13. A person stands on a bathroom scale that is calibrated in newton, in a stationary elevator. The reading on the bathroom scale is W . The elevator now moves with a constant upward acceleration of $$ \frac{1}{4} \mathrm{~g} $$, where g is the gravitational acceleration. What will the reading on the bathroom scale be now? \begin{tabular}{llll} A & B \ \hline\end{tabular}

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1. In a stationary elevator, the reading on the scale is the weight of the person:
   $$
   W = mg.
   $$
2. When the elevator accelerates upward with $$ a = \frac{1}{4}g $$, the net effective acceleration acting on the person becomes:
   $$
   g_{\text{eff}} = g + \frac{1}{4}g = \frac{5}{4}g.
   $$
3. The reading on the scale in the accelerating elevator will be the normal force, which is now:
   $$
   W' = m \cdot g_{\text{eff}} = m \cdot \frac{5}{4}g.
   $$
4. Substituting $$ mg = W $$ into the equation, we get:
   $$
   W' = \frac{5}{4}W.
   $$
The reading on the bathroom scale will be $$ \frac{5}{4}W $$.

A person stands on a bathroom scale that is calibrated in newton, in a stationary elevator.
The reading on the bathroom scale is W .
The elevator now moves with a constant upward acceleration of , where g is the
gravitational acceleration. What will the reading on the bathroom scale be now?

A B

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A person stands on a bathroom scale that is calibrated in newton, in a stationary elevator. The reading on the bathroom scale is W . The elevator now moves with a constant upward acceleration of , where g is the gravitational acceleration. What will the reading on the bathroom scale be now? A B