Physics - NLM Question with Solution | TestHub

M accelerates down the incline with
$$a=g\text{sin}37{}^{\circ }$$
its horizontal and vertical components are $a\text{cos}37{}^{\circ }$ and $a\text{sin}37{}^{\circ }$.
For $m$ on the horizontal top of M: friction
$$f=ma\text{cos}37{}^{\circ }$$
and
$$N=m(g-a\text{sin}37{}^{\circ })$$
so
$$\mu =\frac{a\text{cos}37{}^{\circ }}{g-a\text{sin}37{}^{\circ }}=\frac{g\text{sin}37{}^{\circ }\text{cos}37{}^{\circ }}{g-g{\text{sin}}^{2}37{}^{\circ }}=\text{tan}37{}^{\circ }=\frac{3}{4}.$$
Option C.