Consider a 80.0-kg man standing on a spring scale in an elevator. Starting from

Question

Consider a 80.0-kg man standing on a spring scale in an elevator. Starting from rest, the elevator ascends, attaining its maximum speed of 1.18 m/s in 0.600 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 0.600 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) F = ma = 80*9.81 = 784.8 N

b) a = dv/dt = 1.18/0.60 = 1.9666 m/s^2
F = F_a + ma = 784.8+80*1.9666 = 942.133 N

c) 784.8 N

d) a = dv/dt = -1.18/1.6 = -0.7375 m/s^2
F = F_a + ma = 784.8 N+80*(-0.7375) = 727.275 N

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