On the way to lower floors,an elevator begins its descent from rest at a constant acceleration,descending the first 5.0 m in 0.85 s.What is the apparent weight of a 75 kg man inside the accelerator during this time interval?
A) 0.74 kN
B) 1.8 kN
C) 0.30 kN
D) 1.0 kN

Question

Quizplus · Accepted Answer

Firstly we need to find the acceleration of the elevator:

We know that the elevator starts from rest, so the initial velocity u = 0 m/s

The final velocity v can be found using the equation:

v = u + at

where t = 0.85 s (the time it takes to descend 5.0 m)

v = 0 + a(0.85)
v = 0.85a

We also know that the distance travelled, s = 5.0 m

Using the equation:

s = ut + 1/2at^2

5.0 = 0 + 1/2a(0.85)^2

a = 6.32 m/s^2

Now we can find the apparent weight of the man using the equation:

F = ma

where m = 75 kg (the mass of the man)

F = 75 × 9.81 - 75 × 6.32 (since the acceleration is downwards)

F = 558.75 - 474

F = 84.75 N

Converting this to kN:

F = 0.08475 kN

Therefore, the apparent weight of the man is 0.08475 kN, which is closest to option B (1.8 kN).

On the Way to Lower Floors,an Elevator Begins Its Descent