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 ${}^{\prime }{a}^{\prime }$ 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)

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