Taylor creates a table using the masses of four different objects that travel with the same velocity. She calculates the kinetic energy of the objects and adds it to her table.

Answer 1

To calculate the kinetic energy of objects traveling with the same velocity, Taylor can use the kinetic energy formula:

\[ KE = \frac{1}{2} mv^2 \]

where:

• \( KE \) is the kinetic energy,
• \( m \) is the mass of the object,
• \( v \) is the velocity of the object.

Since the velocity is the same for all objects, the kinetic energy will be directly proportional to the mass of each object. Taylor can choose four different mass values for her objects, then calculate the kinetic energy for each one using a constant velocity.

For example, let's assume the velocity \( v \) is 10 m/s, and she chooses the following masses:

1. Object 1: \( m_1 = 1 \) kg
2. Object 2: \( m_2 = 2 \) kg
3. Object 3: \( m_3 = 3 \) kg
4. Object 4: \( m_4 = 4 \) kg

Now, let's compute the kinetic energy for each object:

1. \( KE_1 = \frac{1}{2} \times 1 , \text{kg} \times (10 , \text{m/s})^2 = \frac{1}{2} \times 1 \times 100 = 50 , \text{J} \)

2. \( KE_2 = \frac{1}{2} \times 2 , \text{kg} \times (10 , \text{m/s})^2 = \frac{1}{2} \times 2 \times 100 = 100 , \text{J} \)

3. \( KE_3 = \frac{1}{2} \times 3 , \text{kg} \times (10 , \text{m/s})^2 = \frac{1}{2} \times 3 \times 100 = 150 , \text{J} \)

4. \( KE_4 = \frac{1}{2} \times 4 , \text{kg} \times (10 , \text{m/s})^2 = \frac{1}{2} \times 4 \times 100 = 200 , \text{J} \)

Thus, Taylor's table of masses and their corresponding kinetic energies would look something like this:

| Mass (kg) | Kinetic Energy (J) | |-----------|---------------------| | 1 | 50 | | 2 | 100 | | 3 | 150 | | 4 | 200 |

This way, she can clearly see that as the mass of an object increases (while maintaining the same velocity), its kinetic energy also increases proportionally.