Calculating Kinetic and Potential Energy of a Moving Object

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Homework Help Overview

The problem involves calculating the kinetic and potential energy of a ball kicked off a porch at a specific height and speed. The scenario includes determining the energies at various points in the ball's trajectory, particularly at the top of its arc and when it reaches the ground.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the calculations for initial kinetic and potential energy, questioning the speed of the ball at the top of the arc and its implications for kinetic energy. There is confusion about whether the vertical component of velocity is zero at the peak and how that affects energy calculations.

Discussion Status

Several participants have provided insights into the energy conservation principles involved, with some suggesting corrections to initial calculations. There is ongoing exploration of the relationship between kinetic and potential energy, particularly regarding the ball's motion at the peak of its trajectory.

Contextual Notes

Participants note that the initial potential energy should not be considered zero, as the ball is kicked from a height of 1.5 m. There are also discussions about the definitions of potential energy and its relevance when an object is in motion.

  • #31
Thanks for correcting me ehild!
 
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  • #32
The solution is almost correct, Janed. But it is convenient to take the potential energy zero at the ground.

janed123 said:
Ok I tried redoing the whole problem...

KE= 36.125 J
PE = 0 J

The ball was kicked from 1.5 m, so the initial potential energy is (0.250) (9.81) (1.50) instead of zero.
Your way of solution is correct otherwise, just correct the total energy according to the initial potential energy.

ehild
 
  • #33
thinkhigh said:
ya if an object is moving means it does not have any P.E energy

It is wrong, Thinkhigh. When you drop a stone, for example, the velocity increases with time t as v=gt, and its height decreases as h=h0-0.5gt2. The potential energy with respect to the ground is PE=mgh, The kinetic energy is KE=0.5mv2. The PE continuously transforms into KE. Substituting v=gt and h=h0-0.5gt2 into the equations for PE and KE, the total energy is

E=0.5mg2t2+mg(h0-0.5g2t2)=mgh0, constant, equal to the initial PE.

ehild
 

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