# How much work would it take to horizontally accelerate an object with a mass of 6 kg to 7 m/s on a surface with a kinetic friction coefficient of 1 ?

May 14, 2016

$147 J$

#### Explanation:

Concept 1
In the problem, it is important to note that there is no friction. In the formula,

${F}_{f} = \mu {F}_{N}$

$\mu$ represents the constant of proportionality. But since it equals to $1$, ${F}_{f} = {F}_{N}$.

Concept 2
To solve the problem, you must use two formulas:

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} W = \Delta {E}_{k} \textcolor{w h i t e}{\frac{a}{a}} |}}}$
where:
$W =$work (joules)
$\Delta {E}_{k} =$change in kinetic energy (joules)

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} {E}_{k} = \frac{1}{2} m {v}^{2} \textcolor{w h i t e}{\frac{a}{a}} |}}}$
where:
$m =$mass (metres)
$v =$velocity (metres per second)

Solving for Work
Start by breaking down $\Delta {E}_{k}$ into final kinetic energy minus the initial kinetic energy.

$W = \Delta {E}_{k}$

$W = {E}_{k , \text{final")-E_(k,"initial}}$

$W = \frac{1}{2} m {v}_{f}^{2} - \frac{1}{2} m {v}_{i}^{2}$

$W = \frac{1}{2} m {\left({v}_{f} - {v}_{i}\right)}^{2}$

$W = \frac{1}{2} \left(6 k g\right) {\left(7 \frac{m}{s} - 0 \frac{m}{s}\right)}^{2}$
$W = \frac{1}{2} \left(6 k g\right) \left(49 {m}^{2} / {s}^{2}\right)$
$W = \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{147 J} \textcolor{w h i t e}{\frac{a}{a}} |}}}$