If an object's kinetic energy is tripled, how may times faster is it moving? V3 times faster 3 times faster V6 times faster 0 times faster
If an object's kinetic energy is tripled, how may times faster is it moving? V3 times faster 3 times faster V6 times faster 0 times faster
Related questions
Question
![**Question:**
If an object's kinetic energy is tripled, how many times faster is it moving?
**Options:**
- \(\sqrt{3}\) times faster
- 3 times faster
- \(\sqrt{6}\) times faster
- 9 times faster
**Explanation:**
This question is about understanding the relationship between kinetic energy and velocity. Kinetic energy is given by the formula:
\[ KE = \frac{1}{2} m v^2 \]
where \( KE \) is the kinetic energy, \( m \) is the mass, and \( v \) is the velocity. If the kinetic energy is tripled, then you solve for the velocity change by setting:
\[ 3 \times KE = \frac{1}{2} m (v_{\text{new}})^2 \]
Solving for \( v_{\text{new}} \), we find:
\[ 3 \left(\frac{1}{2} m v^2\right) = \frac{1}{2} m (v_{\text{new}})^2 \]
\[ 3v^2 = (v_{\text{new}})^2 \]
\[ v_{\text{new}} = \sqrt{3} \times v \]
Therefore, the correct answer is \(\sqrt{3}\) times faster.
This option aligns with the mathematical relationship derived from the change in kinetic energy, reflecting how much faster the object moves when its kinetic energy is tripled.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd144f006-a58b-427c-a6bf-b99617213b05%2F772906b9-2e94-4175-8bb9-50712d22368d%2F3dlu9xl_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Question:**
If an object's kinetic energy is tripled, how many times faster is it moving?
**Options:**
- \(\sqrt{3}\) times faster
- 3 times faster
- \(\sqrt{6}\) times faster
- 9 times faster
**Explanation:**
This question is about understanding the relationship between kinetic energy and velocity. Kinetic energy is given by the formula:
\[ KE = \frac{1}{2} m v^2 \]
where \( KE \) is the kinetic energy, \( m \) is the mass, and \( v \) is the velocity. If the kinetic energy is tripled, then you solve for the velocity change by setting:
\[ 3 \times KE = \frac{1}{2} m (v_{\text{new}})^2 \]
Solving for \( v_{\text{new}} \), we find:
\[ 3 \left(\frac{1}{2} m v^2\right) = \frac{1}{2} m (v_{\text{new}})^2 \]
\[ 3v^2 = (v_{\text{new}})^2 \]
\[ v_{\text{new}} = \sqrt{3} \times v \]
Therefore, the correct answer is \(\sqrt{3}\) times faster.
This option aligns with the mathematical relationship derived from the change in kinetic energy, reflecting how much faster the object moves when its kinetic energy is tripled.
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
Step by step
Solved in 2 steps
