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Chapter 4: Problem 101

You are standing on a bathroom scale inside an elevator. Your weight is 140 lb, but the reading of the scale is \(120 \mathrm{lb} .\) (a) What is the magnitude and direction of the acceleration of the elevator? (b) Can you tell whether the elevator is speeding up or slowing down?

Short Answer

Expert verified
Question: Determine (a) the magnitude and direction of the acceleration of the elevator, and (b) whether the elevator is speeding up or slowing down. Answer: (a) The magnitude of the acceleration of the elevator is approximately 1.425 m/s², and its direction is upward. (b) We cannot determine whether the elevator is speeding up or slowing down with the given information.

Step by step solution

01

Understand the forces acting on the person

The actual weight of the person is given as 140 lb, and the scale reading, which is the normal force or net force, is 120 lb. The difference between these two forces will provide us with the force related to the acceleration of the elevator.
02

Calculate net force exerted on the person

The net force exerted on the person can be calculated as follows: Net force = Actual weight - Measured weight Net force = 140 lb - 120 lb Net force = 20 lb
03

Determine the direction of the acceleration

Since the measured weight is less than the actual weight, the elevator scale experiences a force in the upward direction. Therefore, the acceleration vector will also point upward. This indicates that the elevator is accelerating upwards.
04

Calculate the magnitude of acceleration

We can use Newton's second law of motion, F = ma, to find the magnitude of the acceleration. Rearranging the equation, we get: Acceleration (a) = Net force (F) / Mass (m) However, we don't have the mass of the person, but we have their weight, which is mass multiplied by the acceleration due to gravity (g): Weight = mg So, Mass (m) = Weight / g Now we can calculate the magnitude of the acceleration: a = F / m a = (Net force) / (Weight / g) Given: g = 9.81 m/s^2 (approximate value) Weight = 140 lb = 62.142 kg (1 lb ≈ 0.453592 kg) Net force = 20 lb = 9.07185 kg (1 lb ≈ 0.453592 kg) a = 9.07185 kg / (62.142 kg / 9.81 m/s^2) a ≈ 1.425 m/s^2
05

Answer

(a) The magnitude of the acceleration of the elevator is approximately 1.425 m/s², and its direction is upward. (b) We cannot determine whether the elevator is speeding up or slowing down with the given information.

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