Calculating Kinetic and Potential Energy of a Moving Object

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SUMMARY

The discussion focuses on calculating the kinetic and potential energy of a ball kicked by Drew, with a mass of 0.250 kg, from a height of 1.50 m and an initial speed of 17.0 m/s. The initial kinetic energy is calculated as 36.125 J using the formula KE = 1/2 mv², while the initial potential energy is determined to be 3.679 J using PE = mgh. The participants clarify that at the top of the ballistic arc, the vertical component of velocity is zero, but the horizontal component remains, allowing the ball to continue moving forward. The total energy conservation principle is emphasized, with kinetic energy converting to potential energy at the peak height of 9.50 m.

PREREQUISITES
  • Understanding of kinetic energy (KE = 1/2 mv²)
  • Knowledge of potential energy (PE = mgh)
  • Familiarity with the conservation of energy principle
  • Basic concepts of projectile motion
NEXT STEPS
  • Study the conservation of mechanical energy in projectile motion
  • Learn about the components of velocity in two-dimensional motion
  • Explore the effects of gravitational potential energy on moving objects
  • Investigate the equations of motion for projectiles
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and energy conservation, as well as educators looking for practical examples of kinetic and potential energy calculations in projectile 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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