## Essential University Physics: Volume 1 (3rd Edition)

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: $d_2=\frac{(15.88)^2}{2(1.176)}$ $d_2=110m$