The apparent weight of a person of mass m in an elevator is 2 m g . The elevator is moving

Question

The apparent weight of a person of mass m in an elevator is 2 m g . The elevator is moving (A) up with an acceleration of (g/2) (B) up with an acceleration of g (C) up with an acceleration of 2 g (D) down with an acceleration of g

Tardigrade Admin · Accepted Answer

Apparent weight is the normal force N experienced by the person. N-m g=m a ; N=m(g+a) Here N=2 m g ⇒ 2 m g=m g+m a or a=g <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/3fb73af355a1d2363b27830a5dbe1d86-.png" />

Q. The apparent weight of a person of mass $m$ in an elevator is $2 m g .$ The elevator is moving

up with an acceleration of $\frac{g}{2}$

up with an acceleration of $g$

up with an acceleration of $2 g$

down with an acceleration of $g$

Apparent weight is the normal force $N$ experienced by the person.
$N - m g = ma; N = m (g + a)$
Here $N = 2 m g \Rightarrow 2 m g = m g + ma$
or $a = g$

Q. The apparent weight of a person of mass m in an elevator is 2mg. The elevator is moving

up with an acceleration of 2g​

up with an acceleration of g

up with an acceleration of 2g

down with an acceleration of g