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(B) The block $m$ moves with the same horizontal acceleration $a = 5\,m/s^2$ as the wedge because it does not slip on the wedge.Since the surfaces are

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$50$

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(B) The block $m$ moves with the same horizontal acceleration $a = 5\,m/s^2$ as the wedge because it does not slip on the wedge.Since the surfaces are frictionless,the only force acting on the block in the horizontal direction is the horizontal component of the normal force $N$.Let $N$ be the normal force exerted by the wedge on the block. The horizontal component of $N$ provides the acceleration $a$ to the block: $N \sin 	heta = ma$.The vertical component of $N$ balances the weight: $N \cos 	heta = mg$.However,we can use the Work-Energy Theorem directly. The work done by all forces on the block equals the change in its kinetic energy.The only force doing work on the block in the horizontal direction is the normal force $N$. Since the block moves only horizontally with acceleration $a$,its velocity at time $t = 2\,s$ is $v = at = 5 	imes 2 = 10\,m/s$.The initial velocity is $0$. The change in kinetic energy is $\Delta KE = \frac{1}{2}mv^2 - 0 = \frac{1}{2} 	imes 1 	imes (10)^2 = 50\,J$.Since the normal force is the only force doing work in the direction of motion (as gravity acts vertically and displacement is horizontal),the work done by the normal force is $50\,J$.

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