Question

the compressed spring of a dart gun has potential energy of 50 j. if the spring constant is 200 n/m, what is the displacement of the spring? 0.5 m 0.2 m 0.7 m 0.4 m

Explanation:

Step1: Recall the elastic - potential - energy formula

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

Step2: Rearrange the formula to solve for $x$

Starting from $U=\frac{1}{2}kx^{2}$, we can solve for $x$. First, multiply both sides of the equation by 2 to get $2U = kx^{2}$. Then, divide both sides by $k$: $x^{2}=\frac{2U}{k}$. Taking the square - root of both sides gives $x=\sqrt{\frac{2U}{k}}$.

Step3: Substitute the given values

We are given that $U = 50\ J$ and $k = 200\ N/m$. Substituting these values into the formula $x=\sqrt{\frac{2U}{k}}$, we have $x=\sqrt{\frac{2\times50}{200}}=\sqrt{\frac{100}{200}}=\sqrt{0.5}= 0.5\ m$.

Answer:

A. 0.5 m