Consider a 70.0-kg man standing on a spring scale in an elevator. Starting from

Question

Consider a 70.0-kg man standing on a spring scale in an elevator. Starting from rest, the elevator ascends, attaining its maximum speed of 1.40 m/s in 0.850 s. It travels with this constant speed for the next 5.00 s. The elevator then undergoes a uniform acceleration in the negative y direction for 1.60 s and comes to rest.

(a) What does the spring scale register before the elevator starts to move? _________N

(b) What does the spring scale register during the first 0.850 s? _________N

(c) What does the spring scale register while the elevator is traveling at constant speed? __________N

(d) What does the spring scale register during the time it is slowing down? __________N

Explanation / Answer

a) force = mg = 70*9.81 = 686.7 N

So the spring scale register before the elevator starts to move = 686.7 N

b) intial velocity = 0 m/s

final velocity = 1.4 m/s

time = 0.85 s

acceleration = 1.4/0.85 = 1.647 m/s2

So net force on spring balance during first 0.850 s = mg + ma = 70*9.81+70*1.647 = 802 N

c) spring scale register while the elevator is traveling at constant speed = mg = 70*9.81 = 686.7 N

d) acceleration in negative direction = 1.4/1.6 = 0.875 m/s2

So  the spring scale register during the time it is slowing down = mg - ma = 70*9.81 - 70*0.875 = 625.45 N

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