Question

A block of mass $0.15 \mathrm{~kg}$ rests on an inclined plane of coefficient of friction $0.3$. If the angle of the plane with the horizontal is $30^{\circ}$, the minimum force parallel to the plane necessary to prevent it from slipping is $\left(\mathrm{g}=10 \mathrm{~m} \mathrm{~s}^{-2}\right)$ (a) $0.5 \mathrm{~N}$ (b) $0.36 \mathrm{~N}$ (c) $1.5 \mathrm{~N}$ (d) $0.25 \mathrm{~N}$

          A block of mass $0.15 \mathrm{~kg}$ rests on an inclined plane of coefficient of friction $0.3$. If the angle of the plane with the horizontal is $30^{\circ}$, the minimum force parallel to the plane necessary to prevent it from slipping is $\left(\mathrm{g}=10 \mathrm{~m} \mathrm{~s}^{-2}\right)$
(a) $0.5 \mathrm{~N}$
(b) $0.36 \mathrm{~N}$
(c) $1.5 \mathrm{~N}$
(d) $0.25 \mathrm{~N}$

Added by Nathaniel V.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Dwijendra Rao

01/04/2022

Step 1

There are three forces acting on the block: - The weight of the block, which acts vertically downward and can be calculated as $mg$ (mass x gravity). - The normal force, which acts perpendicular to the plane. - The frictional force, which acts parallel to the Show more…

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A block of mass $0.15 \mathrm{~kg}$ rests on an inclined plane of coefficient of friction $0.3$. If the angle of the plane with the horizontal is $30^{\circ}$, the minimum force parallel to the plane necessary to prevent it from slipping is $\left(\mathrm{g}=10 \mathrm{~m} \mathrm{~s}^{-2}\right)$ (a) $0.5 \mathrm{~N}$ (b) $0.36 \mathrm{~N}$ (c) $1.5 \mathrm{~N}$ (d) $0.25 \mathrm{~N}$