Question

A copper block of mass 1 kg slides down on a rough inclined plane of inclination ${37}^{\circ }$ at a constant speed. Find the increase in the temperature of the block as it slides down through $60cm$ assuming that the loss in mechanical energy goes into the copper block as thermal energy. (Specific heat of copper = $420J{kg}^{-1}{K}^{-1},g=10{ms}^{-2}$) .

• A. $6.6\times {10}^{-3}$${}^{\circ }C$
• B. $7.6\times {10}^{-3}$${}^{\circ }C$
• C. $8.6\times {10}^{-3}$${}^{\circ }C$
• D. $9.6\times {10}^{-3}$${}^{\circ }C$

Answer 1

The correct option is C $8.6\times {10}^{-3}$${}^{\circ }C$

Height lost $=60\sin {37}^0=60\times \frac{3}{5}=36m$

Loss in $PE=1\times 10\times 0.36=3.6J$

$V_i=0m/s$

$V_f=\sqrt{2gh}=\sqrt{2\times 10\times 36\times {10}^{-2}}=\sqrt{7.20}$

${KE}_i=0$

${KE}_f=\frac{1}{2}m{V}_f^0=\frac{1}{2}\times 1\times 7.20=3.6J$

Loss in Mechanical energy $=3.6J$

We know,

$Q=ms\Delta T$

$\Rightarrow 3.6=420\times 1\times \Delta T$

$\Rightarrow \Delta T=\frac{3.6}{420}=\frac{3600}{420}\times {10}^{-3}=8.57\times {10}^{{-3}^0}c$ [C]