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+//A tram car
+//refer fig.16.8 (a),(b) and (c)
+//frictional resistance
+W=120 //kN
+F=5*120/1000 //kN
+v=(20*1000)/(60*60) //m/sec
+// (1) on level track
+P1=F //kN
+//output power Pw1
+Pw1=P1*v //kW
+eta1=0.8
+//input power Ip1
+Ip1=Pw1/0.8 //kW
+// (2) Up the plane
+P2=F+W*(1/300) //kN
+//output power required Pw2
+Pw2=P2*v //kW
+//Input power of engine Ip2
+Ip2=Pw2/0.8 //kW
+// (3) Down the incline plane
+Pd=F-W*(1/300)
+Pwd=0.2*5.5556 //kW
+//Input power
+Ipd=1.1111/0.8 //kW
+printf("\On level track Input Power=%.3f kW\nUp the plane Input Power=%.3f kW\nDown the incline plane Input Power=%.3f kW",Ip1,Ip2,Ipd)
+