#### Answer

When there are two springs connected together, the block's speed is $0.71~v_0$

#### Work Step by Step

When there is one spring, the kinetic energy of the block when it leaves the spring is equal to the initial energy stored in the spring.
We can find the energy stored in the spring when there is just one spring.
$(U_s)_1 = \frac{1}{2}k(\Delta x)^2$
When two springs are compressed a total of $\Delta x$, then each spring is compressed a distance of $\frac{\Delta x}{2}$. We can find the total energy stored in both springs.
$(U_s)_2 = \frac{1}{2}k(\frac{\Delta x}{2})^2+ \frac{1}{2}k(\frac{\Delta x}{2})^2$
$(U_s)_2 = \frac{1}{4}k(\Delta x)^2$
$(U_s)_2 = \frac{1}{2}(U_s)_1$
When there are two springs connected together, the kinetic energy of the block when it leaves the springs is equal to the total energy stored in both springs. The block's kinetic energy in the case of two springs will be half the block's kinetic energy when there was just one spring.
We can find the speed of the block when there are two springs.
$KE_2 = \frac{1}{2}~KE_1$
$\frac{1}{2}mv_2^2 = \frac{1}{2}~(\frac{1}{2}mv_0^2)$
$v_2^2 = \frac{1}{2}~v_0^2$
$v_2 = \sqrt{\frac{1}{2}~v_0^2}$
$v_2 = 0.71~v_0$
When there are two springs connected together, the block's speed is $0.71~v_0$.