Physics - NLM Question with Solution | TestHub

Total force along inclined plane, $\begin{array}{rl}f& =-{\mathrm{F}}_1+6g\sin 30^{\circ }+{\mathrm{F}}_2\\ f& =-60+6g\sin 30^{\circ }+18\\ f& =-60+6\times 10\times \frac{1}{2}+18\\ f& =12\mathrm{N}\text{ (downwards) }\\ \text{acceleration, }a& =\frac{f}{m}=\frac{12}{6}=2\mathrm{m}/{\mathrm{s}}^2\end{array}$As per question, the normal force between $2\mathrm{kg}$ and $1\mathrm{kg}$ is, $\begin{array}{rl}\mathrm{N}-18-10\sin 30^{\circ }& =ma\\ \mathrm{N}-18-10\times \frac{1}{2}& =1\times 2\\ \mathrm{N}& =2+23\\ \mathrm{N}& =25\mathrm{N}\end{array}$