Question

A 10 kg block slides, without acceleration, down a rough inclined plane making an angle of ${20}^{\circ }$ with the horizontal. Find the acceleration when the angle of inclination is increased to ${30}^{\circ }$ (g = 9.8 $m{s}^{-2}$)

Answer 1

Free body diagram:

From free body diagram, along the direction perpendicular to the plane we have,

N = mg cos ${20}^{\circ }$
Since the block is moving without acceleration, therefore along the plane we have,
f = mg sin ${20}^{\circ }$
Also, $f={\mu }_{k}N={\mu }_{k}mgcos{20}^{\circ }$
$\therefore mgsin{20}^{\circ }={\mu }_{k}mgcos{20}^{\circ }$
$\Rightarrow {\mu }_k=tan{20}^{\circ }$
Now, when inclination is made ${30}^{\circ }$,
$a=\frac{mgsin{30}^{\circ }-f^{\prime }}{m}=\frac{mgsin{30}^{\circ }-\mu kmgcos{30}^{\circ }}{m}$
$a=gsin{30}^{\circ }-gtan{20}^{\circ }cos{30}^{\circ }$

$a=9.8(0.5-0.3647\times 0.866)=1.8m/s^2$.