Hi there,

Answer is C -87 N

as it is given weight=300 N

u=0.3

Let p is force and force of friction is f

resolving forces horizontally

F=p cos 25

=p×0.9063

Vertically it is R= W-p sin 25 degree

R= 300-P × 0.4226

Force of friction

0.0963×p=uR= 0.3×(300-0.4226P)

=90-0.1268p

=0.9063p+0.1268p= 90

(1.0331)p=90

P=87.1 N