Physics - NLM Question with Solution | TestHub

The 15 kg block slides down the smooth incline with acceleration $A$along the plane; the 6 kg block rests on its horizontal top face, so its vertical acceleration is $A\sin 37^{\circ }$

For the 15 kg block along the incline: $(15g+N)\sin 37^{\circ }=15A$

For the 6 kg block (vertical): $6g-N=6A\sin 37^{\circ }$

Substituting $A$: $60-N=\frac{2}{5}{\sin }^{2}37^{\circ }(150+N)=\frac{18}{125}(150+N)$

$7500-125N=2700+18N$

$143N=4800$