Average Net Force: Kinetic Energy & Momentum

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SUMMARY

The average net force can be calculated using two distinct methods: one based on kinetic energy and the other on momentum. To find the average force over distance, divide the change in kinetic energy by the distance traveled, represented by the formula Work = ΔE = F_average * d. Conversely, to calculate the average force over time, use the change in momentum divided by the total time, expressed as impulse = ΔP = F_average * T. Understanding these two approaches clarifies the concept of average force in physics.

PREREQUISITES
  • Understanding of kinetic energy and its formula
  • Familiarity with momentum and impulse concepts
  • Basic knowledge of physics equations related to force
  • Ability to manipulate algebraic expressions
NEXT STEPS
  • Study the relationship between work, energy, and force in physics
  • Learn about the principles of momentum and its conservation
  • Explore advanced topics in dynamics, including impulse-momentum theorem
  • Investigate real-world applications of average force calculations in engineering
USEFUL FOR

Physics students, educators, and professionals in engineering or mechanics who seek to deepen their understanding of force calculations in relation to kinetic energy and momentum.

Maiia
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This is just a general question- Given a beginning and ending velocity, and distance that the object travels, would the average net force be the change in kinetic energy divided by the distance? (Using energy methods only)
I think I've also seen avg force calculated with momentum...
 
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There are (at least) two sorts of average you can take:

If you want to calculate a force averaged over *distance*, you can use the change in kinetic energy, divided by distance: Work =\delta E=F_average * d

If you want to calculate an average over *time*, then that would be given by the change in momentum, divided by the total time: impulse=\delta P= F_average * T
 
so Kef-Kei/d right?
 
Yeah.
 
ah i see- "average" is really misleading b/c u don't add the two numbers..
thanks!
 

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