It takes the elevator in a skyscraper 4 0 s to reach its cruising speed of 10 m/

Question

It takes the elevator in a skyscraper 4 0 s to reach its cruising speed of 10 m/s A 60 kg passenger gets aboard on the ground floor. What is the passenger s apparent weight Before the elevator starts moving? While the elevator is speeding up? After the elevator reaches its cruising speed? Zach, whose mass is 80 kg, is in an elevator descending at 10 m/s The elevator takes 3 0s to brake to a stop at the first floor. What is Zach's apparent weight before the elevator starts braking? What is Zach's apparent weight while the elevator is braking? shows the velocity graph of a 75 kg passenger in an elevator. What is the passenger s apparent weight at At t=1s? 5.0 s? At 9 0 s?

Explanation / Answer

mass m = 60 kg

Initial speed u = 0

time t = 4 s

Speed after 4 s is v = 10 m/s

Accleration of the elevator a = (v-u) / t

                                        = (10-0) / 4

                                        = 2.5 m/s 2

(a).Apperent weight before the elevator starts = mg

                                                                  = 60 x9.8 = 588 N

(b).Apperent weight when the elevator speeding up W ' = m(a+g)

W ' = 60 (2.5 + 9.8)

       = 738 N

(c).Apperent weight after reaches cruising speed = mg = 588 N

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