Answer
The magnitude of the average frictional force is $~~50~N$
Work Step by Step
In part (a), we found that the length of the slide is $~~L = 10~m$
We can find the average force of friction $F_f$:
$K_1+U_1+W = K_2+U_2$
$0+U_1+W = K_2+0$
$W = K_2-U_1$
$F_f~L = K_2-U_1$
$F_f = \frac{K_2-U_1}{L}$
$F_f = \frac{\frac{1}{2}mv^2-mgh}{L}$
$F_f = \frac{\frac{1}{2}(25~kg)(6.2~m/s)^2-(25~kg)(9.8~m/s^2)(4.0~m)}{10~m}$
$F_f = -50~N$
The magnitude of the average frictional force is $~~50~N$
