Q46 / 60 · KVPY 2012 (SA) · Physics

A block of mass $\mathrm{m}$ is sliding down an inclined plane with constant speed. At a certain instant $\mathrm{t}_{0}$, its height above the ground is $\mathrm{h}$. The coefficient of kinetic friction between the block and the plane is $\mu$. If the block reaches the ground at a later instant $\mathrm{t}_{\mathrm{g}}$, then the energy dissipated by friction in the time interval $\left(\mathrm{t}_{\mathrm{g}}-\mathrm{t}_{0}\right)$ is- Figure

  • A
    $\mu \mathrm{mgh}$
  • B
    $\mu \mathrm{mgh} / \sin \theta$
  • Correct answer
    C
    $\mathrm{mgh}$ )
  • D
    $\mu \mathrm{mgh} / \cos \theta$

Solution

$\begin{array}{l} \mathrm{W}_{\mathrm{f}}+\mathrm{W}_{\mathrm{mg}}=\Delta \mathrm{K} . \mathrm{E} .,(\Delta \mathrm{K} . \mathrm{E} \cdot=0) \\ \mathrm{W}_{\mathrm{f}}=-\mathrm{W}_{\mathrm{mg}} \\ \mathrm{W}_{\mathrm{f}}=-\mathrm{mgh} \\ \therefore \text { Energy dissipated }=\mathrm{mgh} \end{array}$

Hence, the correct answer is option: 'C'.

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