Physics - NLM Question with Solution | TestHub

For the 14 kg block on the smooth $30^{\circ }$incline: $T=mg\sin \theta =14\times 10\times \frac{1}{2}=70$N

Normal reaction on the block: $N=mg\cos \theta =140\times \frac{\sqrt{3}}{2}=70\sqrt{3}$N

Horizontal equilibrium of the wedge (spring force $F=kx$): $F=T+N\sin \theta -T\cos \theta$

$F=70+70\sqrt{3}\times \frac{1}{2}-70\times \frac{\sqrt{3}}{2}=70$N

$x=\frac{F}{k}=\frac{70}{1000}$m $=7$cm