Question

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. $-\frac{ F }{ m } \cos \theta-\mu_{ K }\left( g -\frac{ F }{ m } \sin \theta\right)$
• B. $\frac{ F }{ m } \cos \theta-\mu_{ K }\left( g -\frac{ F }{ m } \sin \theta\right)$
• C. $\frac{ F }{ m } \cos \theta-\mu_{ K }\left( g +\frac{ F }{ m } \sin \theta\right)$
• D. $\frac{ F }{ m } \cos \theta+\mu_{ K }\left( g -\frac{ F }{ m } \sin \theta\right)$

Answer 1

Answer: (B)

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