A block of mass m is sliding down an inclined | vedclass

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(C) Since the block is sliding down the inclined plane with a constant speed,its acceleration is zero. Therefore,the change in kinetic energy $(\Delta

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$m g h$

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(C) Since the block is sliding down the inclined plane with a constant speed,its acceleration is zero. Therefore,the change in kinetic energy $(\Delta KE)$ of the block as it moves from height $h$ to the ground is zero.According to the work-energy theorem,the total work done on the block is equal to the change in its kinetic energy:$W_{	ext{total}} = \Delta KE = 0$The total work done is the sum of the work done by gravity $(W_{	ext{gravity}})$ and the work done by friction $(W_{	ext{friction}})$:$W_{	ext{gravity}} + W_{	ext{friction}} = 0$The work done by gravity as the block descends a vertical height $h$ is $W_{	ext{gravity}} = mgh$.Substituting this into the equation:$mgh + W_{	ext{friction}} = 0$$W_{	ext{friction}} = -mgh$The energy dissipated by friction is the magnitude of the work done by the friction force:$	ext{Energy dissipated} = |W_{	ext{friction}}| = mgh$.

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