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