Answer
(a) The reading on the scale will be 570 N.
(b) The reading on the scale will be 570 N.
(c) The reading on the scale will be 700 N.
(d) The reading on the scale will be 440 N.
(e) The reading on the scale is 0.
Work Step by Step
The reading on the scale will be equal to the normal force $F_N$ pushing up on the person.
(a) If the speed is constant, then the magnitude of $F_N$ is equal to $mg$.
$F_N = mg = (58.0~kg)(9.80~m/s^2) = 570~N$
The reading on the scale will be 570 N.
(b) If the speed is constant, then the magnitude of $F_N$ is equal to $mg$.
$F_N = mg = (58.0~kg)(9.80~m/s^2) = 570~N$
The reading on the scale will be 570 N.
(c) We can use a force equation to find the normal force $F_N$.
$\sum F = ma$
$F_N - mg = 0.23~mg$
$F_N = 1.23~mg = (1.23)(58.0~kg)(9.80~m/s^2)$
$F_N = 700~N$
The reading on the scale will be 700 N.
(d) We can use a force equation to find the normal force $F_N$. Let down be the positive direction.
$ma = \sum F$
$(m)(0.23g) = mg-F_N$
$F_N = mg - 0.23~mg = 0.77~mg$
$F_N = (0.77)(58.0~kg)(9.80~m/s^2)$
$F_N = 440~N$
The reading on the scale will be 440 N.
(e) We can use a force equation to find the normal force $F_N$. Let down be the positive direction.
$ma = \sum F$
$mg = mg-F_N$
$F_N = mg-mg = 0$
The reading on the scale will be 0.
