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+// Initilization of variables
+W=5000 // N // Total weight of the elevator
+u=0 // m/s
+v=2 // m/s // velocity of the elevator
+s=2 // m // distance traveled by the elevator
+t=2 // seconds // time to stop the lift
+w=600 // N // weight of the man
+g=9.81 // m/s^2 // acc due to gravity
+// Calculations
+// Acceleration acquired by the elevator after travelling 2 m is given by,
+a=sqrt((v^2-u^2)/(2*s)) // m/s^2
+// (a) Let T be the the tension in the cable which is given by eq'n,
+T=W*(1+(a/g)) // N
+// (b) Motion of man
+// Let R be the pressure experinced by the man.Then from the Eq'n of motion of man pressure is given as,
+R=w*(1-(a/g)) // N
+// Results
+clc
+printf('(a) The Tensile force in the cable is %f N \n',T)
+printf('(b) The pressure transmitted to the floor by the man is %f N \n',R)