A copper cube of mass `200g` slides down an a rought inclined plane of inclination `37^(@)` an 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 if slides down through `60cm` Specific head capacity of capper `= 420J kg^(-1) K^(-1)`

Answer

A, C

Explanation

`m = 200g = 0.2kg`
l = 60 cm = 0.6m, f = mg`
Work done by an energy down by the copper block to a distance `60cm`
`W = mg//sin theta`
`= 0.2 xx 10 xx 0.6 sin 37^(@)epsilon `
`= 12 xx((3)/(5)) = 0.75`
The thermal energy gained by block
`= ms Delta theta`
`= 0.2 xx 420 Delta theta = 84 Delta theta `
But `84 Delta theta = 0.72`
`rArr Delta theta = (0.72)/(84) = 0.00957`
= 0.0096 = 8.6 = 10^(-3)^(@)C`