Question

Zach, whose mass is 90 kg , is in an elevator descending at 12 m/s . the elevator takes 3.0 s to brake to a stop at the first floor. part a what is zach's apparent weight before the elevator starts braking?

Answer 1

Answer:

apparent weight will be 1242.9 N

Explanation:

As we know that elevator is moving downwards with speed 12 m/s

now it will comes to stop after t = 3 s

so the acceleration of the elevator is opposite to the direction of its motion

It is given as

$a = \frac{v_f - v_i}{t}$

$a = \frac{0 - 12}{3}$

$a = - 4 m/s^2$

now the normal force on the person who is on the elevator is given as

$N - mg = ma$

$N = m(g + a)$

$N = 90(9.81 + 4)$

$N = 1242.9 N$

so apparent weight will be 1242.9 N

Answer 2

Finding out the acceleration 12/3 = 4m/s^2
thus it is descending so the actual acceleration would be 9.8-4 = 5.8 m/s^2
the weight will be 90*5.8 = 522 N
522/9.8 = 53.2 kg