Finally, the block comes to its equilibrium position, where Kxeq​=mg−veg800 (xeq​ )=(0.5)(10)−veg Energy lost by body due to oscillatins =21​K (xeq​+x )2−21​K(xm​)2−(mg−veg)=21​K (x2 )+ (kxeq​−mg+(0)veg )x=21​kx2=21​(800)(0.02)2=400×4×10−4=0.16JThis is equal to the heat transferred0.16J=(0.5)(400)(ΔT)+(1)(4184)(ΔT)0.16=(4384)ΔT .ΔT=4.38416​×10310−2​∴ It is of the order of 10−5