# How much work would it take to horizontally accelerate an object with a mass of 8 kg to 7 m/s on a surface with a kinetic friction coefficient of 9 ?

Jul 29, 2017

$W = 196$ $\text{J}$

#### Explanation:

We're asked to find the total work done on an object so that it accelerates to $7$ $\text{m/s}$, with a coefficient of kinetic friction of $9$.

Using the equation

${f}_{k} = {\mu}_{k} n$,

we can find the retarding friction force the surface exerts on the object (knowing that the normal force $n = m g$):

f_k = mu_kmg = (9)(8color(white)(l)"kg")(9.81color(white)(l)"m/s"^2) = color(red)(706.32 color(red)("N"

To make this simple, we'll make the net force experienced by the object $1$ $\text{N}$, so the applied force will be $707.32$ $\text{N}$:

$\sum {F}_{x} = 707.32$ $\text{N}$ $- 706.32$ $\text{N}$ $= 1$ $\text{N}$

Now, we can use Newton's second law to find the acceleration of the object:

$\sum {F}_{x} = m {a}_{x}$

${a}_{x} = \frac{\sum {F}_{x}}{m} = \left(1 \textcolor{w h i t e}{l} \text{N")/(8color(white)(l)"kg}\right) = 0.125$ ${\text{m/s}}^{2}$

Now, let's use a kinematics equation to find the distance traveled under this constant acceleration so that the final velocity is $7$ $\text{m/s}$:

${\left({v}_{x}\right)}^{2} = {\left({v}_{0 x}\right)}^{2} + 2 {a}_{x} \left(\Delta x\right)$

Starting from an assumed state of rest:

$\left(7 \textcolor{w h i t e}{l} {\text{m/s")^2 = 0 + 2(0.125color(white)(l)"m/s}}^{2}\right) \left(\Delta x\right)$

$\Delta x = 196$ $\text{m}$

Finally, using the equation

$W = F s$ (one dimensional)

($F$ is the net force), we have

W = (1color(white)(l)"N")(196color(white)(l)"m") = color(blue)(ul(196color(white)(l)"J"