How much energy is dissipated in the fourth bounce?

In summary, a 0.70-kg basketball dropped from a height of 1.4 m on a hardwood floor bounces back up to 66% of its original height. The energy dissipated in the first bounce is 3.3J, and in the fourth bounce it is 7.8J.
  • #1
emily081715
208
4

Homework Statement


A 0.70-kg basketball dropped on a hardwood floor rises back up to 66 % of its original height.If the basketball is dropped from a height of 1.4 m , how much energy is dissipated in the first bounce?( answer =3.3J).
How much energy is dissipated in the fourth bounce?

Homework Equations


U=mgy

The Attempt at a Solution


i know that at the top of the first bounce the kinetic energy is zero and the potential energy is 9.604J. After the first bounce there's 6.304 of the potential energy left, meaning 3.3 J was dissipated. I tried using the same approach as in part one, assuming its losing 66% of the height each bounce since i am not told otherwise.
i got, a height of 0.2656m after the fourth bounce. plugging that into the formula, U=mgy i got
U=(0.7kg)(9.8m/s2)(0.2656m)
=1.8J
then i subtracted that from the original potential energy
9.604-1.8= 7.8J
i got the answer of 7.8J of energy was dissipated. this is incorrect and i am unsure as to why
 
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  • #2
emily081715 said:

Homework Statement


A 0.70-kg basketball dropped on a hardwood floor rises back up to 66 % of its original height.If the basketball is dropped from a height of 1.4 m , how much energy is dissipated in the first bounce?( answer =3.3J).
How much energy is dissipated in the fourth bounce?

Homework Equations


U=mgy

The Attempt at a Solution


i know that at the top of the first bounce the kinetic energy is zero and the potential energy is 9.604J. After the first bounce there's 6.304 of the potential energy left, meaning 3.3 J was dissipated. I tried using the same approach as in part one, assuming its losing 66% of the height each bounce since i am not told otherwise.
i got, a height of 0.2656m after the fourth bounce. plugging that into the formula, U=mgy i got
U=(0.7kg)(9.8m/s2)(0.2656m)
=1.8J
then i subtracted that from the original potential energy
9.604-1.8= 7.8J
i got the answer of 7.8J of energy was dissipated. this is incorrect and i am unsure as to why
i caught my mistake it was looking just from the third bounce potential energy to the fourth bounce, not the system as a whole
 

Related to How much energy is dissipated in the fourth bounce?

1. How is energy dissipated during a bounce?

Energy is dissipated during a bounce due to the transfer of kinetic energy into other forms, such as heat and sound. When an object bounces, the impact with the surface causes some of its kinetic energy to be converted into heat, which dissipates into the surrounding environment.

2. Does the height of the bounce affect the amount of energy dissipated?

Yes, the height of the bounce does affect the amount of energy dissipated. The higher the bounce, the more potential energy is converted into kinetic energy, resulting in a greater impact and more energy being dissipated.

3. How does the material of the bouncing object affect energy dissipation?

The material of the bouncing object can have a significant impact on energy dissipation. Softer materials, such as rubber, are more elastic and can absorb more of the impact energy, resulting in less energy being dissipated. Harder materials, like metal, are less elastic and therefore will dissipate more energy.

4. Is energy dissipated equally in each bounce?

No, energy is not dissipated equally in each bounce. As the bouncing object loses energy with each bounce, the height of the bounce decreases, resulting in less potential energy being converted into kinetic energy. This means that the amount of energy dissipated will decrease with each subsequent bounce.

5. Can the amount of energy dissipated be calculated?

Yes, the amount of energy dissipated can be calculated using the formula for elastic potential energy, which takes into account the height of the bounce, the mass of the object, and the material properties. However, there may be other factors at play, such as air resistance or friction, which can make the calculation more complex.

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