A small block of mass m is placed on a wedge | vedclass

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(A) Let the wedge move with acceleration $A$ to the left. The block $m$ has acceleration $a'$ relative to the wedge along the incline (at $45^\circ$ t

Vedclass · Accepted Answer

$\sqrt {\frac{{2mghk}}{{{M^2}}}} $

Vedclass · Answer

(A) Let the wedge move with acceleration $A$ to the left. The block $m$ has acceleration $a'$ relative to the wedge along the incline (at $45^\circ$ to the horizontal).Using Newton's second law for the wedge in the horizontal direction: $N \sin 45^\circ = MA$,where $N$ is the normal force between the block and the wedge.For the block $m$,the forces are $mg$ (downward) and $N$ (perpendicular to the incline). The acceleration of the block is the vector sum of the wedge's acceleration $A$ and the relative acceleration $a'$.Using the work-energy theorem,the potential energy lost by the block $mgh$ is converted into the elastic potential energy of the spring $\frac{1}{2}kx^2$ at maximum compression.Thus,$mgh = \frac{1}{2}kx_{max}^2$,which gives $x_{max} = \sqrt{\frac{2mgh}{k}}$.The maximum force exerted by the spring is $F_{max} = k x_{max} = k \sqrt{\frac{2mgh}{k}} = \sqrt{2mghk}$.Since the wedge is retarded by the spring force,the maximum retardation $a_{max} = \frac{F_{max}}{M} = \frac{\sqrt{2mghk}}{M} = \sqrt{\frac{2mghk}{M^2}}$.Comparing this with the given options,option $A$ is correct.

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