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A lift starts from rest with a constant upward acceleration It moves 1.5 m in the first 0.4 A person standing in the lift holds a packet of 2 kg by a string Calculate the tension in the string during the motion .A. 5.89 NB. 57.1 NC. 6.71 ND. none of these |
Answer» Correct Answer - B (b) Given, `u=0ms=15 m,t=0.4s` From `s=ut(1)/(2) at^(2) implies 1.5 =0+(1)/(2)a(0.4)^(2)` `implies a=(1.5xx2)/((0.4)^(2))=18.75 ms^(-2)` As the string is moving upwards with in acceleration `therefore T=m(g+a)=2(9.8+18.75)=57.1 N` |
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