Problem 11 - Physics

Question

An airtight box has a removable lid of area \(1.3 \times 10^{-2} \mathrm{~m}^{2}\) and negligible weight. The box is taken up a mountain where the air pressure outside the box is \(0.85 \times 10^{5} \mathrm{~Pa}\). The inside of the box is completely evacuated. What is the magnitude of the force required to pull the lid off the box?

Step-by-Step Solution

Answer

1105 N

Step 1: Understand the problem

We need to find the force required to pull the lid off an airtight box where the inside pressure is zero (evacuated) and the outside pressure is given. The force depends on the pressure difference and the area of the lid.

Step 2: Write the formula for force

The force exerted by the pressure difference on the lid can be calculated using the formula: \[ F = P imes A \]where \( P \) is the pressure difference and \( A \) is the area of the lid.

Step 3: Calculate the pressure difference

Since the inside of the box is completely evacuated, the internal pressure is effectively zero. Therefore, the pressure difference \( P \) is equal to the external pressure, which is \( 0.85 \times 10^{5} \) Pa.

Step 4: Substitute values in the formula

Given the area \( A = 1.3 \times 10^{-2} \) m² and pressure difference \( P = 0.85 \times 10^{5} \) Pa, substitute these values into the force equation:\[ F = 0.85 \times 10^{5} \, \text{Pa} \times 1.3 \times 10^{-2} \, \text{m}^2 \]

Step 5: Perform the multiplication

Calculate the force by multiplying:\(F = (0.85 \times 10^{5}) \times (1.3 \times 10^{-2})\)\(F = 1.105 \times 10^{3} \, \text{N}\)

Step 6: Conclusion

The magnitude of the force required to pull the lid off the box, in the given conditions, is approximately 1105 N.

Key Concepts

Understanding Pressure Difference

Pressure difference is a fundamental concept in physics, often affecting how objects interact in various environments. In this problem, the pressure difference is between the inside and outside of the airtight box.
The inside of the box is evacuated, meaning there's effectively no air or pressure inside. The pressure on the outside, however, is a known value, which in this case is 0.85 x 10⁵ Pa.
Since the internal pressure is zero, the pressure difference equals the external pressure entirely. This difference is crucial since it determines the force needed to remove the box lid. By understanding pressure difference, we can see how crucial it is in calculating the forces acting on surfaces.

Force Calculation

To determine the force needed to remove the lid of the box, we employ the formula for force involving pressure and area:

• Formula: \( F = P \times A \)
• Where: \( F \) is the Force, \( P \) is the Pressure difference, and \( A \) is the Area of the lid.

By plugging in the values -- the pressure difference \( P = 0.85 \times 10^5 \) Pa and the area \( A = 1.3 \times 10^{-2} \) m² -- we solve for force:
\[ F = 0.85 \times 10^5 \text{ Pa} \times 1.3 \times 10^{-2} \text{ m}^2 \]
This calculation yields a force of approximately 1105 N. The concept here is using known figures to derive an unknown, using basic physics principles.

The Airtight Box Concept

An airtight box is a container designed to prevent the exchange of air between its inside and the outside environment. This characteristic is critical in various applications, including this physics problem.
For the task, knowing the box is airtight tells us that when it's evacuated, it maintains a vacuum inside without letting external air in. This setup allows us to make accurate calculations about forces, as the pressure inside is consistently negligible or zero. This simple feature is key to many scientific experiments and protective storage solutions.

External Pressure Influence

External pressure plays a significant role in this scenario. It is the pressure exerted by the surrounding environment—in this case, the air outside the box.
When moving to higher altitudes, like on a mountain, air pressure decreases, which aligns with the given 0.85 x 10⁵ Pa at the location in the problem.
Understanding how external pressure impacts objects helps in analyzing real-world situations, such as weather patterns, aircraft functioning, and more. Here, it's the force that contributes to the pressure difference, ultimately affecting the lid's removal.

Evacuated Space Importance

An evacuated space in the context of this exercise refers to creating a vacuum inside the box.
Evacuation means removing air and consequently the pressure within the box setting it to nearly zero. This condition is key for calculating forces as there’s no counter-pressure inside to balance the external pressure.
Vacuum environments are essential in many scientific and industrial activities where certain conditions or reactions need to be controlled without atmospheric interference. Recognizing how an evacuated space changes pressure dynamics is critical for solving problems like this one.