Question

A copper cube of mass 200g slides down on a rough inclined plane of inclination ${37}^{\circ }$ at a constant speed. Assume that any loss in mechanical energy goes into the copper block as thermal energy. Find the increases in the temperature of the block as it slides down through 60 cm. Specific heat capacity of copper = $420Jk{g}^{-1}{K}^{-1}$.

• A. $4.6\times {10}^{\circ }C$
• B. $4.8\times {10}^{\circ }C$
• C. $8.6\times {10}^{\circ }C$
• D. $7.8\times {10}^{\circ }C$

Answer 1

The correct option is C

$8.6\times {10}^{\circ }C$

As block slides along $x$ only $mgsin\theta$ and friction F do work, let them be $mglsin\theta$ and $wF$.

Now, speed is constant

$\therefore$ All the loss in potential energy is dissipated as heat, no. kinetic energy being gained.

(s is the specific heat) $\Rightarrow$ $mglsin\theta$ = $wF$ = Heat lost = $ms\Delta T$

$\Rightarrow \Delta T$ = $\frac{glsin\theta }{5}$ = $\frac{10\times \frac{60}{100}\times \frac{3}{5}}{420}$

= $8.6\times {10}^{-3}{}^{\circ }C$