# Zach, whose mass is 73 kg, is in an elevator descending at 12 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 and what is Zach's apparent weight while the elevator is braking?

## Expert Answers

Zach has a mass of 73 kg and is in an elevator. Initially the elevator is descending at a constant speed of 12 m/s. Zach's apparent weight during this period is equal to the acceleration due to gravity. It is equal to 73*9.8 = 715.4 N

When the elevator starts...

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Zach has a mass of 73 kg and is in an elevator. Initially the elevator is descending at a constant speed of 12 m/s. Zach's apparent weight during this period is equal to the acceleration due to gravity. It is equal to 73*9.8 = 715.4 N

When the elevator starts to brake it comes to a halt in 4 s. The acceleration of the elevator is 12/4 = 3 m/s^2 acting upwards. The net force exerted by Zach on the floor of the elevator is his weight during this period. This gives Zach's weight during this period of 4 seconds as 73*(9.8 + 3) = 934.4 N

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