Answer:
1.07m/s²
Explanation:
From the question we are given the following
applied force Fapp = 450N
mass of the safe m = 100kg
Weight of the safe W = 980N
coefficient of kinetic friction n = 0.35
Required
Acceleration of the safe
According to Newton's second law
\sum Fx = ma
Fapp - Ff = ma
Ff is the frictional force
Fapp - nmg = ma
450 - 0.35(100)(9.8) = 100a
450 - 343= 100a
107 = 100a
a = 107/100
a = 1.07m/s²
Hence the acceleration of the safe is 1.07m/s²
