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+//Train along an inclined plane
+//refer fig. 15.14 and 15.15
+u=0
+v=(36*1000)/(60*60)  //m/sec^2
+s=1000  //m
+//From kinematic equation
+a=100/2000 //m/sec^2
+//Tractive resistance (Tr)
+Tr=5*1500  //N
+//Component of weight of train (Wt1)
+Wt=1500/100  //kN
+//Inertia force (I1)
+I=(1500*0.05)/(9.81)  //kN  (Down the plane)
+//Dynamic equilibrium equation gives
+T=7.5+15+7.645  //kN
+//Consider dynamic equilibrium of train
+//Total tractive resistance (Rt)
+Rt=5*2000  //N
+//Inertia force (I2)
+I2=(2000*0.05)/(9.81)  //kN  (Down the plane)
+//Component of weight down the plane (Wt2)
+Wt2=(2000)/(100)  //kN
+//Dynamic equilibrium equation gives
+P=10+10.194+20  //kN
+printf("\nT=%.3f kN\nP=%.3f kN",T,P)
+
+ 
+
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+