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Question

A copper block of mass 1 kg slides down on a rough inclined plane of inclination 37 at a constant speed. Find the increase in the temperature of the block as it slides down through 60cm assuming that the loss in mechanical energy goes into the copper block as thermal energy. (specific heat of copper 420JKg1K1.

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Solution

Specific heat of copper 420JKg1K1

Heat energy absorbed:

Heat energy absorbed:

Q=mcΔT....(1)

Work done is equal to the loss in mechanical energy

ΔE=W

=(mgsinθ)L........(2)

From (1) and (2)

mcΔT=mgsinθL

ΔT=gsinθLc

ΔT=10sin(37)×0.6420

ΔT=8.5×1030C


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