How much energy is dissipated in the fourth bounce?

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SUMMARY

The discussion centers on calculating the energy dissipated by a 0.70-kg basketball dropped from a height of 1.4 m, which rises to 66% of its original height after each bounce. The energy dissipated in the first bounce is confirmed to be 3.3 J, calculated using the potential energy formula U=mgy, yielding an initial potential energy of 9.604 J. The incorrect calculation for the fourth bounce resulted in 7.8 J of energy dissipated, as the user mistakenly considered only the potential energy change between the third and fourth bounces rather than the total energy lost throughout the bounces.

PREREQUISITES
  • Understanding of gravitational potential energy (U=mgy)
  • Basic knowledge of energy conservation principles
  • Familiarity with the concept of energy dissipation in elastic collisions
  • Ability to perform calculations involving mass, height, and gravitational acceleration
NEXT STEPS
  • Review the principles of energy conservation in bouncing objects
  • Learn about the effects of energy loss in multiple bounces
  • Explore advanced topics in elastic and inelastic collisions
  • Study the mathematical modeling of energy dissipation in real-world scenarios
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Physics students, educators, and anyone interested in understanding energy dynamics in mechanical systems, particularly in the context of sports physics and material interactions.

emily081715
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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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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
 

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