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Arrange the examples in order, starting with the object that has the least amount of energy at the top and the most amount of energy at the bottom. In each case, assume there's no friction. Use [tex]g=9.8 \, \text{m/s}^2[/tex], [tex]PE=m \times g \times h[/tex], and [tex]KE=\frac{1}{2} m v^2[/tex].

1. A book with a mass of 0.75 kilograms resting on a shelf at a height of 1.5 meters
2. A ball with a mass of 0.25 kilograms rolling on flat ground with a velocity of 10 meters/second
3. A brick with a mass of 2.5 kilograms falling with a velocity of 10 meters/second when it's 4 meters above the ground

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Answer :

GinnyAnswer
Sure! Let's solve the problem step by step. Our goal is to arrange the objects based on their total energy from least to greatest. We need to calculate the potential energy (PE) and kinetic energy (KE) for each object and then compare these energies.

### Step 1: Calculate the energies for each object

#### 1. The book
- Mass [tex]\( m = 0.75 \)[/tex] kg
- Height [tex]\( h = 1.5 \)[/tex] meters

Potential Energy (PE):
[tex]\[ PE_{\text{book}} = m \times g \times h = 0.75 \times 9.8 \times 1.5 = 11.025 \text{ joules} \][/tex]

Since the book is resting, it has no kinetic energy.

#### 2. The brick
- Mass [tex]\( m = 2.5 \)[/tex] kg
- Height [tex]\( h = 4 \)[/tex] meters
- Velocity [tex]\( v = 10 \)[/tex] meters/second

Potential Energy (PE):
[tex]\[ PE_{\text{brick}} = m \times g \times h = 2.5 \times 9.8 \times 4 = 98.0 \text{ joules} \][/tex]

Kinetic Energy (KE):
[tex]\[ KE_{\text{brick}} = \frac{1}{2} \times m \times v^2 = \frac{1}{2} \times 2.5 \times 10^2 = 125.0 \text{ joules} \][/tex]

Total energy for the brick:
[tex]\[ \text{Total energy}_{\text{brick}} = PE_{\text{brick}} + KE_{\text{brick}} = 98.0 + 125.0 = 223.0 \text{ joules} \][/tex]

#### 3. The ball
- Mass [tex]\( m = 0.25 \)[/tex] kg
- Velocity [tex]\( v = 10 \)[/tex] meters/second

Kinetic Energy (KE):
[tex]\[ KE_{\text{ball}} = \frac{1}{2} \times m \times v^2 = \frac{1}{2} \times 0.25 \times 10^2 = 12.5 \text{ joules} \][/tex]

Since the ball is rolling on flat ground, it has no potential energy.

### Step 2: Arrange the objects by their total energy

Now that we have calculated the energy for each object, we can sort them:

- Book: [tex]\( 11.025 \)[/tex] joules (all potential energy)
- Ball: [tex]\( 12.5 \)[/tex] joules (all kinetic energy)
- Brick: [tex]\( 223.0 \)[/tex] joules (sum of potential and kinetic energy)

So, the arrangement from least amount of energy to the most is:

1. Book: [tex]\( 11.025 \)[/tex] joules
2. Ball: [tex]\( 12.5 \)[/tex] joules
3. Brick: [tex]\( 223.0 \)[/tex] joules

Therefore, the energy distribution is:

[tex]\[ \begin{array}{c} \boxed{\text{Book}} \\ \boxed{\text{Ball}} \\ \boxed{\text{Brick}} \end{array} \][/tex]

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