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A body starts from rest on a long inclined plane of slope $45^{\circ}$ . The coefficient of friction between the body and the plane varies as $\mu =0.3 x$ , where x is distance travelled down the plane . The body will have maximum speed ( for $g=10m/s^{2}$ ) when x= :
Option: 1
$9.8m$

Option: 2
$27m$

Option: 3
$12m$

Option: 4
$3.33m$

Answers (1)

best_answer

$N=m g \cos 45^{\circ}$

acceleration = g.sin45

$\text { deceleration }=\mu g \cos 45^{\circ}$

$\\ Net \ acceleration $=\left(g \sin 45^{\circ}-4 g \cos 45^{\circ}\right)$ \\ \\ max \ speed \ will \ be \ reached \ \ when \ $g \sin 45^{\circ}=4 g \cos 45^{\circ}$$

$\begin{aligned} &g\left(\frac{1}{5^{2}}\right)=(0.3 x) \frac{g}{5}\\ &\frac{10}{3} m=x\\ &x=3.33 m \end{aligned}$

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