Calculating Total Energy of Object

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SUMMARY

The total energy of an object with a mass of 5 kg, at a height of 3 m with a horizontal velocity of 2 m/s, is calculated by summing its potential energy (PE) and kinetic energy (KE). The potential energy is determined using the formula PE = mgΔh, resulting in 147.15 J. The kinetic energy is calculated using KE = 0.5mv², yielding 10 J. Therefore, the total energy is 157.15 J, confirming that the calculations are correct and complete.

PREREQUISITES
  • Understanding of gravitational potential energy (PE = mgΔh)
  • Knowledge of kinetic energy formula (KE = 0.5mv²)
  • Basic principles of physics related to energy conservation
  • Familiarity with units of measurement (Joules, kg, m/s)
NEXT STEPS
  • Study the conservation of mechanical energy in physics
  • Learn about the effects of air resistance on energy calculations
  • Explore advanced topics in energy transformations
  • Investigate real-world applications of energy calculations in engineering
USEFUL FOR

Students studying physics, educators teaching energy concepts, and anyone interested in understanding the principles of energy calculations in mechanics.

Sixty3
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Homework Statement


If a question asks to calculate total energy of an object that does have a horizontal velocity (constant, air resistance ignored), and it is at it's highest point so it's vertical velocity is zero.

I need to know how you'd calculate the total energy of this object.

Let's take mass as 5kg. And a highest point of 3m. Horizontal velocity of 2m/s. (I just made them up).


Homework Equations


Potential energy = mgΔh
Kinetic energy = 0.5mv^2


The Attempt at a Solution


Pe= 5*9.81*3 = 147.15J
Ke= 5*2^2= 20/2 = 10J

10+147.15= 157.15J

I was just wondering if I am not leaving anything out, or if I'm even doing it correct! Thanks
 
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Hi Sixty3 :smile:

(try using the X2 tag just above the Reply box :wink:)
Sixty3 said:
Let's take mass as 5kg. And a highest point of 3m. Horizontal velocity of 2m/s. (I just made them up).

Pe= 5*9.81*3 = 147.15J
Ke= 5*2^2= 20/2 = 10J

10+147.15= 157.15J

I was just wondering if I am not leaving anything out, or if I'm even doing it correct! Thanks

Yes, that's fine …

at launch, and when on return to the ground, its KE is 157.15J.

(btw, we normally put a capital 'E' in KE and PE :wink:)
 

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