You are going up in an elevator. Initially the elevator accelerates for 2.0 s to 6.0 m/s. Then it moves for 4.0 s, and then it stops uniformly in 3.0 s. What are the magnitudes of the normal forces exerted by the floor on you, if your weight is 784 N, in the three situations?

 

i) While accelerating: FN?=1024 N

ii) At constant velocity: FN?=784 N

iii) While decelerating: FN?=624 N

Step-by-step explanation

i) Accelerating:

Find the acceleration:

a=ΔtΔv?

a=2.0 s(6.0−0.0) m/s?

a=3.0 m/s2

Find your mass:

FW?=mg

784 N=m⋅9.81 m/s2

m=79.9 kg

Use Newton's 2nd Law:

Fnet?=ma

Fnet?=79.9 kg⋅3.0 m/s2

Fnet?=240 N

Calculate the vertical vector sum:

Fnet?=FN?−FW?

FN?=240 N+784 N

FN?=1024 N

ii) At constant velocity: 

From Newton's 1st Law, movement at constant velocity implies Fnet?=0 N

Calculate the vertical vector sum:

Fnet?=FN?−FW?

FN?=784 N

iii) Decelerating:

Find the acceleration:

a=ΔtΔv?

a=3.0 s(0.0−6.0) m/s?

a=−2.0 m/s2

Use Newton's 2nd Law:

Fnet?=ma

Fnet?=79.9 kg⋅−2.0 m/s2

Fnet?=−160 N

Calculate the vertical vector sum:

Fnet?=FN?−FW?

FN?=−160 N+784 N

FN?=624 N