A 72 kg man stands on a spring scale in an elevator. Starting from rest, the ele

Question

A 72 kg man stands on a spring scale in an elevator. Starting from rest, the elevator ascends, attaining its maximum speed of 1.2 m/s in 0.73 s. It travels with this constant speed for 5.0 s, undergoes a uniform negative acceleration for 1.2 s, and comes to rest. What does the spring scale register in each of the following time intervals?

(a) before the elevator starts to move
_____ N
(b) during the first 0.73 s
_____N
(c) while the elevator is traveling at constant speed
_____N
(d) during the negative acceleration
_____N

Explanation / Answer

Master equation:
Fnet = ma
where the 2 forces are the Normal force (spring scale reading) and weight of man (mg)

a) Before the elevator moves,
Fnet = ma = 0 because it's not accelerating nor moving
N - mg = 0
N = mg
N = 72*9.8 = 705.6 N
------------------------
b)During those .73s, the elevator is accelerating to its max speed. Find that acceleration:

a = (vf - vo) / t
a = (1.2 - 0) / .73 = 1.64
Now since it is ascending, the Normal force is going in the positive y direction, while gravity is going in the negative y direction:

N - mg = ma
N = mg + ma
N = m(g+a)
Plug in the acceleration found and solve for N

N = 824.68 N
-------------------------------
c)constant speed means no acceleration.
Fnet = 0
N -mg = 0
N = mg = 706.3 N
-----------------------
d) since it's undergoing a negative acceleration, you want to find this acceleration:

a = (vf - vo) /t
a = (0 - 1.2)/1.2 = -1  

Now plug that a into this equation:
Fnet = ma
N - mg = ma
N = m(g+a) = 634.3 N

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