Question

On the way to lower floors, an elevator begins its descent from rest at a constant acceleration, ascending the first 0.5 m in 0.85s. What is the apparent weight of a 75 kg man inside the accelerator during this time interval?

.1 kn

.18 kn

.74 kn

.3 kn

Answer 1

By Newton's equation of motion

X = ut + 1/2 at²

u (initial speed) = 0

x (displacement) = 0.5 m

t = 0.85

Hence acceleration

                             

Apparent weight = mxg - m x a = m(g-a) = 75 x (9.8 - 1.38) = 631.5 N

                                                                                        = 0.6315 kN