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-


- A$\mu \mathrm{mgh}$
- B$\mu \mathrm{mgh} / \sin \theta$
- Correct answerC$\mathrm{mgh}$ )
- 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'.