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An inclined plane is inclined at an angle $\theta$ when the block placed on it is just at the point of moving down the plane. What can be the minimum acceleration with which the block can be moved up the inclined plane.

\[(a)\;g \; \sin \theta \quad (b)\;2g \sin \theta \quad (c)\;3g \sin \theta \quad(d)\;4g \sin \theta \]

1 Answer

When the angle of plane is just the block just begins to slide down $\tan \theta=\mu$
Also the force needed to push a mass upward $= mg \sin \theta+\mu mg \cos \theta$
acceleration $=\large\frac{Force}{mass}$
$\qquad= \large\frac{mg \;\sin \theta+\mu \cos \theta}{m}$
$\qquad= g \sin \theta+ \mu g \cos \theta$
$\qquad= g \sin \theta+ \tan \theta g \cos \theta$
$\qquad= g \sin \theta+ g \sin \theta$
$\qquad= 2g \sin \theta$
Hence b is the correct answer
answered Jul 15, 2013 by meena.p
edited May 26, 2014 by lmohan717

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