Given that:
W = weight of the block = mg = 70 N
mass of the block is given as
m = W/g
m = 70/9.8 = 7.14 kg
a)
N = normal force on the block
Perpendicular to incline , force equation is given as
N = m g cos30
inserting the values
N = (70) Cos30
N = 60.6 N
b)
f = frictional force acting on the block to keep it from sliding
Parallel to incline force equation is given as
f = m g Sin30
f = (70) Sin30
f = 35 N
c)
F = pushing force
f = frictional force = 18 N
Parallel to incline force equation is given as
F = f + mg Sin30
F = 18 + (70) Sin30
F = 53 N