# How much work would it take to horizontally accelerate an object with a mass of 8 kg to 3 m/s on a surface with a kinetic friction coefficient of 6 ?

Aug 18, 2017

$W = 36$ $\text{J}$ $+ 471 \textcolor{w h i t e}{l} \text{N" xx "distance}$

#### Explanation:

We're asked to find the necessary work to accelerate an $8$-$\text{kg}$ object from $0$ to $3$ $\text{m/s}$ on a surface with a coefficient of kinetic friction of $6$.

The equation for work according to the work-energy theorem is given by

ul(W = DeltaE_k = 1/2m(v_2)^2 - 1/2m(v_1)^2

The final velocity is $3$ $\text{m/s}$, and its mass is $8$ $\text{kg}$, so we have

W = 1/2(8color(white)(l)"kg")(3color(white)(l)"m/s" - 0)^2 = color(red)(ul(36color(white)(l)"J"

However, this does not include the work done to counteract the friction, so we have to use the equation

ul(W = Fs

The necessary force $F$ is

$F = {f}_{k} = {\mu}_{k} m g = \left(6\right) \left(8 \textcolor{w h i t e}{l} {\text{kg")(9.81color(white)(l)"m/s}}^{2}\right) = 471$ $\text{N}$

And so we have

$W = \left(471 \textcolor{w h i t e}{l} \text{N}\right) s$

The distance it travels $s$ can be any value because the applied force can be any number greater than the necessary force. Thus, we can leave it in terms of $s$:

W_"necessary" = color(blue)(ulbar(|stackrel(" ")(" "36color(white)(l)"J" + (471color(white)(l)"N")s" ")|)