Question

assuming all of the springs energy is transferred to the 3.0 - kilogram block, calculate the speed, $v_{1}$, of the 3.0 - kilogram block immediately after it is propelled by the spring.

Explanation:

Step1: Set up energy - conservation equation

The elastic potential energy of the spring $E_{p}$ is transferred to the kinetic energy of the block $E_{k}$. The formula for elastic potential energy is $E_{p}=\frac{1}{2}kx^{2}$ (not given here, but we know $E_{p}=E_{k}$), and the formula for kinetic energy is $E_{k}=\frac{1}{2}mv^{2}$. Since $E_{p} = E_{k}$, we have $\frac{1}{2}mv^{2}_{1}=\frac{1}{2}kx^{2}$. Canceling out $\frac{1}{2}$ on both sides, we get $mv^{2}_{1}=kx^{2}$. We assume the spring - energy is given as some value equivalent to the kinetic energy of the block. Let's assume the spring - energy transferred $E = E_{k}=\frac{1}{2}mv^{2}_{1}$.

Step2: Solve for $v_{1}$

We know $m = 3.0\ kg$ and $E_{k}=\frac{1}{2}mv^{2}_{1}$. Rearranging the formula $v_{1}=\sqrt{\frac{2E_{k}}{m}}$. If we assume the spring - energy transferred $E_{k}$ is known (let's say $E_{k}=15\ J$ for example purposes, since it's not given in the problem statement. In a complete problem, this value would be provided). Substituting $m = 3.0\ kg$ and $E_{k}=15\ J$ into the formula: $v_{1}=\sqrt{\frac{2\times15}{3}}$. First, calculate the value inside the square - root: $\frac{2\times15}{3}=\frac{30}{3}=10$. Then, $v_{1}=\sqrt{10}\approx 3.16\ m/s$.

Answer:

If we assume the kinetic energy transferred from the spring to the block $E_{k}$ is known, the speed $v_{1}=\sqrt{\frac{2E_{k}}{m}}$. Without the value of the spring - energy transferred, we can only give the formula. If $E_{k}$ is given (for example, if $E_{k}=15\ J$ and $m = 3.0\ kg$), then $v_{1}\approx3.16\ m/s$.