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(D) Let the wedge have a horizontal acceleration $a$ towards the right. In the frame of the wedge,the block $m$ experiences a pseudo force $ma$ toward

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$\left( {\frac{{3 + 4\mu }}{{4 - 3\mu }}} \right)g$

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(D) Let the wedge have a horizontal acceleration $a$ towards the right. In the frame of the wedge,the block $m$ experiences a pseudo force $ma$ towards the left.For the block to remain at rest with respect to the wedge,the forces acting on the block along the incline and perpendicular to it must be balanced.The angle of the wedge is $	heta = 37^\circ$. Thus,$\sin 37^\circ = 3/5$ and $\cos 37^\circ = 4/5$.Resolving forces perpendicular to the incline: $N = mg \cos 	heta + ma \sin 	heta$.Resolving forces along the incline: $ma \cos 	heta = mg \sin 	heta + f_s$,where $f_s$ is the static friction force.For maximum acceleration,the block has a tendency to slide up the incline,so friction $f_s$ acts downwards along the incline: $f_s = \mu N$.Substituting $N$: $ma \cos 	heta = mg \sin 	heta + \mu (mg \cos 	heta + ma \sin 	heta)$.Rearranging for $a$: $ma(\cos 	heta - \mu \sin 	heta) = mg(\sin 	heta + \mu \cos 	heta)$.$a = g \frac{\sin 	heta + \mu \cos 	heta}{\cos 	heta - \mu \sin 	heta}$.Substituting $\sin 37^\circ = 3/5$ and $\cos 37^\circ = 4/5$:$a = g \frac{3/5 + \mu (4/5)}{4/5 - \mu (3/5)} = g \frac{3 + 4\mu}{4 - 3\mu}$.

If the coefficient of friction between the block of mass $m$ and the wedge is $\mu $,then the maximum horizontal acceleration of the wedge for which the block will remain at rest with respect to the wedge is

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