A 100.0 kg safe is pushed across a floor with a force of 450 N. The coefficient of kinetic friction is 0.35. What is the acceleration of the safe?

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

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d. violin, clarinet, french horn, and tuba