Question

A passenger of mass m stands on a weighing scale in an elevator accelerating upward with an acceleration of $g/2$. The scale would read ($g=$ acceleration due to gravity)

• A. $\frac{3mg}{2}$
• B. $\frac{mg}{2}$
• C. $\frac{2mg}{2}$
• D. mg
• E. $\frac{mg}{3}$

Hint:

When an elevator accelerates \textbf{upward}, you feel heavier ($N = m(g+a)$). When it accelerates \textbf{downward}, you feel lighter ($N = m(g-a)$). If the elevator is in free fall ($a=g$), you feel weightless ($N=0$).

Answer 1

The Correct Option is A

Concept:
Physics - Newton's Laws of Motion (Apparent Weight).
Step 1: Identify the forces acting on the passenger.
When a person stands on a scale in an elevator, two primary forces act on them:

• Weight ($W = mg$) acting vertically downward.
• Normal force ($N$) from the scale acting vertically upward.

The reading on the weighing scale is equal to this normal force ($N$).
Step 2: Apply Newton's Second Law.
The elevator is accelerating upward with acceleration $a = g/2$. According to $F_{net} = ma$: $$ N - mg = ma $$
Step 3: Solve for the normal force (N).
Rearrange the equation to isolate $N$: $$ N = m(g + a) $$ Substitute the given acceleration $a = g/2$: $$ N = m\left(g + \frac{g}{2}\right) $$
Step 4: Calculate the final value.
$$ N = m\left(\frac{2g + g}{2}\right) = \frac{3mg}{2} $$ The scale reading will be $\frac{3mg}{2}$.