Answer
$\mu_k = 0.51$
Work Step by Step
Let $x$ be the distance the child slides. Therefore:
$v^2 = 2ax$
$v = \sqrt{2ax}$
If the speed at the bottom (with friction) is half the speed of the frictionless case, then the acceleration in the frictionless case must be greater by a factor of 4.
We can use a force equation to find acceleration $a_1$ in the friction-less case:
$ma_1 = mg~sin(\theta)$
$a_1 = g~sin(\theta)$
We can use a force equation to find acceleration $a_2$ in the case with friction:
$ma_2 = mg~sin(\theta) - mg~cos(\theta)\cdot \mu_k$
$a_2 = g~sin(\theta) - g~cos(\theta)\cdot \mu_k$
We know that the the acceleration $a_1$ in the friction-less case must be greater by a factor of 4. Therefore:
$a_1 = 4a_2$
$gsin(\theta) = 4g~sin(\theta) - 4g~cos(\theta)\cdot \mu_k$
$4~cos(\theta)\cdot \mu_k = 3~sin(\theta)$
$\mu_k = \frac{3~sin(\theta)}{4~cos(\theta)} = \frac{3~sin(34^{\circ})}{4~cos(34^{\circ})}$
$\mu_k = 0.51$
