Question

An 80 kg person is standing in a n elevator that is travelling downward at 4 m/s. Upon reaching the bottom floor, the elevator slows to a stop over 6 seconds time. What is the normal force acting on the person while the elevator is stopping?

Answer 1

Solution:-

Given –

Mass – 80 kg

Velocity or speed – 4 m/s

Time – 6 second

According to newton second law

F = ma

Force = mass*acceleration

First calculate acceleration

Acceleration = velocity / time

But in our problem given quantity is speed so,

Acceleration = speed / time

Acceleration (a) = 4m/s / 6s

Acceleration (a) = 0.6666 m/s²

Then calculate force

F = ma

F = 80 * 0.6666           

F = 53.33 N

The normal force acting on the person while the elevator is stopping is 53.33 N