Answer:
False: only when the person accelerating downward
Explanation:
We can measure the "apparent weight" of a person by considering the following situation:
the person is standing on a scale. The normal reaction exerted by the scale (and so, by the floor) is actually what we perceive as "apparent weight". We call this force N (normal force), and we know that it acts upward.
At the same time, there is another force acting on the person: the force of graviity, also called (true) weight, which acts downward, and we label it with W.
Therefore, the equation of the forces on the person is:
[tex]N-W = ma[/tex]
where we have chosen upward as positive direction, and where
m is the mass of the person
a is its acceleration
We notice the following observations:
- If the person is moving at constant speed (no matter if upward or downward), the acceleration is zero: a = 0, so the apparent weight is the same as the true weight:
N = W
- If the person is accelerating upward, then [tex]a>0[/tex], and so
[tex]N=ma+W \rightarrow N>W[/tex]
So, the apparent weight will be larger than the true weight
- If the person is accelerating downward, then [tex]a<0[/tex], and so
[tex]N=ma-W \rightarrow N<W[/tex]
So, the apparent weight will be less than the true weight
So, the initial statement is wrong.
