It takes the elevator in a skyscraper 4.0 s to reach its cruising speed of 10 m/s. a 60 kg passenger gets aboard on the ground floor. determine the passenger's weight, before the elevator starts moving, while the elevator is speeding up, and after the elevator reaches its cruising speed

Refer to the diagram shown below.

State A: Before the elevator starts moving.
Initially, the weight of the 60 kg passenger is
(60 kg)*(9.8 m/s²) = 588 N

State A to B: Elevator accelerates
The acceleration between states A and B is
a = (10 m/s)/(4 s) = 2.5 m/s²
The inertial force induced on the passenger is
(60 kg)*(2.5 m/s²) = 150 N
The effective weight of the passenger is
588 + 150 = 738 N

After state B: Dynamic equilibrium
After state B, the elevator moves at constant speed and the passenger is in dynamic equilibrium.
Therefore the passenger's weight is  588 N.

Weight = 588 N, before the elevator starts;
             = 738 N, while the elevator is speeding up;
             =  588 N, when the elevator reaches cruising speed.

Answer///air resistance,drag;

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