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
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?
1 answer
1 answer