Question

018 (part 1 of 2)
An elevator accelerates upward at 1.2 m/s.
The acceleration of gravity is 9.8 m/s.
What is the upward force exerted by the
floor of the elevator on a(n) 71 kg passenger?
Answer in units of N.
019 (part 2 of 2)
If the same elevator accelerates downwards
with an acceleration of 1.2 m/s, what is the
upward force exerted by the elevator floor on
the passenger?
Answer in units of N.

Answer 1

Answer:

When elevator goes upward F = 781 N

When elevator goes downward F = 610.6 N

Explanation:

In both cases the free body diagram of forces on the passenger consists of a force pointing upwards ($F_e$ exerted by the elevator's floor on the passenger), and a force pointing downwards: force exerted by gravity ($F_g$) on the passenger.

a) When the elevator goes upward with an acceleration of 1.2 $\frac{m}{s^2}$ the net force ($F_N$) is pointing upwards and equal to the product of the passenger's mass time its acceleration (1.2 tex]\frac{m}{s^2}[/tex]) satisfying the equation shown below and that can be solved for the force exerted by the elevator ($F_e$) :

$F_N=F_e - F_g\\71kg *1.2\frac{m}{s^2} =F_e-71kg*9.8\frac{m}{s^2} \\71(1.2+9.8)N=F_e\\F_e=71*11N=781N$

b) When the elevator is going down, with acceleration 1.2 $\frac{m}{s^2}$, now the equation for the net force looks like (since the net force points in the same direction as the force of gravity):

$-F_N=F_e-F_g\\-71kg*1.2\frac{m}{s^2} =F_e-71kg*9.8 \frac{m}{s^2}\\71(9.8-1.2)N=F_e\\F_e=71*8.6N=610.6N$