Graph of Kinetic Energy v. Displacement of a Falling Object

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SUMMARY

The discussion focuses on graphing the kinetic and potential energy of an object in free fall, specifically one that starts with a horizontal speed of 30 m/s from a height of 80 m. The potential energy is represented as a linear graph due to the constants in the equation U=mgh, while the kinetic energy, derived from the equation k=1/2mv^2, is confirmed to be directly proportional to displacement. The relationship between velocity squared and displacement is established using the kinematic equation v^2=v0^2+2a(x-x0), leading to the conclusion that kinetic energy varies quadratically with displacement.

PREREQUISITES
  • Understanding of gravitational potential energy (U=mgh)
  • Knowledge of kinetic energy formula (k=1/2mv^2)
  • Familiarity with kinematic equations
  • Basic concepts of energy conservation in physics
NEXT STEPS
  • Study the derivation of the kinematic equation v^2=v0^2+2a(x-x0)
  • Explore energy conservation principles in free-fall scenarios
  • Learn about graphing techniques for energy versus displacement
  • Investigate the effects of initial velocity on kinetic and potential energy graphs
USEFUL FOR

Students studying physics, educators teaching energy concepts, and anyone interested in the dynamics of falling objects and energy transformations.

brendan3eb
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[solved] Graph of Kinetic Energy v. Displacement of a Falling Object

Homework Statement


Draw and label a graph with energy for the y-axis and displacement for the x-axis of the kinetic energy and potential energy of an object in free-fall that started with a horizontal speed of 30 m/s from a height of 80 m.


Homework Equations


U=mgh
k=1/2mv^2


The Attempt at a Solution


I drew a straight line with negative slope for potential energy because m and g are constants in U=mgh, so it should be a linear graph, as for kinetic energy, I am sitting on the fence on whether or not to draw a straight line or a curve The v^2 makes me want to draw a curve, however, if the the potential energy is a straight line and K = E - U, then shouldn't k be directly proportional to displacement as well?
 
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brendan3eb said:
The v^2 makes me want to draw a curve, however, if the the potential energy is a straight line and K = E - U, then shouldn't k be directly proportional to displacement as well?
Good thinking!

Why not use a bit of kinematics to double check? What's the relationship between v (or v^2) and displacement for a falling object?
 
ahh..

v^2=v0^2+2a(x-x0)
but since initial velocity in the y direction is 0
v^2=-2g(x-x0)

and since -2g is a constant, v^2 is directly proportional to displacement, and thus kinetic energy is directly proportional to displacement!

Thanks Doc Al!
 

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