Normal Force of Elevator given vertical distance above ground?

Question

"The vertical distance of an elevator above the ground floor is given by y=0.4t^2 (where length is in meters and time is in seconds). What is the normal force of the elevator floor on an 85kg person riding inside? (HINT: First find the acceleration of the elevator.)"

Answer 1

The vertical distance of an elevator above the ground floor is given by y=0.4t² .

The mass/weight of the person is 85 kg .

The velocity of elevator is elevator rate change in the position of elevator .

So now derivative the position function .

Velocity function v(t) = dy / dt .

v(t) = d/dt ( 0.4t² )

v(t) = 0.4*2 t .                   [ since the derivative of  xⁿ = n x^n-1  ] 

v(t) = 0.8 t . 

The velocity function of elevator at time t is v(t) = 0.8 t .

The acceleration of elevator is rate change in the velocity of elevator .

So now derivative the velocity function .

Velocity function v(t) = dv / dt .

a(t) = d/dt ( 0.8 t )

a(t) = 0.8 m/s² .

So the acceleration of elevator is 0.8 m/s² .

case (1) 

Normal force of an elevator when moving upward 

                                      N = mg + ma 

                                      N = 85 * 9.8 + 85 * 0.8 

                                      N = 833 + 68

                                      N =  901 N .

So the Normal force of an elevator when moving upward is 901 N .

case (2) 

Normal force of an elevator when moving downward 

                                      N = mg - ma 

                                      N = 85 * 9.8 - 85 * 0.8 

                                      N = 833 - 68

                                      N =  765 N .

So the Normal force of an elevator when moving downward is 765 N .