Question

if the springlike system of experiment #3 is stretched across the width of the lab bench thru a distance $\delta x$ and then released, what is the impact velocity?
assume m = 0.313 kg, $\delta x$ = 0.5 m, and k = 20 n/m
\bigcirc 4 m/s^2
\bigcirc 4 kg
\bigcirc 4 j
\bigcirc 4 m/s
\bigcirc 4 n/m

Explanation:

Step1: Identify Energy Conservation

The spring - mass system involves conservation of energy. The elastic potential energy of the spring is converted into kinetic energy of the mass. The formula for elastic potential energy is $U=\frac{1}{2}k(\Delta x)^{2}$ and the formula for kinetic energy is $K = \frac{1}{2}mv^{2}$. By conservation of energy, $U = K$, so $\frac{1}{2}k(\Delta x)^{2}=\frac{1}{2}mv^{2}$.

Step2: Simplify the Energy Equation

Cancel out the $\frac{1}{2}$ from both sides of the equation $\frac{1}{2}k(\Delta x)^{2}=\frac{1}{2}mv^{2}$, we get $k(\Delta x)^{2}=mv^{2}$. Then solve for $v$: $v=\sqrt{\frac{k(\Delta x)^{2}}{m}}=\Delta x\sqrt{\frac{k}{m}}$.

Step3: Substitute the Given Values

We are given that $k = 20\ N/m$, $\Delta x=0.5\ m$, and $m = 0.313\ kg$. First, calculate $\frac{k}{m}=\frac{20}{0.313}\approx63.9$. Then $\sqrt{\frac{k}{m}}\approx\sqrt{63.9}\approx7.99$. Then $v=\Delta x\sqrt{\frac{k}{m}}=0.5\times7.99\approx4\ m/s$. The unit of velocity is $m/s$, and among the options, the option with unit $m/s$ and value $4$ is the correct one.

Answer:

4 m/s (the option with "4 m/s" as its content)