Answer
(a) The coefficient of kinetic friction on the horizontal surface is 0.392.
(b) The work done by friction as the package slides down the circular arc is -0.832 J.
Work Step by Step
$K_2 = K_1+W$
$0 = \frac{1}{2}mv^2 - mg~\mu_k~d$
$mg~\mu_k~d = \frac{1}{2}mv^2$
$\mu_k = \frac{v^2}{2gd}$
$\mu_k = \frac{(4.80~m/s)^2}{(2)(9.80~m/s^2)(3.00~m)}$
$\mu_k = 0.392$
The coefficient of kinetic friction on the horizontal surface is 0.392.
(b) $K_1+U_1+W = K_2+U_2$
$0+mgh + W = \frac{1}{2}mv^2+0$
$W = \frac{1}{2}mv^2 -mgh$
$W = \frac{1}{2}(0.200~kg)(4.80~m/s)^2 -(0.200~kg)(9.80~m/s^2)(1.60~m)$
$W = -0.832~J$
The work done by friction on the circular arc is -0.832 J.
