Question

read through the scenarios below and calculate the predicted change in kinetic energy of the object compared to a 50 kg ball traveling at 10 m/s. a 50 kg ball traveling at 20 m/s would have kinetic energy. a 50 kg ball traveling at 5 m/s would have kinetic energy. a 50 kg person falling at 10 m/s would have kinetic energy.

Explanation:

To solve this, we use the kinetic energy formula \( KE = \frac{1}{2}mv^2 \), where \( m \) is mass and \( v \) is velocity. The reference is a 50 kg ball at 10 m/s, so we calculate its \( KE \) and then compare with the other scenarios.

Step 1: Calculate reference \( KE \)

For the reference (50 kg, 10 m/s):
\( KE_{\text{ref}} = \frac{1}{2} \times 50 \, \text{kg} \times (10 \, \text{m/s})^2 \)
\( KE_{\text{ref}} = 25 \times 100 = 2500 \, \text{J} \)

Scenario 1: 50 kg ball at 20 m/s

Step 1: Calculate \( KE \) for 20 m/s

\( KE_1 = \frac{1}{2} \times 50 \times (20)^2 \)
\( KE_1 = 25 \times 400 = 10000 \, \text{J} \)

Step 2: Compare to reference

\( \frac{KE_1}{KE_{\text{ref}}} = \frac{10000}{2500} = 4 \). So it has 4 times more kinetic energy.

Scenario 2: 50 kg ball at 5 m/s

Step 1: Calculate \( KE \) for 5 m/s

\( KE_2 = \frac{1}{2} \times 50 \times (5)^2 \)
\( KE_2 = 25 \times 25 = 625 \, \text{J} \)

Step 2: Compare to reference

\( \frac{KE_2}{KE_{\text{ref}}} = \frac{625}{2500} = \frac{1}{4} \). So it has 1/4 (or 25%) of the reference kinetic energy (or "4 times less").

Scenario 3: 50 kg person at 10 m/s

Step 1: Calculate \( KE \) for 10 m/s (same as reference speed)

\( KE_3 = \frac{1}{2} \times 50 \times (10)^2 = 2500 \, \text{J} \)

Step 2: Compare to reference

Since \( KE_3 = KE_{\text{ref}} \), it has the same kinetic energy.

Final Answers:

• A 50 kg ball traveling at 20 m/s would have \(\boldsymbol{4}\) times more kinetic energy.
• A 50 kg ball traveling at 5 m/s would have \(\boldsymbol{\frac{1}{4}}\) (or 25%) of the reference kinetic energy.
• A 50 kg person falling at 10 m/s would have \(\boldsymbol{\text{the same}}\) kinetic energy.

Answer:

To solve this, we use the kinetic energy formula \( KE = \frac{1}{2}mv^2 \), where \( m \) is mass and \( v \) is velocity. The reference is a 50 kg ball at 10 m/s, so we calculate its \( KE \) and then compare with the other scenarios.

Step 1: Calculate reference \( KE \)

For the reference (50 kg, 10 m/s):
\( KE_{\text{ref}} = \frac{1}{2} \times 50 \, \text{kg} \times (10 \, \text{m/s})^2 \)
\( KE_{\text{ref}} = 25 \times 100 = 2500 \, \text{J} \)

Scenario 1: 50 kg ball at 20 m/s

Step 1: Calculate \( KE \) for 20 m/s

\( KE_1 = \frac{1}{2} \times 50 \times (20)^2 \)
\( KE_1 = 25 \times 400 = 10000 \, \text{J} \)

Step 2: Compare to reference

\( \frac{KE_1}{KE_{\text{ref}}} = \frac{10000}{2500} = 4 \). So it has 4 times more kinetic energy.

Scenario 2: 50 kg ball at 5 m/s

Step 1: Calculate \( KE \) for 5 m/s

\( KE_2 = \frac{1}{2} \times 50 \times (5)^2 \)
\( KE_2 = 25 \times 25 = 625 \, \text{J} \)

Step 2: Compare to reference

\( \frac{KE_2}{KE_{\text{ref}}} = \frac{625}{2500} = \frac{1}{4} \). So it has 1/4 (or 25%) of the reference kinetic energy (or "4 times less").

Scenario 3: 50 kg person at 10 m/s

Step 1: Calculate \( KE \) for 10 m/s (same as reference speed)

\( KE_3 = \frac{1}{2} \times 50 \times (10)^2 = 2500 \, \text{J} \)

Step 2: Compare to reference

Since \( KE_3 = KE_{\text{ref}} \), it has the same kinetic energy.

Final Answers:

• A 50 kg ball traveling at 20 m/s would have \(\boldsymbol{4}\) times more kinetic energy.
• A 50 kg ball traveling at 5 m/s would have \(\boldsymbol{\frac{1}{4}}\) (or 25%) of the reference kinetic energy.
• A 50 kg person falling at 10 m/s would have \(\boldsymbol{\text{the same}}\) kinetic energy.