Question #aeffb

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Question #aeffb

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Answer 1 · 2017-03-17T06:31:01

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

So its initial KE $E_{k} = \frac{1}{2} m v^{2}$ $E_k=1/2mv^2$

Given that the coefficient of rolling fricion is $μ_{r} = 0.02$ $mu_r=0.02$

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

So $\frac{1}{2} m v^{2} = μ_{r} m g \times s$ $1/2mv^2=mu_rmgxxs$

$\Rightarrow s = \frac{v^{2}}{2 μ_{r} g}$ $=>s=v^2/(2mu_rg)$

Now $v = 70 km/hr = \frac{70 \times 10^{3}}{3600} = \frac{700}{36} m/s$ $v=70"km/hr"=(70xx10^3)/3600=700/36"m/s"$

$g = 9.8 m/ s^{2}$ $g=9.8"m/"s^2$

Therefore

$s = \frac{{(\frac{700}{36})}^{2}}{2 \times 0.02 \times 9.8} \approx 964.5 m$ $s=(700/36)^2/(2xx0.02xx9.8)~~964.5m$

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