A copper cube of mass `200g` slides down an 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 `60cm`. Specific heat capacity of copper `= 420J kg^(-1) K^(-1)`

Answer

`(3)/(350) = 8.6 xx 10^(-3).^(@)C`

Explanation

`mg sin 37^(@)d = ms Delta theta`
`Delta theta = (10 xx3xx 0.6)/(5 xx 420) = (3 xx 6)/(5 xx 420) = (3)/(350)`
`= 8.6 xx 10^(-3).^(@)C`.