Question

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?

Answer 1

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)

Answer 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