ANSWER:
The maximun velocity is 16.07 m/s
At x = 0.26
The velocity is 8.36 m/s
The accelearion is 286.67 m/s^2
The resorting force is 86 N
STEP-BY-STEP EXPLANATION:
Given:
k = 310 N / m
Max distance = 0.5 m
Mass of block = 0.3 kg
Max velocity:
Using conservation of energy:
[tex]\begin{gathered} \frac{1}{2}kx^2=\frac{1}{2}mv^2 \\ v^2=\frac{kx^2}{m} \\ \text{ replacing} \\ v^2=\frac{310\cdot0.5^2}{0.3} \\ v=\sqrt[]{258.33} \\ v=16.07\text{ m/s} \end{gathered}[/tex]
At x = 0.26 m:
[tex]\begin{gathered} v^2=\frac{kx^2}{m} \\ v^2=\frac{310\cdot0.26^2}{0.3} \\ v=\sqrt[]{69.85} \\ v=8.36\text{ m/s} \end{gathered}[/tex]
Acceleration:
[tex]\begin{gathered} F=k\cdot x \\ F=m\cdot a \\ \text{ therefore} \\ m\cdot a=k\cdot x \\ a=\frac{k\cdot x}{m} \\ \text{ replacing} \\ a=\frac{310\cdot0.26}{0.3} \\ a=286.67\text{ }\frac{m}{s^2} \end{gathered}[/tex]
The resorting force:
[tex]\begin{gathered} F=m\cdot a \\ \text{ replacing} \\ F=0.3\cdot286.67 \\ F=86\text{ N} \end{gathered}[/tex]
