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 gravitation pul on it, `mg sin theta`. Thus
`f=mg sin theta=0.2xx10xxsin 37^(@)`
`=2xx3/5=1.2N`
As block slides 60 cm distance, work done by it against friction is
`W=f.l`
`=1.2xx0.6=0.72J`
This energy is only in increasing the temperature of copper block, thus
`0.72= m.s. DeltaT`
or `DeltaT=0.72/(ms)`
`=0.72/(0.2xx420)=0.00857^(@)C`.