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Chapter 4: Problem 25

4.25 You have just joined an exclusive health club, located on the top floor of a skyscraper. You reach the facility by using an express elevator. The elevator has a precision scale installed so that members can weigh themselves before and after their workouts. A member steps into the elevator and gets on the scale before the elevator doors close. The scale shows a weight of 183.7 lb. Then the elevator accelerates upward with an acceleration of \(2.43 \mathrm{~m} / \mathrm{s}^{2},\) while the member is still standing on the scale. What is the weight shown by the scale's display while the elevator is accelerating?

Short Answer

Expert verified
Answer: The weight shown by the scale's display while the elevator is accelerating is 229.3 lb.

Step by step solution

01

Convert the weight to mass in SI units

The weight of the person is given in lb. To calculate the person's mass, we need to convert the weight to SI units (Newtons) and then divide by the acceleration due to gravity (9.81 m/s²). To convert pounds to Newtons, use the following conversion factor: 1 lb = 4.44822 Newtons. So, the person's weight in Newtons is: \(W = 183.7 \,\text{lb} \times 4.44822 \,\frac{\text{N}}{\text{lb}} = 817.64834\, \text{N}\) Now, we will find the person's mass (m) using the equation \(W = m * g\), where g is the acceleration due to gravity (9.81 m/s²): \( m = \frac{W}{g} = \frac{817.64834\, \text{N}}{9.81\, \mathrm{m}/\mathrm{s}^2} = 83.41738\, \mathrm{kg}\)
02

Calculate the net force acting on the person

When the elevator accelerates upward, there will be a net force acting on the person. According to Newton's second law, the net force is given by \(F_\text{net} = m \times a\), where a is the acceleration of the elevator. Given the elevator's acceleration is 2.43 m/s², the net force is: \(F_\text{net} = 83.41738\, \mathrm{kg} \times 2.43\, \mathrm{m}/\mathrm{s}^2 = 202.7052\, \text{N}\)
03

Calculate the total force acting on the person

When the elevator is accelerating, the scale measures the normal force (N) acting on the person, which is the sum of the person's actual weight (W) and the net force (F_net) acting on them due to the elevator's acceleration: \(N = W + F_\text{net} = 817.64834\, \text{N} + 202.7052\, \text{N} = 1020.35354\, \text{N}\)
04

Convert the total force back to pounds

Now, we need to convert the total force in Newtons back to pounds to get the apparent weight shown by the scale: \(\text{Apparent Weight} = \frac{1020.35354\, \text{N}}{4.44822\, \frac{\text{N}}{\text{lb}}} = 229.3\, \text{lb}\) Therefore, the weight shown by the scale's display while the elevator is accelerating is 229.3 lb.

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