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an elevator starts from rest with a constant upward accelaration. it moves 2m in thefirst 0.6 sec. a passanger in the elevator is holding a 3kg package by a verticle string. when the elevator is moving, what is the tension in the string?

9 years ago

let the acceleration of the lift be a    then  2=0*t +(1/2)at2   so, a=4/t2 = 400/36  =100/9      now as the lift is moving upwards ,so a pseudo force acts on the package in the downward direction and its magnitude =3*(100/3) =100/3
also the weight of the package acts downwards =3*g = 30newton     ia have taken g=10m/s2
so, tension is balanced by pseudo force +weight     so, tension = 30 +(100/3) newton

5 years ago
T-mg=maT=m(g+a)T=3(g+a).......1S=ut+1/2at^22=1/2at^2a=(2*2)/(0.6 *0.6)a=100/9Now substituting a in equation .....1,we haveT=3(9.8+100/9)T=3((88.2+100)/9) ( taking 9 as LCM)T=(188.2/9)*3T=188.2/3T=62.733333333333333333333333......N