Question

a 1000 n person is standing on an ordinary bathroom scale in an elevator.
a. what should the scale read as the elevator accelerates upward at a rate of 1 m/s² before arriving at constant velocity?
b. what should the scale read as the elevator accelerates downward at a rate of 1 m/s² before arriving at constant velocity?

Explanation:

Step1: Find the mass of the person

We know that weight \(W = mg\), where \(W = 1000\space N\) and \(g= 9.8\space m/s^{2}\). So, \(m=\frac{W}{g}=\frac{1000}{9.8}\space kg\approx102.04\space kg\)

Step2: Analyze the forces when the elevator accelerates upward (for part a)

Using Newton's second law \(F_{net}=ma\). The normal force \(N\) (which is the scale reading) and the weight \(W\) act on the person. \(F_{net}=N - W\). Since \(a = 1\space m/s^{2}\) upward, \(N=W+ma\). Substitute \(W = 1000\space N\) and \(m=\frac{1000}{9.8}\space kg\), \(a = 1\space m/s^{2}\)

$$
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Step3: Analyze the forces when the elevator accelerates downward (for part b)

Using Newton's second law \(F_{net}=ma\). Now \(F_{net}=W - N\) (because the acceleration is downward). So \(N=W - ma\). Substitute \(W = 1000\space N\), \(m=\frac{1000}{9.8}\space kg\) and \(a = 1\space m/s^{2}\)

$$
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$$

Answer:

a. The scale should read approximately \(1102\space N\)
b. The scale should read approximately \(898\space N\)