Consider a 79.0-kg man standing on a spring scale in an elevator. Starting from

Question

Consider a 79.0-kg man standing on a spring scale in an elevator. Starting from rest, the elevator ascends, attaining its maximum speed of 1.50 m/s in 0.550 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.

(b) What does the spring scale register during the first 0.550 s?
N

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



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

Explanation / Answer

b) during first 0.55 s,

a = 1.50 / 0.55 = 2.73 m/s^2

and applying Fnet = ma

N - mg = ma

N = mg + ma = 79 ( 9.81 + 2.73) = 990.4 N ........Ans

c) for constant speed,

a = 0

hence N = mg = 79 x 9.81 = 775 N

d) now, a = - 1.5 / 1.6 = - 0.9375 m/s^2

N = mg - ma

N= 79(9.81 - 0.9375) = 700.9 N

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.