Question

A $5\text{kg}$ block slides down an inclined plane, having angle of inclination ${30}^{\circ }$ from horizontal. Find the acceleration (${\text{m/s}}^2$) of block, if the coefficient of kinetic friction is ${\mu }_k=\frac{1}{2\sqrt{3}}$. Take $g=10{\text{m/s}}^2$ .

• A. $2.5{\text{m/s}}^2$
• B. $3.5{\text{m/s}}^2$
• C. $10{\text{m/s}}^2$
• D. $7.5{\text{m/s}}^2$

Answer 1

The correct option is A $2.5{\text{m/s}}^2$


Since the block is sliding down the inclined plane with acceleration $a$, kinetic friction will act:
$f_k={\mu }_{k}N$
From the equilibrium condition for block, $\perp$ to inclined plane:
$N=mg\cos \theta .......i$

Applying newton's $2^{nd}$ law along the inclined plane in direction of acceleration:
$mg\sin \theta -{\mu }_{k}N=ma...........ii$

From Eq $i$ and $ii$:

$mg\sin 30-{\mu }_{k}mg\cos 30=ma$

$\Rightarrow a=g\sin 30-{\mu }_{k}g\cos {30}^{\circ }$
$\therefore a=\frac{g}{2}-\frac{1}{2\sqrt{3}}\frac{\sqrt{3}g}{2}$
$\therefore a=\frac{5}{2}=2.5{\text{m/s}}^2$