An elevator moves downward in a tall building at a constant speed of 6.05 m/s. E

Question

An elevator moves downward in a tall building at a constant speed of 6.05 m/s. Exactly 5.58 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 does the bolt hit the top of the still-descending elevator? (Assume the bolt is dropped at t=0 s)

___________ s

b) Estimate the highest floor from which the bolt can fall if the elevator reaches the ground floor before the bolt hits the top of the elevator. (Assume 1 floor = 3m.)

____________

Explanation / Answer

a) distance covered by top of elevator in 5.85s (before reaching the bolt) is

d1 = (speed)*(time) = (6.05)*(5.85) = 35.39 m
After 5.85s, top of elevator passes the bolt. Let d2 be the distance travelled by the top of the elevator after passing the bolt and t be the time taken to cover distance d2.

d2 = (speed)*(time) = (6.05)*t m

Total distance travelled by top of elevator is (that is from t=0 s)

d_{e} = d1 + d2 = 35.39 + (6.05*t) ------------------(1)

Now, distance travelled by bolt is (that is falls down)

d_{b} = ut+(1/2)gt^2

  d_{b} = (1/2)(9.8)t^2 = (4.9)t^2 ----------------(2)

here, initial velocity of bolt is zero, g is acceleration due to gravity = 9.8 m/s^2

Equating eq:(1)&(2) we get

  35.39 + (6.05*t) = (4.9)t^2

  (4.9)t^2 - (6.05)t - 35.39 = 0

On solving the quadratic equation. the roots we get are

t = 3.37 or -2.14 .

Since t is always positive. Therefore, time taken by the bolt to hit top of still-descending elevator is

t = 3.37s

b) total distanc travelled by elevator is (from eq.1)

d_{e} = 35.39 + (6.05*t) = 35.39 + 6.05*(3.37) = 55.78 m

given, 1 floor is 3m. Therfore, no. of floors is

55.78/3 = 18.59333 = 18th floor

The highest floor is 18th floor..

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