Question
How does the total mechanical energy of the bowling ball change as it falls in a vacuum?
Answers
Answered by
GPT-4o mini
In a vacuum, when a bowling ball falls, the total mechanical energy of the system remains constant, assuming that no external forces (such as air resistance) are acting on it. The total mechanical energy is the sum of kinetic energy (KE) and potential energy (PE).
1. **Potential Energy (PE)**: When the bowling ball is at a height \( h \), it has gravitational potential energy given by:
\[
PE = mgh
\]
where \( m \) is the mass of the bowling ball, \( g \) is the acceleration due to gravity, and \( h \) is the height above the ground.
2. **Kinetic Energy (KE)**: As the ball falls, its height decreases, causing its potential energy to decrease. As it falls, its speed increases, which increases its kinetic energy given by:
\[
KE = \frac{1}{2} mv^2
\]
where \( v \) is the velocity of the bowling ball.
As the ball falls, the potential energy is converted into kinetic energy. Therefore, while the potential energy decreases, the kinetic energy increases, but the sum of the two remains constant:
\[
PE + KE = \text{constant}
\]
At the start of the fall (when the bowling ball is at its initial height), all the energy is potential. As it falls and reaches the ground, all that potential energy has been converted into kinetic energy. Just before it hits the ground, its potential energy is zero, and kinetic energy will be at its maximum.
In summary, in a vacuum, the total mechanical energy of the bowling ball remains constant throughout its fall, as energy is transformed from potential to kinetic without any losses.
1. **Potential Energy (PE)**: When the bowling ball is at a height \( h \), it has gravitational potential energy given by:
\[
PE = mgh
\]
where \( m \) is the mass of the bowling ball, \( g \) is the acceleration due to gravity, and \( h \) is the height above the ground.
2. **Kinetic Energy (KE)**: As the ball falls, its height decreases, causing its potential energy to decrease. As it falls, its speed increases, which increases its kinetic energy given by:
\[
KE = \frac{1}{2} mv^2
\]
where \( v \) is the velocity of the bowling ball.
As the ball falls, the potential energy is converted into kinetic energy. Therefore, while the potential energy decreases, the kinetic energy increases, but the sum of the two remains constant:
\[
PE + KE = \text{constant}
\]
At the start of the fall (when the bowling ball is at its initial height), all the energy is potential. As it falls and reaches the ground, all that potential energy has been converted into kinetic energy. Just before it hits the ground, its potential energy is zero, and kinetic energy will be at its maximum.
In summary, in a vacuum, the total mechanical energy of the bowling ball remains constant throughout its fall, as energy is transformed from potential to kinetic without any losses.