Refer to Example 4.19. What is the apparent weight of the same passenger (weighing \(598 \mathrm{N}\) ) in the following situations? In each case, the magnitude of the elevator's acceleration is $0.50 \mathrm{m} / \mathrm{s}^{2} .$ (a) After having stopped at the 15 th floor, the passenger pushes the 8 th floor button; the elevator is beginning to move downward. (b) The elevator is moving downward and is slowing down as it nears the 8 th floor.

The true weight of the passenger is already given, which is \(598N\). Now let's find the apparent weight in both situations.

During the start of moving downward, the elevator has a downward acceleration of \(0.50m/s^2\). For an accelerating reference frame, the apparent weight is given by: Apparent_weight = True_weight ± mass × acceleration In this case, since the elevator is moving downward, we need to subtract the force due to the downward acceleration: \(W_{apparent} = W_{true} - ma\) To find the apparent weight, we first need the mass of the passenger. We can find it using: \(W_{true} = mg\) Solving for mass 'm': \(m = \frac{W_{true}}{g} = \frac{598N}{9.8m/s^2} = 61kg\) Now, we can find the apparent weight: \(W_{apparent} = 598N - 61kg \times 0.50m/s^2 = 598N - 30.5N = 567.5N\) In the first situation, the apparent weight of the passenger is \(567.5N\).

As the elevator is moving downward but slowing down, its acceleration is now directed upwards i.e., \(-0.50m/s^2\). Therefore, when calculating the apparent weight, we will add the force due to the upward acceleration: \(W_{apparent} = W_{true} + ma\) Using the same mass as calculated in situation (a): \(W_{apparent} = 598N + 61kg \times (-0.50m/s^2) = 598N - 30.5N = 628.5N\) In the second situation, the apparent weight of the passenger is \(628.5N\).