waecE-LEARNING
Physics Paper 2, Nov/Dec. 2012  
Questions: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Main
General Comments
Weakness/Remedies
Strength
















Question 2

Question 2
          The upward acceleration of a lift of total mass 2500 kg is 0.5 ms-2.  The lift is
          supported by a steel cable which has a maximum safe working stress of

  1. x 107 Nm-2.  Determine the cross sectional area of the cable.

( g = 10ms-2 )

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comment

            This question was not popular among the candidates and it was poorly attempted 
          Most candidates did not realize that for a body inside a upward moving lift it
          experiences a weight increase hence they were unable to determine the tension
          in the cable.  Some candidates stated that stress =   instead of stress =       

          Some candidates used  ‘g’ instead of (a+g).
          The expected answer is
            Tension in the cable = (mass x acceleration) + (weight of the mass)                         
                        Stress, S  =            Tension
                                                Cross-sectional area                                                               
                                               OR
Cross-sectional
area                  =          mass (acceleration + acceleration of free fall)
                                                          stress
                                   OR
A                     =          m (a+g)
                                        A
                        =          2500 (0.5 + 10)
                                    1.0 x 107                     

                        =          26.25 cm2        OR   2.625 x 10-3 m2              


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