A person stands on a weighing machine placed on the floor of an elevator. The elevator starts moving upwards with a certain acceleration, proceeds with a uniform velocity for a while, and finally decelerates to a stop. The maximum and minimum weights registered are 72 kg and 60 kg . Given that the magnitudes of the acceleration and the deceleration are identical, determine the magnitude of the acceleration. Take g =

Step-by-step solution

Let the true mass of the person be m . When the elevator accelerates upwards with acceleration a , the apparent weight is maximum. W_ max = m(g + a) When the elevator decelerates upwards with acceleration a , the apparent weight is minimum. W_ min = m(g - a) Taking the ratio of the maximum and minimum apparent weights: W_ max W_ min = m(g + a) m(g - a) = 72 60 g + a g - a = 6 5 5g + 5a = 6g - 6a 11a = g a = g 11 Substituting g = 9.9 m/s ^2 : a = 9.9 11 = 0.9 m/s ^2