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 increase in the temperature of the block as it slides $60cm$ along the inclined plane. Specific heat capacity of copper = $420Jk{g}^{-1}{K}^{-1}$. Take $sin\left({37}^{\circ }\right)=\frac{3}{5}$

• A. $4.6\times {10}^3{}^{\circ }C$
• B. $4.8\times {10}^{-3}{}^{\circ }C$
• C. $8.6\times {10}^{-3}{}^{\circ }C$
• D. $7.8\times {10}^2{}^{\circ }C$

Answer 1

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

Given:
Slope of inline, $\theta ={37}^{\circ }$
Length along incline on which block slides, $L=60cm=0.6m$
Mass of cube, $m=200g=0.2kg$
Specific heat capacity of copper, $s=420Jk{g}^{-1}{K}^{-1}$
Speed is constant and all loss in mechanical energy is converted to heat.

Loss in potential energy is:
$\Delta U=mgH$
Using trigonometry, $H=L\sin \theta$
$\Delta U=mgL\sin \theta$

Heat gained, $Q=ms\Delta T$

Loss in potential energy (as kinetic energy remains the same) is dissipated as heat.
$\Delta U=Q$
$mgL\sin \theta =ms\Delta T$
$\Delta T=\frac{gL\sin \theta }{s}$
$\Delta T=\frac{10\times 0.6\times \frac{3}{5}}{420}$
$\Delta T=8.6\times {10}^{-3}{}^{\circ }C$