The spring has a stiffness k=200n/m and is unstretched when the 25 kg block is at a. Determine the acceleration of the block when s=0.4m. The contact surface between the block and the plane has

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

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

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

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