How would I solve this physics problem?
You are running on an icy sidewalk. Suddenly, you stop running and let yourself slide. You slide at an initial speed of 10 m/s. You slide for a distance of 16 meters and your final speed, when you stop sliding, is 7 m/s. With a mass of 60 kilograms, what is the force of kinetic friction of the ice?
Any help is appreciated. Thanks
On
$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} v_{\mrm{f}}^{2} & = v_{\mrm{i}}^{2} - 2\,\mu_{\mrm{k}}\,g\,d \implies \mu_{\mrm{k}} = {v_{\mrm{i}}^{2} - v_{\mrm{f}}^{2} \over 2\,g\,d} = {10^{2} - 7^{2} \over 2\times 9.8 \times 16} = \bbx{\ds{{255 \over 1568} \approx0.1626}} \end{align}
You can use: \begin{equation} \Delta s=\frac{v_f ^2-v_i ^2}{2a}\\ a=\frac{v_f^2-v_i^2}{2\Delta s} \end{equation} where $\Delta s$ is the space covered by the man/woman, $v_i=10m/s, v_f=7m/s$.
Then you can apply the Newton's law getting the friction:
\begin{equation} F=ma=60Kg(-1,59m/s^2)=-95,4N \end{equation}