# Question #aeffb

Mar 17, 2017

Let the mass of the car be $m$ and its initial velcity of rolling be $v$

So its initial KE ${E}_{k} = \frac{1}{2} m {v}^{2}$

Given that the coefficient of rolling fricion is ${\mu}_{r} = 0.02$

Rolling frictionoal force on the car ${F}_{r} = {\mu}_{r} m g$. If the car coasts $s$ m before stopping, then Initial KE will be equal in magnitude of the work done against friction.

So $\frac{1}{2} m {v}^{2} = {\mu}_{r} m g \times s$

$\implies s = {v}^{2} / \left(2 {\mu}_{r} g\right)$

Now $v = 70 \text{km/hr"=(70xx10^3)/3600=700/36"m/s}$

$g = 9.8 \text{m/} {s}^{2}$

Therefore

$s = {\left(\frac{700}{36}\right)}^{2} / \left(2 \times 0.02 \times 9.8\right) \approx 964.5 m$