A block of mass $M = 5 \, \mathrm{kg}$ is resting on a rough horizontal surface for which the coefficient of friction is 0.2. When a force $F = 40 \, \mathrm{N}$ is applied, the acceleration of the block will be $\left(g = 10 \, m / s^{2}\right)$
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Kinetic friction = $\mu_k R$ $= 0.2(mg - F \sin 30^\circ)$
$= 0.2 \left( 5 \times 10 - 40 \times \frac{1}{2} \right)$ $=0.2(50-20)=6\,N$
Acceleration of the block $\frac{F \cos 30^\circ - \text{Kinetic friction}}{\text{Mass}}$
$40 \times \frac{\sqrt{3}}{2} - 6$ $= \frac{2}{5} = 5.73 \, m/s^{2}$
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