Question

While on an elevator, Jaden's apparent weight is \(550 \mathrm{N}\) When he is on the ground, the scale reading is \(600 \mathrm{N}\) What is Jaden's acceleration?

Short answer

Jaden's acceleration on the elevator is approximately 0.82 m/s² downward.

Step-by-step solution

Understand the problem

Jaden's apparent weight changes while he is on an elevator. His weight on the ground is 600 N, but while on the elevator, it reads 550 N. We need to determine Jaden's acceleration due to the change in apparent weight.

Apply the concept of apparent weight

Apparent weight is the normal force exerted by the scale and can vary due to acceleration. When the elevator accelerates, this apparent weight differs from the true weight. The force measured by the scale can be given by: \[ F_{ ext{apparent}} = mg + ma \] where \( F_{\text{apparent}} \) is the apparent weight, \( m \) is the mass, \( g \) is the acceleration due to gravity (9.8 m/s²), and \( a \) is the acceleration of the elevator.

Set up the equation for the apparent weight on the elevator

When Jaden is on the elevator, his apparent weight is 550 N, and on the ground, it is 600 N. Since the apparent weight on the elevator is less, it means the elevator is accelerating downwards. The equation becomes: \[ 550 = mg - ma \] Since \( mg = 600 \), substitute \( mg \) into the equation:\[ 550 = 600 - ma \]

Solve for acceleration \(a\)

Rearrange the equation from Step 3:\[ ma = 600 - 550 \] \[ ma = 50 \] Now, solve for \( a \): \[ a = \frac{50}{m} \] Given that \( mg = 600 \), we can find the mass \( m \):\[ m = \frac{600}{9.8} \approx 61.22 \] Substitute \( m \) into the equation for \( a \):\[ a = \frac{50}{61.22} \approx 0.82 \, \text{m/s}^2 \]

Conclusion

Jaden experiences a downward acceleration of approximately 0.82 m/s² while on the elevator. This results in a decreased apparent weight of 550 N.

Key concepts

Elevator Physics

When you're in an elevator, you might have experienced a brief feeling of weightlessness or extra heaviness as it moves. This is due to changes in your apparent weight, which is the force exerted on you by the floor or whatever you're standing on.
In simple terms, apparent weight is what a scale registers when you stand on it. If an elevator accelerates upwards, you might feel heavier as the scale reads a weight larger than your usual due to upward acceleration adding to gravity's pull. Conversely, if the elevator accelerates downwards, you may feel lighter since the elevator's motion partially counteracts gravity.
In Jaden's case, his actual weight on the ground is 600 N, but in the elevator, it's 550 N. This decrease means the elevator is accelerating downwards. Understanding this change helps us grasp the bigger picture of elevator physics and apparent weight adjustment.

Newton's Second Law

To solve problems involving elevators and apparent weight, we rely heavily on Newton's Second Law of Motion. This fundamental principle connects force, mass, and acceleration in a simple formula: \[ F = ma \] where:

• \( F \) is the net force exerted on an object.
• \( m \) is the mass of the object.
• \( a \) is the acceleration of the object.

In an elevator scenario, the force exerted on Jaden by the scale is a combination of his weight due to gravity and the elevator's acceleration. This adjustment results in a different figure displayed on the scale. In mathematical terms, the scale or apparent weight can be expressed as \[ F_{\text{apparent}} = mg + ma \] for upward acceleration. When moving downward, the elevator's acceleration is subtracted.Utilizing Newton's Second Law allows us to break down the physics so we can solve for quantities such as acceleration and understand practical changes like the fluctuation in scale readings.

Acceleration Calculation

Determining the acceleration of an elevator based on apparent weight changes involves several key steps. First, recognize the change in the force the scale registers, which is tied to acceleration. In Jaden's exercise, we know his on-ground apparent weight is 600 N, while in the elevator it's 550 N.
Notice the decrease indicates downward acceleration. With the formula given through Newton's Second Law, we have: \[ F_{\text{apparent}} = mg - ma \] Starting with the known values:

• Ground weight: \( mg = 600 \, \text{N} \)
• Elevator apparent weight: \( 550 \, \text{N} \)

Substitute to find:\[ 550 = 600 - ma \]\[ ma = 600 - 550 \]\[ ma = 50 \]Next, calculate mass \( m \) using gravity's constant, 9.8 m/s²:\[ m = \frac{600}{9.8} \approx 61.22 \]Finally, solve for acceleration \( a \):\[ a = \frac{50}{61.22} \approx 0.82 \, \text{m/s}^2 \]This systematic breakdown lets us compute Jaden's downward acceleration at 0.82 m/s², demonstrating how physical laws clarify real-world observations.