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# A particle of mass m slides a distance d down a plane inclined at $\theta$ to the horizontal. The work done by the normal reaction R is

$(a)\;0 \quad (b)\;Rd \quad (c)\; mgd\;\cos \theta \quad (d)\;mgd\;\sin \theta$

Since the normal reaction R is perpendicular to the direction of displacement 'd'
$W=F.d$
$W=R. d \cos \theta$
$\quad=Rd \cos 90$
$\quad=0$
Hence a is the correct answer.

edited Feb 10, 2014 by meena.p