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. $\frac{F}{m}\cos \theta -{\mu }_K(g-\frac{F}{m}\sin \theta )$
• B. $\frac{F}{m}\cos \theta -{\mu }_K(g+\frac{F}{m}\sin \theta )$
• C. $\frac{F}{m}\cos \theta +{\mu }_K(g-\frac{F}{m}\sin \theta )$
• D. $-\frac{F}{m}\cos \theta -{\mu }_K(g-\frac{F}{m}\sin \theta )$

Answer 1

The correct option is A $\frac{F}{m}\cos \theta -{\mu }_K(g-\frac{F}{m}\sin \theta )$

Drawing the FBD of the block.


Along vertical direction:
$N=mg-F\sin \theta$ ...(1)
& along horizontal direction,
$F\cos \theta -{\mu }_{K}N=ma$ ...(2)

Substituting the value of $N$ from eq. (1) in eq. (2),
$\Rightarrow F\cos \theta -{\mu }_K(mg-F\sin \theta )=ma$
$\Rightarrow a=\frac{F}{m}\cos \theta -{\mu }_K(g-\frac{F}{m}\sin \theta )$
Hence, option (a) is correct.