Conservation of energy of ball with drag

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SUMMARY

The discussion focuses on calculating the energy dissipated by air drag on a .63 kg ball thrown upwards at an initial speed of 14 m/s, reaching a maximum height of 8.1 m. The kinetic energy (KE) is calculated using the formula KE = 1/2 MV², resulting in 61.74 J. The energy dissipated due to air drag is determined to be -12 J, which is found by comparing the change in kinetic energy with the change in potential energy (mgh). This highlights the impact of air resistance on the ball's ascent.

PREREQUISITES
  • Understanding of kinetic energy and potential energy concepts
  • Familiarity with the equation KE = 1/2 MV²
  • Knowledge of gravitational potential energy (mgh)
  • Basic principles of physics related to motion and forces
NEXT STEPS
  • Study the effects of air resistance on projectile motion
  • Learn about energy conservation principles in physics
  • Explore advanced topics in fluid dynamics related to drag forces
  • Investigate real-world applications of energy dissipation in sports physics
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Students studying physics, educators teaching mechanics, and anyone interested in understanding the effects of air resistance on moving objects.

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Homework Statement


A .63 kg ball is thrown up with an initial speed of 14 m/s and reaches a maximum height of 8.1m. How much energy is dissipated by the air drag acting on the ball during the ascent?


Homework Equations


KE= 1/2MV^2


The Attempt at a Solution


I'm not very good at physics. I think I have to find the joules by plugging in the numbers accordingly in the equation, but I don't know what to do from there...

KE = .5(.63)(14)^2 = 61.74

The actual answer is -12 J >.<
 
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Compare the change in kinetic energy with the change in potential energy. Which is mgh. The difference is the loss to friction.
 

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