A boy is standing on a weighing machine placed on the floor of an elevator. The elevator starts going down with some acceleration, moves with constant velocity for some time and finally decelerates to stop. The minimum and maximum weight recorded are `480N` and `840N` If the magnitude of reardation is double of acceleration, find (a) true weight of the boy and (b) magnitude of acceleration.

Explanation

(A) When the elevator is under acceleration

`mg-N_(0)=ma=ma_(0)`
`N_(0)=m(g-a_(0))=W_(min)480`
(b) When the wlevator is under retardation

`mg-N_(0)'=m(-2a_(0))`
`N_(0)'=m(g+3a_(0))=W_(max)=840` (ii)
Dividing (ii) by (i), we get
`(m(g+a_(0)))/(m(g+a_(0)))=(840)/(480)rArr(g+3a_(0))/(g+a_(0))=(7)/(4)`
`rArra_(0)=0.6g=6m//s^(2)`
`m(g+a_(0))=480`
`m(10+6)=480rArrm=30kg`
True weight of the boy `=mg=30xx10=300N`
Magnitude of acceleration `6m//s^(2)`