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(C) According to the Work-Energy Theorem,the total work done on an object is equal to the change in its kinetic energy: $W_{	ext{all}} = \Delta K$.He

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$-m g(x+y)$

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(C) According to the Work-Energy Theorem,the total work done on an object is equal to the change in its kinetic energy: $W_{	ext{all}} = \Delta K$.Here,the forces acting on the knife are gravity and the resistive force exerted by the wooden block. Since the knife starts from rest and comes to rest,the change in kinetic energy $\Delta K = 0$.Thus,$W_{	ext{gravity}} + W_{	ext{block}} = 0$.The total vertical displacement of the knife is $(x + y)$. Therefore,the work done by gravity is $W_{	ext{gravity}} = m g (x + y)$.Substituting this into the equation: $m g (x + y) + W_{	ext{block}} = 0$.Solving for the work done by the block: $W_{	ext{block}} = -m g (x + y)$.

$A$ knife of mass $m$ is at a height $x$ from a large wooden block. The knife is allowed to fall freely,strikes the block and comes to rest after penetrating distance $y$. The work done by the wooden block to stop the knife is ..............

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