Question

A department store elevator is moving downward in a multistory building at a constant speed of 6.30 m/s. Exactly 2.04 s after the top of the elevator car passes a bolt loosely attached to the wall of the elevator shaft, the bolt falls from rest.

(a)

At what time (in s) does the bolt hit the top of the still-descending elevator? (Assume the bolt is dropped at

t = 0 s.)

(b) Estimate the number of floors through which the bolt can fall if the elevator reaches the ground floor before the bolt hits the top of the elevator. (Assume

1 floor = 3 m.)

Answer 1

Gravitational acceleration = g = 9.81 m/s²

Initial speed of the bolt = V₁ = 0 m/s

Constant speed of the elevator = V₂ = 6.3 m/s

Time the bolt travels before hitting the top of the elevator = T₁

The bolt falls from rest exactly 2.04 sec after the top of the elevator car passes the bolt.

Time the elevator travels before the bolt hits the top of the elevator = T₂

T₂ = T₁ + 2.04

The distance traveled by the elevator and the bolt are equal.

Distance traveled = D

D = V₂T₂

D = V₁T₁ + gT₁²/2

V₂T₂ = V₁T₁ + gT₁²/2

(6.3)(T₁ + 2.04) = (0)T₁ + (9.81)T₁²/2

4.905T₁² - 6.3T₁ - 12.852 = 0

T₁ = 2.383 sec or -1.1 sec

Time cannot be negative.

T₁ = 2.383 sec

D = V₁T₁ + gT₁²/2

D = (0)(2.383) + (9.81)(2.383)²/2

D = 27.85 m

Distance between 1 floor = D₀ = 3 m

Number of floors through which the bolt falls = n

n = D/D₀

n = 27.85/3

n = 9.28

a) The bolt hits the top of the elevator 2.383 sec after it drops.

b) Number of floors through which the bolt falls = 9.28