Finding kinetic force of friction, given mass, speed, and distance 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
 A: 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}
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,}
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 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\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}
