Question

An object with a mass of `5 kg` is on a surface with a kinetic friction coefficient of `4`. How much force is necessary to accelerate the object horizontally at `9 m/s^2`?

Answer 1

`F_x=5(9-.4*10) ~~25N`

Explanation:

Draw a free body diagram, FBD:
x-axis:
`sumF_x=ma; F_x-F_(mu_k) = ma` Newton Law ===>(1)
Summation of all Forces `=> ma`

y-axis:
`sumF_y=ma; F_w-N = 0; F_w=mg=N` Newton Law
`F_(mu_k)=mu_k *N= mu_k *mg` ===>(2)

Insert (2) into (1):

`F_x - mu_k *mg = ma` Solve for `F_x`
`F_x = ma-mu_k *mg = m(a-mu_kg)`

Now `m=5kg; mu_k = .4; a=9m/s^2`
`F_x=5(9-.4*10) ~~25N`

A word of caution, a kinetic friction, `mu_k = 4` is very large, I suspect you meant to say `.4` which reasonable. I assumed .4. If you want `mu_k = 4` just correct for it.