Question

In the figure shown, find the acceleration in $m/s^2$ of the body kept on the rough horizontal surface. Take $g=10m/s^2$.

Answer 1

The FBD of the block can be shown as

From the FBD we have
$\Sigma F_y=0$
$\Rightarrow N+F\sin {30}^{\circ }=10g$
$\Rightarrow N=100-50\sin {30}^{\circ }=75N$
And we have
$f_{max}={\mu }_{s}N=0.3\times 75=22.5N$

Now, total applied horizontal force $F^{{}^{\prime }}=F\cos {30}^{\circ }=\frac{50\sqrt{3}}{2}=43.3N$
Since, $F^{{}^{\prime }}>f_{max},$ the block will move

Let acceleration of the block be $a$
$\therefore F^{\prime }-{\mu }_{k}N=ma$
$\Rightarrow 43.3-\left(0.1\right)\times 75=10a$
$\Rightarrow a=\frac{43.3-7.5}{10}=3.58m/s^2$