A constant eastward horizontal force of 70. newtons is applied to a 20.-kilogram crate moving toward the east on a level floor. If the frictional force on the crate has a magnitude of 10. newtons, what is the magnitude of the crate's acceleration?

From Newton Law F = Ma: Where the net force F is the difference between the eastward horizontal force - frictional force on the crate. So F = 70 - 10 = 60N and the mass m = 20Kg. Hence we have 60 = 20a. Our acceleration, a, = 3.0 ms^ (-2)

Given: Force acting along x axis or horizontal due eastward Fₓ = 70 N

Mass m = 20 Kg; Frictional Force Ff = 10 N; coefficient of friction = ?

Normal force N = mg; g = 9.8 m/s²

Required: acceleration "a"

Solve for coefficient of friction Ff = μN or Ff = μmg

μ = 10 N / (20 Kg) (9.8 m/s²) μ = 0.05

Formula to follow: Fₓ - Ff = ma; a = Fₓ - μmg/m

a = 70 N - 0.05 (20 Kg) (9.8 m/s²) / 20 Kg Cancel the unit of Kg

a = 3.01 m/s²