A block of mass 3.5 kg is released from rest at point A on a plane inclined at angle $\alpha$ to the horizontal, where $\tan\alpha = \dfrac{3}{4}$. It slides down to point B at the base of the plane. The coefficient of friction between the block and the plane is $\dfrac{1}{4}$. Take $g = 9.81$ ms$^{-2}$.

From $\tan\alpha=\dfrac{3}{4}$ the 3, 4, 5 triangle gives $\sin\alpha=\dfrac{3}{5}=0.6$ and $\cos\alpha=\dfrac{4}{5}=0.8$.

Perpendicular to the plane the block does not move, so $R=mg\cos\alpha$ and the friction, which opposes the sliding and so acts up the plane, is $\mu mg\cos\alpha$.

Along the plane, $mg\sin\alpha-\mu mg\cos\alpha=ma$. The mass cancels from every term:

$a=g(\sin\alpha-\mu\cos\alpha)=9.81\left(0.6-\tfrac{1}{4}(0.8)\right)=9.81(0.6-0.2)=9.81(0.4)=3.924$ ms$^{-2}$,

which is 3.9 ms$^{-2}$ correct to 2 significant figures. Notice the 3.5 kg never enters the answer.