Asked by Mariah
                A 91 kg man stands on a spring scale in an elevator. Starting from rest, the eleva- tor ascends, attaining its maximum speed of 1.7 m/s in 1.07 s. It travels with this constant speed for the next 5 s, undergoes a uniform negative acceleration for 1.67 s, and comes to rest.
Scale
What does the spring scale register before the elevator starts to move?
The acceleration of gravity is 9.8 m/s2 . Answer in units of N.
009 (part 2 of 4) 10.0 points
What does the spring scale register during the first 1.07 s?
Answer in units of N.
010 (part 3 of 4) 10.0 points
What does the spring scale register while the elevator is traveling at constant speed?
Answer in units of N.
011 (part 4 of 4) 10.0 points
What does the spring scale register during the negative acceleration?
            
        Scale
What does the spring scale register before the elevator starts to move?
The acceleration of gravity is 9.8 m/s2 . Answer in units of N.
009 (part 2 of 4) 10.0 points
What does the spring scale register during the first 1.07 s?
Answer in units of N.
010 (part 3 of 4) 10.0 points
What does the spring scale register while the elevator is traveling at constant speed?
Answer in units of N.
011 (part 4 of 4) 10.0 points
What does the spring scale register during the negative acceleration?
Answers
                    Answered by
            Damon
            
    if a = 0
F = m g = (91)(9.8) = 892 N
a = 1.7 m/s / 1.07 s = 1.59 m/s^2
Fup - mg down = m a = 91(1.59)
Fup = 892+145 = 1037 N
constant speed, no acceleration so back to
m g = 892
892 -91(1.67/t)
and
0 = 1.7 - 1.67 t
t = 1.02 seconds to stop
so
892 - 91 (1.67/1.02)
= 743 N
    
F = m g = (91)(9.8) = 892 N
a = 1.7 m/s / 1.07 s = 1.59 m/s^2
Fup - mg down = m a = 91(1.59)
Fup = 892+145 = 1037 N
constant speed, no acceleration so back to
m g = 892
892 -91(1.67/t)
and
0 = 1.7 - 1.67 t
t = 1.02 seconds to stop
so
892 - 91 (1.67/1.02)
= 743 N
                                                    There are no AI answers yet. The ability to request AI answers is coming soon!
                                            
                Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.