d
(d) Net force along the plane = $P - mg\sin \theta $= $750 - 500$= $250\;N$
Limiting friction = ${F_l} = {\mu _s}R = {\mu _s}mg\cos \theta $
$= 0.4 × 102 × 9.8 × cos\,30 = 346\, N$
As net external force is less than limiting friction therefore friction on the body will be $250 N.$
