JEE Main · 2019 · 09 Jan 2019 · Shift 1 · Laws of Motion

From the figure the plane is inclined at $45^{\circ }$, the $3$ N force acts down the plane and $P$ acts up the plane. For the minimum $P$ the block is on the verge of sliding down, so limiting friction $f=\mu mg\cos \theta$ also acts up the plane. Balancing forces along the incline, the two up-slope forces hold the two down-slope forces: $P+\mu mg\cos \theta =mg\sin \theta +3$. With $m=10$ kg, $g=10\mathrm{m}{\mathrm{s}}^{-2}$, $\mu =0.6$ and $\theta =45^{\circ }$: $P=mg\sin 45^{\circ }+3-\mu mg\cos 45^{\circ }=70.7+3-42.4\approx 32$ N.