91Ó°ÊÓ

The given data can be listed below,

• The mass of the block is, m = kg
• The velocity of the block is, ${\mathrm{v}}_0=6.00\mathrm{m}/\mathrm{s}$
• The force constant of the spring is, k = 500 N/m

The spring constant, which has a value of N/m and is the ratio of a spring's force to its additional length, is a measurement of the elasticity of the spring force.

According to the work-energy theorem, the maximum compressed distance is given by,

$$\begin{gathered} W\mathit{=}{K}_{\mathit{i}\mathit{n}\mathit{i}\mathit{t}\mathit{i}\mathit{a}\mathit{l}}\mathit{-}{K}_{\mathit{f}\mathit{i}\mathit{n}\mathit{a}\mathit{l}} \\ \frac{1}{2}k{\left(∆l\right)}^{\mathit{2}}\mathit{=}\frac{\mathit{1}}{\mathit{2}}m{v}_{\mathit{0}}^{\mathit{2}} \\ \mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{∆}l\mathit{=}\sqrt{\frac{\mathit{m}}{\mathit{k}}}{v}_{\mathit{0}} \end{gathered}$$

Here, m is the mass of the block, k is the force constant of the spring, and is the velocity of the block.

Substitute all the values in the above,

$$\begin{gathered} ∆\mathrm{l}=\sqrt{\frac{5\mathrm{kg}}{500\mathrm{N}/\mathrm{m}}}\left(6.0\mathrm{m}/\mathrm{s}\right) \\ =0.6\mathrm{m} \end{gathered}$$

Thus, the maximum distance compressed by the spring is 0.6 m.

The velocity of the block after a distance compressed by spring is given by,

$v_0=\sqrt{\frac{k}{m}}∆l$

Here, m is the mass of the block, k is the force constant of the spring, and is the distance compressed of spring whose value is 0.150 m.

Substitute all the values in the above,

$v_0=\sqrt{\frac{500N/m}{5kg}}\left(0.150m\right)$

Thus, the velocity of the block after spring compressed to 0.150 m is 1.5 m/s .