Gravitational Potential Energy of a jump

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SUMMARY

The discussion centers on calculating the gravitational potential energy (Ug) of a high jumper with a mass of 60.0 kg who consumes 3,000 kcal before a jump. The energy conversion efficiency is noted as 3.3%, leading to a potential energy of 414,414 J. Using the formula Ug = mgh with gravity set at 9.8 m/s², the calculated jump height is 705 m, which is slightly higher than the book's answer of 700 m. The discrepancy is attributed to rounding errors in the calculations.

PREREQUISITES
  • Understanding of gravitational potential energy (Ug)
  • Familiarity with energy conversion efficiency
  • Basic knowledge of physics equations, specifically Ug = mgh
  • Ability to perform unit conversions (kcal to Joules)
NEXT STEPS
  • Review the principles of energy conversion efficiency in physical activities
  • Learn about the impact of mass and gravity on potential energy calculations
  • Explore rounding techniques and their effects on scientific calculations
  • Investigate the relationship between dietary energy intake and athletic performance
USEFUL FOR

This discussion is beneficial for physics students, sports scientists, and nutritionists interested in the relationship between energy intake and athletic performance, particularly in high jump events.

liz_p88
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Homework Statement



A high jumper of mass 60.0 kg consumes a meal of 3.00 x 10^3 kcal prior to a jump.
If 3.3% of the energy from the food could be converted to gravitational potential energy in a single jump, how high could the athlete jump?

Homework Equations



Ug = mgh

gravity = 9.8 m/s/s

The Attempt at a Solution



3000 kcal = (4186 J)(3000) = 12,558,000

3.3% of 12,558,000 = 414,414

414,414 J (kg/m^2/s/s) = (60kg)(9.8 m/s/s)(h)

705 m = h

My book says the answer should be 700 m. I got pretty close but I don't know where I might have made a mistake
 
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I get a similar answer to you using Wolfram|Alpha for the calculations. Most likely the discrepancy with the book is due to rounding along some point.
 
That's what I figured. Thank you for double checking for me.
 

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