Answer
Please see the work below.
Work Step by Step
We know that the acceleration of the hill is given as
$a_1=gsin(\theta)-\mu gsin(\theta)$
We plug in the known values to obtain:
$a_1=(9.8)sin(25)-(0.12)(9.8)sin(25)=3.0758\frac{m}{s^2}$
The speed of the child at the bottom is given as:
$v=\sqrt{2a_1d_1}$
We plug in the known values to obtain:
$v=\sqrt{(2)(3.0758)(41)}=15.88\frac{m}{s}$
The acceleration on the level ground is:
$a_2=\mu g$
We plug in the known values to obtain:
$a_2=(0.12)(9.8)$
$a_2=1.176\frac{m}{s^2}$
The child slides a distance of $d_2$ on the ground level:
$d_2=\frac{v^2}{2a_2}$
We plug in the known values to obtain:
$d_2=\frac{(15.88)^2}{2(1.176)}$
$d_2=110m$
