Question

A 72 kg man stands on a bathroom scale in an elevator. Starting from rest at the top floor, the elevator descends, attaining its maximum speed of 1.20 m/s in 0.80 s. It travels with this constant speed for the next 5.00 s. The elevator then slows down for 1.50 s and comes to rest at ground level. a)What is the acceleration of the man during the first 0.80 s? Answer in units of m/s2 and use the upward direction as positive. b) How many forces act on the Man during this elevator ride? no need to solve I know A and B c) What does the bathroom scale register in N during the first 0.80 s? d) What does the bathroom scale register in N during the 5.00 s period? e) What does the bathroom scale register in N during the 1.50 s period? Use 10 N/kg for g and the upward direction as positive.

Answer 1

a) v = u + at

1.20 = 0 + a0.80

a = 1.5 m/s2

C) in vertical:
mg - N = ma

N = m(g - a ) = 72(9.81 - 1.5) =598.32 N

D) during 5 s

a =0

mg - N = 0

N =mg = 72 x 9.8 = 705.6 N

e) during this period :

0 = 1.20 + a (1.50)

a = - 0.8 m/s2

now,

mg - N= ma

72 x 9.8 - N = 72 x -0.8

N = 648 N