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+//example 3

+//analysis of energy transfer

+clear

+clc

+g=9.806 //acceleration due to gravity in m/s^2

+m=10 //mass of stone in kg

+H1=10.2 //initial height of stone above water in metres

+H2=0 //final height in metres

+dKE1=-m*g*(H2-H1) //change in kinetic energy when stone enters state 2 in J

+dPE1=-1 //change in potential energy when stone enters state 2 in J

+printf("\n hence,when stone is about to enter state 2, dKE = %.3f J. \n",dKE1)

+printf("\n and dPE = %.3f J. \n",dPE1)

+dPE2=0 //change in potential energy when stone enters state 3 in JQ2=0 //no heat transfer when stone enters state 3 in J

+W2=0 //no work done when stone enters state 3 in J

+dKE2=-1 //change in kinetic energy when stone enters state 3

+dU2=-dKE2 //change in internal energy when stone enters state 3 in J

+printf("\n hence,when stone has just come to rest in the bucket i.e. state 3, W=0, dPE=0, dKE1 = %.3f J. \n",dKE2)

+printf("\n and dU = %.3f J. \n",dU2)

+dKE3=0 //change in kinetic energy when stone enters state 4

+dPE=0 //change in potential energy when stone enters state 4 in J

+W3=0 //no work done when stone enters state 4 in J

+dU3=-1 //change in internal energy when stone enters state 4 in J

+Q3=dU3 //heat transfer when stone enters state 4 in J

+printf("\n hence,when stone has entered state 4, dPE=0, W3=0,dKE=0, dU= %.3f J. \n",dU3)

+printf("\n and Q3= %.3f J. \n",Q3)
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