91Ó°ÊÓ

The concept of the center of mass is crucial for understanding how objects behave when they are moved or rotated. Think of the center of mass as the "balance point" of the object, where all of its mass can be thought of as being concentrated.
For a uniform brick, the center of mass is at its geometric center. When lying flat on its largest base, the center of mass of the brick in this exercise is halfway through its thickness. To find this, you divide the thickness (1.5 cm or 0.015 m) by two, resulting in a height of 0.0075 m from the ground.
When the brick is turned to stand with its length vertical, its center of mass shifts. Now it is halfway along its length (5 cm or 0.05 m), so it's 0.025 m high from the ground. Understanding the new height of the center of mass is important as it affects the gravitational potential energy.

Gravitational potential energy relates to the energy stored in an object due to its position in a gravitational field—essentially, how high the object is above the ground. The formula to determine this potential energy is:

• \[ U = m \cdot g \cdot h \]

Here, \(m\) is the mass of the object, \(g\) the acceleration due to gravity (here, 10 \(\text{m/s}^2\)), and \(h\) the height of the object above a reference point.
The gravitational potential energy increases as the height of the center of mass increases. Initially, the energy is based on the initial height of 0.0075 m. After standing the brick up, the height goes up to 0.025 m, leading to a higher potential energy. Knowing the change in height helps us in calculating how much energy has been added (or work done) in moving the brick.

The work done in moving an object relates to the energy required to change its position, particularly against forces like gravity. In lifting or tilting objects, the work done can be calculated using:

• \[ W = m \cdot g \cdot \Delta h \]

Where \(W\) is work done, \(m\) the mass, \(g\) the gravitational acceleration, and \(\Delta h\) the change in height of the center of mass.
In this exercise, the mass of the brick is 2 kg, \(g\) is 10 \(\text{m/s}^2\), and \(\Delta h\) is the difference in height of the center of mass, calculated as 0.0175 m after rearranging the brick.
When you plug these values into the formula, you find that the work done is:

• \[ W = 2 \times 10 \times 0.0175 = 0.35\ \text{Joules} \]

While calculating, it’s crucial to ensure steps and calculations are precise to have accurate results, and mistakes in measurements or calculations should be double-checked.