(b) What is the passenger\'s \"apparent weight\" before the elevator starts movi

Question

(b) What is the passenger's "apparent weight" before the elevator starts moving? It isn't necessary to have a deep understanding of the concept of apparent weight for this problem. Just know that the normal force is what the person perceives to be their apparent weight. So if you find the value of the normal force, then that is the apparent weight.
N

(c) Which free body diagram is correct for the passenger while the elevator is speeding up? (You will have TWO submissions.) HINT: Which direction is the acceleration of the person (and the elevator)? Which free body diagram has magnitudes of forces that are consistent with the net force that results in this acceleration?

(d) What is the passenger's apparent weight while the elevator is speeding up? Remember that the person perceives the normal force to be his/her apparent weight. Make sure to reread the information that is given in the problem.
N

(e) Which free body diagram is correct for the passenger while the elevator is slowing down? (You will have TWO submissions.) HINT: Which direction is the acceleration of the person (and the elevator)?

(f) Assuming the elevator slows down at the same rate that it began moving (it takes 4.8 s to come to a stop) what is the passenger's apparent weight while the elevator is slowing down?
N

It takes the elevator in a skyscraper 4.8 s to reach its cruising (constant) speed of 10 m/s. A 75 kg passenger gets aboard on the ground floor. Free body diagrams for an elevator passenger under various circumstances are shown below. Which free body diagram is correct for the passenger before the elevator starts moving? This is the same situation as a person standing on the floor. (You will have TWO submissions.) A B C D What is the passenger's "apparent weight" before the elevator starts moving? It isn't necessary to have a deep understanding of the concept of apparent weight for this problem. Just know that the normal force is what the person perceives to be their apparent weight. So if you find the value of the normal force, then that is the apparent weight. N Which free body diagram is correct for the passenger while the elevator is slowing down? (You will have TWO submissions.) HINT: Which direction is the acceleration of the person (and the elevator)? A B C D Assuming the elevator slows down at the same rate that it began moving (it takes 4.8 s to come to a stop) what is the passenger's apparent weight while the elevator is slowing down? N

Explanation / Answer

a) before,

N =mg

C .

b) N = mg = 75 x 9.8 = 735 N

c) N - mg = ma

N = mg + ma

so D.

d) v = u + at

10 = 0 + a4.8

a = 2.08 m/s2 upwards

N = mg + ma = 75 x (9.8 + 2.08) =891 N

e) slowing down

mg - N = ma

N = mg -ma

so A

f) N = mg - ma = 75 x (9.8 - 2.08) = 578.75 N

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