# 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?

Mar 29, 2016

${F}_{x} = 5 \left(9 - .4 \cdot 10\right) \approx 25 N$

#### 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 $\implies m a$

y-axis:
sumF_y=ma; F_w-N = 0; F_w=mg=N  Newton Law
${F}_{{\mu}_{k}} = {\mu}_{k} \cdot N = {\mu}_{k} \cdot m g$ ===>(2)

Insert (2) into (1):

${F}_{x} - {\mu}_{k} \cdot m g = m a$ Solve for ${F}_{x}$
${F}_{x} = m a - {\mu}_{k} \cdot m g = m \left(a - {\mu}_{k} g\right)$

Now m=5kg; mu_k = .4; a=9m/s^2
${F}_{x} = 5 \left(9 - .4 \cdot 10\right) \approx 25 N$

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.