Why Does Dropping a Ball into a Well Result in Different Energy Sums?

[bigsmile]

A 2.00-kg ball has zero kinetic and potential energy. John drops the ball into a 10.0-m-deep well.
1) Just before the ball hits the bottom, the sum of its kinetic and potential energy is: Zero J
2) After the ball comes to a stop in the mud, the sum of its potential and kinetic energy is: -196 J

On #1, why is "sum of kinetic and potential energy" turns to zero? I just guess it is due to conservation of energy, but I don't know how to use Math to prove it.

I know in theory Sum of all work equals the change of KE. I am not sure how to apply here. Before the ball hits the bottom, it should still have velocity and KE >0,

In addition, I really don't understand why the answer in both #1 and #2 are different. I thought the answer to #2 is also zero too due to conservation of energy. Why is the sum equal to a) a negative 196 instead of +196 and b) final potential energy of -196 J?

I am very confused. Please help.

 

[mfb]

On #1, why is "sum of kinetic and potential energy" turns to zero? I just guess it is due to conservation of energy, but I don't know how to use Math to prove it.

0=0? You can calculate the initial sum of 0+0=0, and you know the sum is conserved.

it should still have velocity and KE >0,

Sure, its kinetic energy is positive. Its potential energy is negative. The sum is zero.

For (b), just calculate the potential energy of a ball 10 meter below zero.
The kinetic energy is lost due to friction in the mud.

 

[bigsmile]

0=0? You can calculate the initial sum of 0+0=0, and you know the sum is conserved.
Sure, its kinetic energy is positive. Its potential energy is negative. The sum is zero.

For (b), just calculate the potential energy of a ball 10 meter below zero.
The kinetic energy is lost due to friction in the mud.

After it goes into the mud and KE is lost due to friction, is energy still conserved at that point?
Initial: 0 + 0 = 0 ( initial PE + initial KE = 0)
In the mud: KE (final) + PE(final) = Friction + (- mgh) = 0= KE (initial) + PE (initial)?
I am confused on how to use the Conservation of Energy equation to show all the relationship among the numbers in the mud. Any guidance?
Is my reasoning incorrect somewhere?

 

[mfb]

After it goes into the mud and KE is lost due to friction, is energy still conserved at that point?

Overall energy is conserved, but the sum of kinetic and potential energy changes as this sum doesn't include friction in the mud. Ignore energy conservation for (2), it doesn't help there.

 

[Chandra Prayaga]

A 2.00-kg ball has zero kinetic and potential energy. John drops the ball into a 10.0-m-deep well.
1) Just before the ball hits the bottom, the sum of its kinetic and potential energy is: Zero J
2) After the ball comes to a stop in the mud, the sum of its potential and kinetic energy is: -196 J

On #1, why is "sum of kinetic and potential energy" turns to zero? I just guess it is due to conservation of energy, but I don't know how to use Math to prove it.

I know in theory Sum of all work equals the change of KE. I am not sure how to apply here. Before the ball hits the bottom, it should still have velocity and KE >0,

In addition, I really don't understand why the answer in both #1 and #2 are different. I thought the answer to #2 is also zero too due to conservation of energy. Why is the sum equal to a) a negative 196 instead of +196 and b) final potential energy of -196 J?

I am very confused. Please help.

What is the question? What are youn required to find?

 

[CrazyNinja]

I am guessing the problem is with your understanding of the energy equation.The law of conservation of energy states that "The total energy of a system remains a constant till external work is done on it."

Basically, final energy = initial energy + external work.

So for your questions, the answers can now be figured out. In 1) there are no external forces doing work on the ball. Hence total energy remains conserved = 0.
In 2) the normal reaction by the mud does work on the ball-ground system (just take it to be on the ball). This work appears in the RHS of the above equation. From the work energy theorem, work done = change in KE=196 J.

Note that by energy I mean kinetic +potential energy and by external work I mean work done by any force not part of system.

 

[Chandra Prayaga]

At the top of the well, kinetic energy Ktop = 0J, potential energy Utop = 0. This is given. So total mechanical energy Etop = 0
1. Just before the ball hits the bottom, the total mechanical energy must still be the same, by conservation of energy. Ebot = 0. Calculate the potential energy Ubot at the bottom. It will come out to be - 196 J. The means the kinetic energy must be + 196 J.
2. Once it comes to a stop in the mud, the potential energy is still the same, since the height has not changed, so it is still - 196 J. The kinetic energy is now zero since the ball has come to rest. So the total energy is - 196 J.
Mechanical energy is conserved only for an isolated system. As long as the ball is still in flight, in steps 1 and 2, there is no force other than gravity, acting on the ball. So the mechanical energy in step 2 is still zero. Once the ball hits the bottom, there is an extra force exerted by the mud. So the mechanical energy is not conserved.

 

[Chandra Prayaga]

At the top of the well, kinetic energy Ktop = 0J, potential energy Utop = 0. This is given. So total mechanical energy Etop = 0
1. Just before the ball hits the bottom, the total mechanical energy must still be the same, by conservation of energy. Ebot = 0. Calculate the potential energy Ubot at the bottom. It will come out to be - 196 J. The means the kinetic energy must be + 196 J.
2. Once it comes to a stop in the mud, the potential energy is still the same, since the height has not changed, so it is still - 196 J. The kinetic energy is now zero since the ball has come to rest. So the total energy is - 196 J.
Mechanical energy is conserved only for an isolated system. As long as the ball is still in flight, in steps 1 and 2, there is no force other than gravity, acting on the ball. So the mechanical energy in step 2 is still zero. Once the ball hits the bottom, there is an extra force exerted by the mud. So the mechanical energy is not conserved.

Sorry about the sentence: So the mechanical energy in step 2 is still zero". It should have been mechanical energy before step 2, that is, before the ball hits the mud.

 

[CrazyNinja]

At the top of the well, kinetic energy Ktop = 0J, potential energy Utop = 0. This is given. So total mechanical energy Etop = 0
1. Just before the ball hits the bottom, the total mechanical energy must still be the same, by conservation of energy. Ebot = 0. Calculate the potential energy Ubot at the bottom. It will come out to be - 196 J. The means the kinetic energy must be + 196 J.
2. Once it comes to a stop in the mud, the potential energy is still the same, since the height has not changed, so it is still - 196 J. The kinetic energy is now zero since the ball has come to rest. So the total energy is - 196 J.
Mechanical energy is conserved only for an isolated system. As long as the ball is still in flight, in steps 1 and 2, there is no force other than gravity, acting on the ball. So the mechanical energy in step 2 is still zero. Once the ball hits the bottom, there is an extra force exerted by the mud. So the mechanical energy is not conserved.

Same thing I said. Well put though.

 

[Chandra Prayaga]

Same thing I said. Well put though.

Thanks. Looks like our responses arrived at nearly the same time.

 

[bigsmile]

0=0? You can calculate the initial sum of 0+0=0, and you know the sum is conserved.
Sure, its kinetic energy is positive. Its potential energy is negative. The sum is zero.

For (b), just calculate the potential energy of a ball 10 meter below zero.
The kinetic energy is lost due to friction in the mud.

Thank you.