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+clear;
+clc;
+printf("\t\t\tProblem Number 2.9\n\n\n");
+// Chapter 2: Work, Energy, and Heat
+// Problem 2.9 (page no. 70)
+// Solution
+
+m=1500; //Unit:kg //m=mass
+V1=50; //Km/hour V1=Velocity before it is slowed down
+//V1=(50*1000 m/hour)^2/(3600 s/hour)^2
+KE1=(m*(V1*1000)^2/3600^2)/2; //KE1=Initial kinetic energy //Unit:Joule
+
+//After slowing down
+V2=30; //Unit:KM/hour //V2=Velocity after it is slowed down
+//V2=(30*1000 m/hour)^2/(3600 s/hour)^2
+KE2=(m*(V2*1000)^2/3600^2)/2; //KE2=After slowing down, the kinetic energy //Unit:Joule
+
+KE=KE1-KE2; //KE=Change in kinetic energy //Unit:Joule
+printf("Change in kinetic energy is %f kJ",KE/1000);