Physics - NLM Question with Solution | TestHub

M accelerates down the incline with
$$a=g\text{sin}45{}^{\circ }$$
its horizontal and vertical components are $a\text{cos}45{}^{\circ }$ and $a\text{sin}45{}^{\circ }$.
For $m$ on the horizontal top of M: friction
$$f=ma\text{cos}45{}^{\circ }$$
and
$$N=m(g-a\text{sin}45{}^{\circ })$$
so
$$\mu =\frac{a\text{cos}45{}^{\circ }}{g-a\text{sin}45{}^{\circ }}=\frac{g\text{sin}45{}^{\circ }\text{cos}45{}^{\circ }}{g-g{\text{sin}}^{2}45{}^{\circ }}=\text{tan}45{}^{\circ }=1.$$
Option A.