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+//Block being pulled
+//refer fig. 16.10 (a) and (b)
+//when pull P is acting
+W=2500  //N
+P=1000  //N
+N=W-P*sind(30)
+mu=0.2
+F=mu*N  //N
+//Initial velocity=0
+//Let final velocity be v
+s=30  //m
+//Applying work energy equation for the horizontal motion
+v=sqrt((0.866*1000-400)*30*2*9.81/2500)
+printf("\nv=%.3f m/sec",v)
+//Now if the 1000 N force is removed,let the distance moved before rest be s
+//Initial velocity=10.4745  //m/sec
+//Final velocity=0
+s=(2500*(10.4745^2))/(400*2*9.81)  //m
+printf("\ns=%.3f m",s)
+//The answer provided in the textbook is wrong