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?

Respuesta :

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

[tex]a = \frac{v_f - v_i}{t}[/tex]

[tex]a = \frac{0 - 12}{3}[/tex]

[tex]a = - 4 m/s^2[/tex]

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

[tex]N - mg = ma[/tex]

[tex]N = m(g + a)[/tex]

[tex]N = 90(9.81 + 4)[/tex]

[tex]N = 1242.9 N[/tex]

so apparent weight will be 1242.9 N

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