Answer
$3.47\frac{m}{s}$
Work Step by Step
As we know that change in $K.E$ is equal to net work done i.e $W_{net}$.
Mathematically
$K_f-K_i=W_{net}$
$W_{net}$ is the sum of work done by gravitational force that is $W_{mg}$ and work done by spring that is $W_{spring}$
Hence $W_{net}=W_{mg}+W_{spring}$
Put the values from part 'a' and 'b'
$W_{net}=0.294-1.8=-1.506J$
$K_f-K_i=W_{net}$
$K_f=0$ as finally the spring comes to rest and thus its final velocity is zero, while $K.E=\frac{1}{2}mv^2$. So the above equation becomes
$0-\frac{1}{2}mv^2=W_{net}$
$v^2=\frac{-2W_{net}}{m}$
Taking square roots on both sides
$v=\sqrt {\frac{-2W_{net}}{m}}$
Put the values
$v=\sqrt {\frac{-2\times(-1.506)}{0.25}}$
$v=3.47\frac{m}{s}$
