Answer
(a) $a = 1.78~m/s^2$ (downward)
(b) $a = 2.14~m/s^2$ (upward)
(c) The elevator is in freefall, so the student should probably worry.
(d) In part (a), the tension in the cable is 6820 N.
In part (c), the tension in the cable is zero. This would explain why the elevator is in freefall.
Work Step by Step
We can find the mass of the student.
$mg = 550~N$
$m = \frac{550~N}{9.80~m/s^2} = 56.12~kg$
(a) We can set up a force equation for the student. Let down be the positive direction.
$\sum F = ma$
$mg - F_N = ma$
$a = \frac{mg-F_N}{m} = \frac{550~N-450~N}{56.12~kg}$
$a = 1.78~m/s^2$ (downward)
(b) We can set up a force equation for the student. Let up be the positive direction.
$\sum F = ma$
$F_N - mg = ma$
$a = \frac{F_N-mg}{m} = \frac{670~N-550~N}{56.12~kg}$
$a = 2.14~m/s^2$ (upward)
(c) We can set up a force equation for the student. Let down be the positive direction.
$\sum F = ma$
$mg - F_N = ma$
$a = \frac{mg-0}{m} = g$ (downward)
The elevator is in freefall, so the student should probably worry. Hopefully the elevator has some kind of emergency braking system.
(d) Let's consider the system of the student and the elevator. Let down be the positive direction.
$\sum F = ma$
$mg - T = ma$
$T = mg - ma = (850~kg)(9.80~m/s^2)-(850~kg)(1.78~m/s^2)$
$T = 6820~N$
In part (a), the tension in the cable is 6820 N.
In part (c):
$T = mg - ma = mg -mg = 0$
In part (c), the tension in the cable is zero. This would explain why the elevator is in freefall.
