Question 164 of 210

A block of mass $m$ slides along a floor while a force of magnitude $F$ is applied to it at an angle $\theta$ as shown in figure. The coefficient of kinetic friction is ${\mu }_{K}$. Then, the block's acceleration $a$ is given by : ($g$ is acceleration due to gravity)

A$-\dfrac{F}{m}\cos \theta -{\mu }_{K}\left(g-\dfrac{F}{m}\sin \theta \right)$B$\dfrac{F}{m}\cos \theta -{\mu }_{K}\left(g-\dfrac{F}{m}\sin \theta \right)$C$\dfrac{F}{m}\cos \theta -{\mu }_{K}\left(g+\dfrac{F}{m}\sin \theta \right)$D$\dfrac{F}{m}\cos \theta +{\mu }_{K}\left(g-\dfrac{F}{m}\sin \theta \right)$

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Answer: (B) $\dfrac{F}{m}\cos \theta -{\mu }_{K}\left(g-\dfrac{F}{m}\sin \theta \right)$

$N=mg-f\sin \theta$
$F\cos \theta -{\mu }_{k}N=ma$
$F\cos \theta -{\mu }_{k}\left(mg-F\sin \theta \right)=ma$
$a=\dfrac{F}{m}\cos \theta -{\mu }_{K}\left(g-\dfrac{F}{m}\sin \theta \right)$