Given:

Natural length of the spring: 2 meters

Initial compression of the spring ($ x_i $): 1 meter

Final compression of the spring when the block is at distance $ x $ ($ x_f $): $ (2 - x) $ meters

Solution:

To find the block's speed, we'll use energy conservation. Initially, the spring's potential energy is fully converted to kinetic energy and spring potential energy at the new position $ x $.

Conservation of Energy:

Initial kinetic energy ($ K_i $) is 0 since the block starts from rest.

Initial potential energy ($ U_i $) due to spring compression is $\frac{1}{2} K x_i^2$.

At a distance $ x $, the kinetic energy ($ K_f $) is $\frac{1}{2} mv^2$ and the potential energy ($ U_f $) is $\frac{1}{2} K x_f^2$.

$ K_i + U_i = K_f + U_f $

$ 0 + \frac{1}{2} K x_i^2 = \frac{1}{2} m v^2 + \frac{1}{2} K x_f^2 $

Rearrange to find $ v^2 $:

$ \frac{1}{2} m v^2 = \frac{1}{2} K (x_i^2 - x_f^2) $

Substitute known values:

$ \frac{1}{2} \times 2 \times v^2 = \frac{1}{2} \times 200 \times \left(1^2 - (2-x)^2 \right) $

Simplify and solve for $ v^2 $:

$ v^2 = 100 \left[ 1 - (2-x)^2 \right] $

Finally, take the square root to find $ v $:

$ v = 10 \left[ 1 - (2-x)^2 \right]^{1/2} $

Therefore, the speed of the block at distance $ x $ from the wall is given by the expression above.