Simple question on Ratio of Kinetic to rest energy

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SUMMARY

The discussion focuses on calculating the ratio of kinetic energy to rest energy for a baseball thrown at 44 m/s. The relevant equations include rest energy (E=mc²) and kinetic energy (K=1/2mv²). Given that the velocity is significantly less than the speed of light, using K=1/2mv² is appropriate. The conclusion emphasizes that the rest energy is substantially larger than the kinetic energy in this scenario.

PREREQUISITES
  • Understanding of kinetic energy formula (K=1/2mv²)
  • Knowledge of rest energy formula (E=mc²)
  • Familiarity with the concept of relativistic effects (gamma factor)
  • Basic physics principles related to energy and motion
NEXT STEPS
  • Calculate the kinetic energy of the baseball using K=1/2mv²
  • Determine the rest energy using E=mc² for the baseball's mass
  • Compute the ratio of kinetic energy to rest energy
  • Explore the implications of relativistic effects at high velocities
USEFUL FOR

This discussion is beneficial for physics students, educators, and anyone interested in understanding the relationship between kinetic and rest energy in classical mechanics.

MillerGenuine
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Homework Statement



A pitcher can throw a baseball at 44 m/s. what is the ratio of kinetic energy to the rest energy E=mc^2. (can you use k=1/2mv^2 ?)

Homework Equations



rest energy=mc^2
K=gamma*mc^2 - mc^2
k=1/2mv^2

The Attempt at a Solution


I am just not sure how to go about this problem. It seems i can use k=1/2mv^2 because the velocity is well below c. From there i would take me kinetic energy and divide it by the rest enregy to get the ratio..??
 
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MillerGenuine said:

Homework Statement



A pitcher can throw a baseball at 44 m/s. what is the ratio of kinetic energy to the rest energy E=mc^2. (can you use k=1/2mv^2 ?)

Homework Equations



rest energy=mc^2
K=gamma*mc^2 - mc^2
k=1/2mv^2

The Attempt at a Solution


I am just not sure how to go about this problem. It seems i can use k=1/2mv^2 because the velocity is well below c. From there i would take me kinetic energy and divide it by the rest enregy to get the ratio..??

Yep. They just want to show how huge the rest energy is...
 

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