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)`

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`