Given:
The spring constant of the spring is k = 220 N/m
The compression is x = 0.175 m
The mass of the object is m = 0.75 kg
To find the speed at which an object will leave the spring.
Explanation:
According to the conservation energy, elastic potential energy is equal to kinetic energy.
The speed can be calculated by the formula
[tex]\begin{gathered} \frac{1}{2}kx^2=\frac{1}{2}mv^2 \\ v=\sqrt{\frac{kx^2}{m}} \end{gathered}[/tex]
On substituting the values, the speed will be
[tex]\begin{gathered} v=\sqrt{\frac{220\times(0.175)^2}{0.75}} \\ =2.997\text{ m/s} \end{gathered}[/tex]
Thus, the speed is 2.997 m/s
