A copper cube of mass 200 g slides down on a rough inclined plane of inclination 37° 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 down through 60 cm. Specific heat capacity of copper = 420 J kg–1 K–1.

Given:


Heat capacity of copper=420J/Kg/K


Formula used:


Potential energy=mgh


M=mass


H=height


ΔQ=MCΔT


ΔQ=change in energi


ΔT=change in temperature


C=heat capacity


M=mass


We know that loss in mechanical energy is due to frictional force which is equal to mgsinα


So


Loss in potential energy=thermal energy


0.2× g× .36=0.2× 420× ΔT


ΔT=0.00857


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