Question

It takes the elevator in a skyscraper 4.5s  to reach its cruising speed of 11

m/s  . A 70 kg  passenger gets aboard on the ground floor. A- What is the passenger's weight before the elevator starts moving?
B-What is the passenger's weight while the elevator is speeding up?
C-What is the passenger's weight after the elevator reaches its cruising speed?

Answer 1

A and C are very simple. Since there is no acceleration in both cases, the weight of the passenger is just
`mg=(70kg)(9.8(m)/(s^2))=686 N` (about 690 N or 150 lbs with correct sig-figs). For B, we must first find the acceleration of the elevator. Since
`v=v_(0)+at`, we can write that
`11(m)/(s)=0(m)/(s)+a(4.5s)`. Solving for a, we get that
`a\approx2.44(m)/(s^2)`. Plugging this into Newton's second law, we get that
`F=ma=(70kg)(2.44(m)/(s^2))\approx171 N`. When calculating the apparent weight, this gets added onto the 686 N from A and C, so the final apparent weight is about 857 N, or 860 N with the correct number of sig-figs.

tl;dr:
A - 690 N
B - 860 N
C - 690 N