Answer
a)
work done on the package by friction is
       W_f = -μk R d
          = -μk(mg cos 53°)(2)
          =-(0.4)(8.0)(9.8)(cos53°)(2.0)
          =-37.75 J
b)
work done on the package by gravity is
       W_g = m(g sin 53°)d
           = (8.0)(9.8)(sin53°)(2.0)
           = 125.23 J
c) the work done on the package by the normal force is
      W_n = 0
d)
the net work done on the package is
      W = W_f + W_g + W_n
      W = -37.75 + 125.23 + 0
      W = 87.48 J