Understanding Work and Kinetic Energy: Examples and Calculations Explained

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SUMMARY

The discussion clarifies the relationship between work and kinetic energy using a practical example involving a 100 kg jogger named James. When James jogs at a constant velocity of 2.2 m/s, the net force acting on him is zero, resulting in no work done (W=Fs=0). Consequently, while his kinetic energy is calculated to be 242J (KE=0.5mv²), there is no change in energy since he is not accelerating. The key takeaway is that work is only done when there is a change in velocity, which requires acceleration.

PREREQUISITES
  • Understanding of Newton's Second Law (F=ma)
  • Basic knowledge of kinetic energy formula (KE=0.5mv²)
  • Concept of work-energy principle
  • Familiarity with constant velocity and acceleration
NEXT STEPS
  • Study the work-energy theorem in classical mechanics
  • Explore examples of work done in varying acceleration scenarios
  • Learn about the implications of friction and air resistance on work and energy
  • Investigate real-world applications of kinetic energy calculations in sports science
USEFUL FOR

Students of physics, educators teaching mechanics, and anyone interested in understanding the principles of work and energy in motion.

Joseph812
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I am confused of how Work basically equals to Kinetic Energy...I will give an example just as way to help you help me get an idea about how it works...the example I thought of is... '' James (100 kg) jogs on a straight road at a constant velocity of 2.2 m/s for 600 seconds, how much energy does Jame consume?''...How I thought of it is...since he has constant velocity..his Force is 0 (F= ma), therefore work done is 0 (W= Fs), but KE=0.5mv^2...so his kinetic energy is 242J.

What did I do wrong?
 
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If James is sitting on a rolling chair without any friction and air resistance, moving at a constant velocity of 2.2 m/s, then there is indeed no work done and the energy difference equals zero. His kinetic energy is positive (242J I assume), but he started already moving and ends moving in your example at a constant speed, so there is no energy gained and no energy lost, because no work is done.

If you let him start at zero speed, you have to calculated the acceleration (a > 0) and the energy balance is a different one.
 
Oh, I get it now...thank you very much :)
 

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