Discussion
If lift is going up with acceleration, the apparent weight of a body is
Answer: B
The apparent weight is more than the true weight.
For a lift accelerating upwards with acceleration \(a\), the floor must both support the body and accelerate it: \[N - mg = ma \Rightarrow N = m(g+a)\] The reading is that normal force, so it exceeds \(mg\). Going down with acceleration \(a\) it would read \(m(g-a)\).
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