Graph of Kinetic Energy v. Displacement of a Falling Object

AI Thread Summary
The discussion focuses on creating a graph of kinetic and potential energy versus displacement for a falling object with an initial horizontal speed of 30 m/s from a height of 80 m. The potential energy is represented as a linear graph due to the constant values of mass and gravity. The kinetic energy's relationship to displacement is clarified using kinematic equations, revealing that kinetic energy is directly proportional to displacement. The conclusion is that while potential energy decreases linearly, kinetic energy increases in a manner that reflects its dependence on displacement. This understanding solidifies the graph's representation of energy dynamics in free fall.
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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