The spring in $FIGUREEX10.23a$is compressed by $∆x$. It launches the block across a frictionless surface with speed$v_0$. The two springs in $FIGUREEX10.23b$are identical to the spring of $FigureEX10.23a$. They are compressed the same total $∆x$and used to launch the same block. What is the block’s speed now?

Speed refers to how rapidly something is going or being done, or something that is moving swiftly.

The equation gives the amount of energy stored in the spring,

$E=\frac{1}{2}K{x}^2$

$E$= Energy,$K$= spring stiffness,$x$= stretch in the spring.

The block's kinetic energy is given by,

$E=\frac{1}{2}m{v}^2$

$E$= energy,$m$= block's mass,$v$= block's velocity.

(a) Consider the figure as spring block system,

In the spring, elastic energy stored = kinetic energy. Hence, block velocity equation can be,

$$\begin{gathered} \frac{1}{2}K{\left(∆x\right)}^2=\frac{1}{2}m{\left(v_0\right)}^2 \\ {v}_0=∆x\times \sqrt{\frac{K}{m}}......\left(1\right) \end{gathered}$$

(b) Consider the figure as spring block system,

Two springs are connected in series in the illustration above. As a result, the equivalent spring stiffness of these two springs is determined as follows:

$$\begin{gathered} \frac{1}{K_{eq}}=\frac{1}{K}+\frac{1}{K} \\ {K}_{eq}=\frac{K}{2} \end{gathered}$$

In the spring, elastic energy stored = kinetic energy. Hence, block velocity equation can be,

$$\begin{gathered} \frac{1}{2}\times \frac{K}{2}{\left(∆x\right)}^2=\frac{1}{2}m{\left(v_b\right)}^2 \\ {v}_b=∆x\times \sqrt{\frac{K}{2m}}......\left(2\right) \\ =\frac{1}{\sqrt{2}}\times \left(∆x\times \sqrt{\frac{K}{m}}\right) \end{gathered}$$

Using the value from the $Eq.\left(1\right)$

$$\begin{gathered} v_b=\frac{1}{\sqrt{2}}\times v_0 \\ =0.707v_0 \end{gathered}$$

Hence, block speed is$0.707v_0$.