Question

Zach, whose mass is 90 kg , is in an elevator descending at 11 m/s . The elevator takes 4.0 s to brake to a stop at the first floor. What is Zach's apparent weight before the elevator starts braking?

Answer 1

The apparent weight of Zach before the elevator starts braking is approximately 1089 N.

Step-by-step explanation:

The apparent weight of an object in an elevator can vary based on the motion of the elevator. In this case, Zach's apparent weight before the elevator starts braking can be calculated using the equation:

Apparent weight = Mass × (Gravity acceleration - Acceleration of the elevator)

Given that Zach's mass is 90 kg and the elevator is descending at 11 m/s, we need to calculate the acceleration of the elevator. The change in velocity is -11 m/s (since it is descending), and the time taken to brake is 4.0 s.

Using the equation acceleration = (Change in velocity) / (Time taken), we have:

Acceleration of the elevator = -11 m/s / 4.0 s = -2.75 m/s²

Now we can calculate the apparent weight:

Apparent weight = 90 kg × (9.8 m/s² - (-2.75 m/s²)) = 90 kg × (9.8 + 2.75) m/s² ≈ 1089 N