A copper cube of mass 200 g slides down a rough inclined plane of inclination `37^@C` at a constant speed. Assuming that the loss in mechanical energy goes into the copper block as thermal energy. Find the increase in temperature of the block as it slidese down through 60 cm. Specific heat capacity of copper is equal to `420 J//kg-K. (Take, g = 10m//s^2)`.

Explanation

As block slides at constant speed, friction on block exactly balances the gravitational pull on it. `mgsintheta`. Thus `f=mgsintheta=0.2xx10xxsin37^@`
`=2xx(3)/(5)=2N`
As block slides 60 cm, work done by it against friction is
`W=fl`
`=1.2xx0.6=0.72J`
This energy is only used in increasing the temperature of copper block. Thus
`0.72=msDeltaT`
or `DeltaT=(0.72)/(ms)=(0.72)/(0.2xx420)=0.00857^@C`