# A spring with a constant of 12 (kg)/s^2 is lying on the ground with one end attached to a wall. An object with a mass of 8 kg and speed of 12 m/s collides with and compresses the spring until it stops moving. How much will the spring compress?

So,we can say, $\frac{1}{2} K {x}^{2} = \frac{1}{2} m {v}^{2}$ (where, $K$ is the spring constant, $m$ is the mass of the object, $v$ is the velocity of the object and $x$ is the amount of compression)
${x}^{2} = 96$
or, $x = 9.8 m$