A 2 ~kg brick begins to slide over a surface which is inclined at...

2 \mathrm{~kg}

brick begins to slide over a surface which is inclined at an angle of

45^{\circ}

with respect to horizontal axis. The co-efficient of static friction between their surfaces is: A 2 ~kg brick begins to slide over a surface which is inclined at an angle of 45^ with respect to horizontal axis. The

A 2 ~kg brick begins to slide over a surface which is inclined at an angle of 45^ with respect to horizontal axis. The

Options

A $1.7$
B $\frac{1}{\sqrt{3}}$
C $0.5$
D $1$

Correct answer

D. $1$

Step-by-step solution

\begin{aligned} & \mathrm{mg} \sin 45=\mathrm{f}_{\mathrm{L}} \\ & \mathrm{mg} \cos 45=\mathrm{N} \\ & \mathrm{f}_{\mathrm{L}}=\mu_{\mathrm{s}} N \\ & \mu_{\mathrm{s}}=\tan 45=1 \end{aligned}

\tan \theta=\mu_s

(

\theta

is angle of repose)

\tan 45=\mu_{\mathrm{s}}=1

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