3A. A 100-N block on a rough 30degrees-incline is acted upon by a force P as shown causing it to accelerate at 2 m/s^2 u...

$$\begin{gathered} \mathrm{Solution}:-\mathrm{Given} \mathrm{that} \\ \mathrm{W}=100 \mathrm{N} \\ \mathrm{a}=2 \mathrm{m}/{\mathrm{s}}^2 \\ {\mathrm{\mu }}_{\mathrm{k}}=0.3 \end{gathered}$$

$\left(\mathrm{a}\right) \mathrm{FBD} \mathrm{of} \mathrm{block} \mathrm{is} \mathrm{shown} \mathrm{below}:$

$$\begin{gathered} \left(\mathrm{b}\right) \mathrm{Now}, \mathrm{applying} \mathrm{Newton}\text{'}\mathrm{s} \mathrm{first} \mathrm{law} \mathrm{in} \mathrm{y} \mathrm{direction}; \\ \sum _{ }{\mathrm{F}}_{\mathrm{y}}=0 \\ \mathrm{N}-\mathrm{W} \mathrm{Cos}30-\mathrm{PSin}50=0 \\ \mathrm{N}=\mathrm{W} \mathrm{Cos}30+\mathrm{PSin}50 -\left(1\right) \\ \mathrm{Again}, \mathrm{applying} \mathrm{Newton}\text{'}\mathrm{s} \mathrm{second} \mathrm{law} \mathrm{in} \mathrm{x} \mathrm{direction}; \\ \sum _{ }{\mathrm{F}}_{\mathrm{x}}=\mathrm{ma} \\ \mathrm{P} \mathrm{Cos}50-{\mathrm{f}}_{\mathrm{k}}-\mathrm{WSin}30=\frac{\mathrm{W}}{\mathrm{g}}\mathrm{a} \\ \mathrm{where} {\mathrm{f}}_{\mathrm{k}}=\mathrm{kinetic} \mathrm{friction}=\mathrm{\mu N}=\mathrm{\mu }(\mathrm{W} \mathrm{Cos}30+\mathrm{PSin}50) \\ \mathrm{P} \mathrm{Cos}50-\mathrm{\mu }(\mathrm{W} \mathrm{Cos}30+\mathrm{PSin}50)-\mathrm{WSin}30=\frac{\mathrm{W}}{\mathrm{g}}\mathrm{a} \\ \mathrm{P} \mathrm{Cos}50-0.3(100 \mathrm{Cos}30+\mathrm{PSin}50)-100\mathrm{Sin}30=\frac{100}{9.81}\times 2 \\ \mathrm{P} \mathrm{Cos}50-0.3 \mathrm{PSin}50=\frac{100}{9.81}\times 2+100\mathrm{Sin}30+0.3\times 100 \mathrm{Cos}30 \\ \mathrm{P}=\frac{\frac{100}{9.81}\times 2+100\mathrm{Sin}30+0.3\times 100 \mathrm{Cos}30}{\mathrm{Cos}50-0.3 \mathrm{Sin}50} \\ \mathbf{P}\mathbf{=}\mathbf{233}\mathbf{.}\mathbf{35}\mathbf{ }\mathbf{N} \end{gathered}$$