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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)
$-\frac{ F }{ m } \cos \theta-\mu_{ K }\left( g -\frac{ F }{ m } \sin \theta\right)$
$\frac{ F }{ m } \cos \theta-\mu_{ K }\left( g -\frac{ F }{ m } \sin \theta\right)$
$\frac{ F }{ m } \cos \theta-\mu_{ K }\left( g +\frac{ F }{ m } \sin \theta\right)$
$\frac{F}{m} \cos \theta+\mu_{K}\left(g-\frac{F}{m} \sin \theta\right)$
Solution
$N = mg – f \sin \theta$
$F \cos \theta-\mu_{ k } N = ma$
$F \cos \theta-\mu_{ k }( mg – F \sin \theta)= ma$
$a =\frac{ F }{ m } \cos \theta-\mu_{ K }\left( g -\frac{ F }{ m } \sin \theta\right)$
