.09}=360\ N/m$.

Answer:

C. 180 N/m