Answer
(a) $-314J$
(b) $0.305$
Work Step by Step
(a) We can find the required work done as
$W=E_f-E_i$
$\implies W=(\frac{1}{2}mv_f^2+mgy_f)-(\frac{1}{2}mv_i^2+mgy_i)$
$\implies W=\frac{1}{2}m(v_f^2-v_i^2)+mg(y_f-y_i)$
We plug in the known values to obtain:
$W=(42)[\frac{1}{2}((4.40)^2-(0)^2)+(9.81)(0-1.75)]$
$W=-314J$
(b) We can find the required coefficient of kinetic friction as
$\mu_k=-\frac{Wsin\theta}{(mgcos\theta)h}$
We plug in the known values to obtain:
$\mu_k=-\frac{(-314)tan35.0^{\circ}}{(42)(9.81)(1.75)}$
$\mu_k=0.305$
