Question

The spring has a stiffness k=200n/m and is unstretched when the 25 kg block is at

Answer 1

To solve this we are going to use the formula fro the force applied to a spring: $F=ks$
where
$k$ is the spring constant 
$s$ is the extension 

Since we know the $F=ma$, we can replace that in our formula and solve for $a$ :
$ma=ks$
$a= \frac{ks}{m}$
where
$a$ is the acceleration 
$k$ is the spring constant
$s$ is the extension 
$m$ is the mass

We know for our problem that $k=200$, $s=0.4$, and $m=25$. So lets replace those values in our formula to find $a$:
$a= \frac{ks}{m}$
$a= \frac{(200)(0.4)}{25}$
$a=3.5 \frac{m}{s^2}$

We can conclude that the acceleration of the block when s=0.4m is $3.5 \frac{m}{s^2}$.