Understanding Pressure Exerted by the Box

The question asks for the approximate pressure exerted by a wooden box on a table. Pressure is defined as the force acting perpendicularly on a unit area of the surface. In this case, the force is the weight of the box, and the area is the surface area of the box that is in contact with the table.

The formula for pressure is:

$$ \text{Pressure (P)} = \frac{\text{Force (F)}}{\text{Area (A)}} $$

Calculating the Force (Weight of the Box)

The force exerted by the box on the table is its weight. Weight is calculated using the formula:

$$ \text{Force (F)} = \text{mass (m)} \times \text{acceleration due to gravity (g)} $$

Given:

• Mass of the box (m) = 2 kg
• Acceleration due to gravity (g) $\approx$ 10 m/s² (This value is often used for approximation in physics problems and aligns with the options provided).

So, the force exerted by the box is:

$$ \text{F} = 2 \text{ kg} \times 10 \text{ m/s}^2 = 20 \text{ N} $$

Calculating the Contact Area

The box has dimensions 30 cm × 15 cm × 10 cm. It is placed on the table with sides 30 cm and 10 cm touching the tabletop. This means the area of contact is the rectangle formed by these two dimensions.

Given dimensions of the touching surface:

• Length = 30 cm
• Width = 10 cm

We need to convert these dimensions from centimeters (cm) to meters (m) because pressure is measured in N/m² (Pascals).

• Length = 30 cm = 30/100 m = 0.30 m
• Width = 10 cm = 10/100 m = 0.10 m

The area of contact is:

$$ \text{Area (A)} = \text{Length} \times \text{Width} = 0.30 \text{ m} \times 0.10 \text{ m} $$

$$ \text{A} = 0.030 \text{ m}^2 $$

Calculating the Pressure Exerted

Now we have the force and the area of contact. We can calculate the pressure using the formula $\text{P} = \frac{\text{F}}{\text{A}}$.

• Force (F) = 20 N
• Area (A) = 0.030 m²

$$ \text{P} = \frac{20 \text{ N}}{0.030 \text{ m}^2} $$

To simplify the division:

$$ \text{P} = \frac{20}{0.03} = \frac{20}{\frac{3}{100}} = 20 \times \frac{100}{3} = \frac{2000}{3} $$

Calculating the value:

$$ \text{P} \approx 666.666... \text{ N/m}^2 $$

The approximate pressure exerted on the table is 666.6 N/m².

Statement Analysis and Conclusion

Let's review the steps taken to arrive at the pressure value:

• Calculated the force due to gravity (weight of the box).
• Identified the correct contact area based on how the box is placed.
• Converted units to be consistent (cm to m).
• Applied the pressure formula (Force / Area).

The calculated value of approximately 666.6 N/m² matches one of the given options.

Revision Table: Key Physics Concepts

Additional Information: Pressure Variations

The pressure exerted by an object on a surface depends on both the force applied and the area over which the force is distributed. For a constant force (like the weight of the box), pressure can be changed by varying the contact area.

• If the contact area is larger, the pressure is lower.
• If the contact area is smaller, the pressure is higher.

For example, if the box in this problem was placed on the table such that the 15 cm × 10 cm side touched the table, the area would be 0.15 m × 0.10 m = 0.015 m². The pressure would then be $\frac{20 \text{ N}}{0.015 \text{ m}^2} = \frac{20}{15/1000} = \frac{20 \times 1000}{15} = \frac{20000}{15} \approx 1333.3 \text{ N/m}^2$. This is much higher pressure than when placed on the 30 cm × 10 cm side.

Similarly, if the 30 cm × 15 cm side touched the table, the area would be 0.30 m × 0.15 m = 0.045 m². The pressure would be $\frac{20 \text{ N}}{0.045 \text{ m}^2} = \frac{20}{45/1000} = \frac{20 \times 1000}{45} = \frac{20000}{45} \approx 444.4 \text{ N/m}^2$. This is the lowest pressure possible for this box.

This demonstrates how the orientation of the object affects the pressure it exerts due to its weight.