91Ó°ÊÓ

Given Data:

The mass of the piece of the block is m = 250 kg

The force constant of spring is: k = 840 N/m

The initial compression of spring is: ${\mathrm{x}}_1=0.0100\mathrm{m}$

The final compression of spring is:${\mathrm{x}}_2=0.0300\mathrm{m}$

The coefficient of kinetic friction between block and floor is: ${\mathrm{μ}}_{\mathrm{k}}=0.40$

The distance moved by the block is x = 0.0200 m

Work-Energy theorem:

The change in the kinetic energy of a moving block is equal to the work done by the block to move for the change in kinetic energies of the block.

Apply the work-energy theorem to calculate the speed of the block:

$\frac{1}{2}{\mathrm{mv}}^2=\frac{1}{2}\mathrm{k}\left({\mathrm{x}}_2^2-{\mathrm{x}}_1^2\right)-{\mathrm{μ}}_{\mathrm{k}}\mathrm{mgx}$

Here, m is the mass of the block, g is the gravitational acceleration, v is the speed of the block, k is spring constant, ${\mathrm{x}}_1\mathrm{and}{\mathrm{x}}_2$are initial and final compressions of spring, x is the distance moved by the block.

Substitute all the values in the above equation and we get,

$$\begin{gathered} \frac{1}{2}\left(2.50\mathrm{kg}\right)×{\mathrm{v}}^2=\frac{1}{2}\left(840\mathrm{N}/\mathrm{m}\right)\left[{\left(0.0300\mathrm{m}\right)}^2-{\left(0.0100\mathrm{m}\right)}^2\right] \\ -\left(0.40\right)\left(2.50\mathrm{kg}\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\left(0.200\mathrm{m}\right) \\ \mathrm{v}=0.24\mathrm{m}/\mathrm{s} \end{gathered}$$

Therefore, the speed of the block is 0.24 m/s.