Question

a system that consists of a single spring stores a total elastic potential energy $e_p$ when a load is added to the spring. another identical spring connected in parallel is added to the system. the same load is now applied to the parallel springs. what is the total elastic potential energy stored in the changed system? a. $e_p$ b. $\frac{e_p}{2}$ c. $\frac{e_p}{4}$

Explanation:

Step1: Recall elastic - potential - energy formula

The elastic - potential energy of a spring is given by $E_p=\frac{1}{2}kx^2$, where $k$ is the spring constant and $x$ is the displacement. For a single spring with spring constant $k_1 = k$ and displacement $x_1$ under a load $F$, $F = k_1x_1=kx_1$.

Step2: Analyze parallel - spring system

When two identical springs (each with spring constant $k$) are connected in parallel, the equivalent spring constant $k_{eq}=k + k=2k$. Since the same load $F$ is applied, and $F = k_{eq}x_2$, then $F = 2kx_2$. From $F = kx_1$ and $F = 2kx_2$, we can find the relationship between $x_1$ and $x_2$. We get $kx_1=2kx_2$, so $x_2=\frac{x_1}{2}$.

Step3: Calculate elastic - potential energy for parallel - spring system

The elastic - potential energy of the single - spring system is $E_{p1}=\frac{1}{2}kx_1^2$. The elastic - potential energy of the two - spring parallel system is $E_{p2}=\frac{1}{2}k_{eq}x_2^2$. Substitute $k_{eq}=2k$ and $x_2 = \frac{x_1}{2}$ into the formula for $E_{p2}$: $E_{p2}=\frac{1}{2}(2k)(\frac{x_1}{2})^2=\frac{1}{2}(2k)\frac{x_1^2}{4}=\frac{1}{2}kx_1^2\times\frac{1}{2}$. Since $E_{p1}=\frac{1}{2}kx_1^2 = E_p$, then $E_{p2}=\frac{E_p}{2}$.

Answer:

B. $\frac{E_p}{2}$