Answer
(a) The scale reads 75.0 kg or 735 N.
(b) The scale reads 75.0 kg or 735 N.
(c) The scale reads 75.0 kg or 735 N.
(d) The scale reads 98 kg or 960 N.
(e) The scale reads 52 kg or 510 N.
Work Step by Step
Let $F_N$ be the normal force exerted upwards on the person. Note that $F_N$ will be the reading on the scale.
(a) If the elevator is at rest or moving at a constant speed, then the acceleration is zero.
$\sum F = ma$
$mg - F_N = 0$
$F_N = mg = (75.0 ~kg)(9.80 ~m/s^2) = 735 ~N$
The scale reads 75.0 kg or 735 N.
(b) If the elevator is at rest or moving at a constant speed, then the acceleration is zero.
$\sum F = ma$
$mg - F_N = 0$
$F_N = mg = (75.0 ~kg)(9.80 ~m/s^2) = 735 ~N$
The scale reads 75.0 kg or 735 N.
(c) If the elevator is at rest or moving at a constant speed, then the acceleration is zero.
$\sum F = ma$
$mg - F_N = 0$
$F_N = mg = (75.0 ~kg)(9.80 ~m/s^2) = 735 ~N$
The scale reads 75.0 kg or 735 N.
(d) Let's suppose the elevator is accelerating upward at a rate of $3.0 ~m/s^2$
$\sum F = ma$
$F_N - mg = ma$
$F_N = m(g+a) = (75.0 ~kg)(9.80 ~m/s^2 + 3.0 ~m/s^2)$
$F_N = 960 ~N$
The reading on the scale in kg is $\frac{960 ~N}{9.80 ~m/s^2} = 98 ~kg$
The scale reads 98 kg or 960 N.
(e) Let's suppose the elevator is accelerating downward at a rate of $3.0 ~m/s^2$
$\sum F = ma$
$mg - F_N = ma$
$F_N = m(g-a) = (75.0 ~kg)(9.80 ~m/s^2 - 3.0 ~m/s^2)$
$F_N = 510 ~N$
The reading on the scale in kg is $\frac{510 ~N}{9.80 ~m/s^2} = 52 ~kg$
The scale reads 52 kg or 510 N.
