How does external work affect the height of a bowling ball when pushed?

[Chris914]

Ok so when one drops a bowling ball hanging from a rope, it won't come back and smack him in the face. This I can understand, due to potential energy of gravity and conservation of energy. However when someone pushes the bowling ball, how can you show that it will rise higher than the original height and hit the person in the face? Is there a Work external or is it the Kinetic Energy starting out the Work-Energy Theorem?

 

[tycoon515]

Law of Conservation of Energy

If you push a bowling ball of mass m from some initial height, h_i, so that it starts out with some initial velocity, v, then, applying the Law of Conservation of Energy, we have: PE_i + KE_i = PE_f + KE_f, or equivalently: mgh_i + (1/2)mv_i² = mgh_f + (1/2)mv_f² + E_"lost". If we neglect air drag, we can discard the E_"lost" term.

We want to know what's happening when the bowling ball returns to you at h_i, so our initial and final states are both taken to be at height h_i, the height from which it was pushed.

mgh_i + (1/2)mv_i² = mgh_i + (1/2)mv_f²

Subtracting the gravitational potential energy from both states reveals that the kinetic energy of the bowling ball as it returns to you is the same as that when it leaves your hands, so it will hit you at the same speed you pushed it.

You may wonder how high the bowling ball would swing if you managed to get out of its way. To solve this, acknowledge that the bowling ball's maximum height, h_max, is attained when all of its energy is in the form of gravitational potential energy, with no kinetic energy remaining. So, mgh_i + (1/2)mv_i² = mgh_max. Dividing both sides by m, we have gh_i + (1/2)v_i² = gh_max. To get a nice expression for h_max, divide both sides by g, and then multiply both the top and bottom of the resulting expression by 2, yielding h_max = (2gh_i + v_i²)/(2g).

 

[Chris914]

Thank you so much, this was a great help.

 

[Simon Bridge]

Welcome to PF;
Looks a lot like homework to me - so you are lucky that someone was prepared to do all that work for you.
The short answer is that the energy added from the push has to be higher than the energy shortfall from what is needed to bounce back to the right height.

 

[PJC01]

I can explain the phenomenon you described using the principles of physics. When a bowling ball is dropped from a height, it has potential energy due to gravity. This potential energy is converted into kinetic energy as the ball falls. When the ball hits the ground, some of its kinetic energy is transferred to the ground, causing it to bounce back up. However, due to the conservation of energy, the total energy of the system (ball and Earth) remains the same.

Now, when someone pushes the bowling ball, they are adding external work to the system. This external work increases the total energy of the system, and as a result, the ball will rise higher than its original height. This is because the additional energy from the external work is added to the potential energy of the ball, increasing its total energy. As the ball rises, the potential energy is converted back into kinetic energy, and when it reaches its maximum height, it will have the same amount of kinetic energy as it had before it was pushed.

In summary, the Work-Energy Theorem states that the work done on an object is equal to the change in its kinetic energy. In this case, the external work done on the bowling ball increases its kinetic energy, which in turn increases its potential energy, allowing it to rise higher and potentially hit the person in the face. It is important to note that this only happens if the external work is enough to increase the total energy of the system above the initial potential energy of the ball. Otherwise, the ball will not rise higher than its original height.