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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}

Ask by Gardner Wagner. in South Africa
Mar 19,2025

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Answer

The reading on the bathroom scale will be \( \frac{5}{4}W \).

Solution

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: \[ W' = \frac{5}{4}W. \]

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Bonus Knowledge

When a person is in a stationary elevator, the reading on the scale reflects their weight, \( W = mg \), where \( m \) is the mass of the person and \( g \) is the acceleration due to gravity. When the elevator accelerates upwards at \( \frac{1}{4} g \), the effective weight increases. Hence, the reading on the scale becomes \( W' = m(g + \frac{1}{4} g) = m(1.25g) \). Therefore, the new reading will be \( 1.25W \). During the upward acceleration, you can think of the scale as measuring not just your weight but the “extra push” needed to move up. Be careful! If you mistakenly calculate only using \( W \), you wouldn’t account for that additional \( \frac{1}{4} g \) acceleration. Always remember that when in motion, the forces add up!

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