A 78kg person stands in an elevator. The elevator is at rest. What is the magnit

ID: 1460052 • Letter: A

Question

A 78kg person stands in an elevator. The elevator is at rest. What is the magnitude of the net force on the person?

The elevator is at rest. What is the magnitude of the normal force of the floor on the person?

The elevator and person have an upward acceleration of 1.80m/s^2. What is the magnitude of the force of the floor on the person?

Consider the force of the person on the floor.

i. direction is upward

ii. direction is downward

iii. the magnitude is the less than the magnitude of the force of the floor on the person.

iv. the magnitude is equal to the magnitude of the force of the floor on the person.

v. the magnitude is greater than the magnitude of the force of the floor on the person

Which are true?

here,

mass of person = M = 78 kg

Part A:
From Newton Law of motion : Sum(F) = 0
Normal force - Force due to Gravity = 0
N = mg
N = 78 * 9.8
N = 764.4 N

Part B:
a = acceleration of elevator,

From Newton Second law of motion :
Fnet = mass * acceleration
N - mg = ma
N = m(g+a)
N = 78(9.8 + 1.8)
N = 904.8 N

Part C:
Force of person on floor will always act downwards which is equal to force due to gravity
Fg = mg
Fg = 78*9.8
Fg = 764.4 N

This force will always equal to force applied by floor on person i.e Normal Force

N = Fg
N = 764.4 N

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.