Calculating Kinetic Energy and Height Change

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SUMMARY

The discussion focuses on calculating kinetic energy and height change for a 2.0 kg mass fired vertically with an initial velocity of 60.0 m/s. For part A, the kinetic energy at 20.0 m above the ground is derived using the formula Ek = mv²/2, leading to a calculation of 3600 J at launch, but the correct kinetic energy at 20 m requires considering both kinetic and potential energy. Part B involves determining the height change when the speed decreases from 50 m/s to 40 m/s, which can be solved using kinematic equations.

PREREQUISITES
  • Understanding of kinetic energy formula (Ek = mv²/2)
  • Knowledge of potential energy concepts
  • Familiarity with kinematic equations
  • Basic physics principles regarding energy conservation
NEXT STEPS
  • Learn how to calculate potential energy at a specific height
  • Study kinematic equations for vertical motion
  • Explore energy conservation principles in physics
  • Practice problems involving changes in kinetic and potential energy
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Students studying physics, particularly those focusing on mechanics, as well as educators looking for examples of energy calculations in vertical motion scenarios.

soulja101
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Homework Statement


A 2.0 kg mass is fired straight up with a intial velocity of 60.0m/s
A) wat is the kinetic energy when its 20.0m above the ground
B)wat is the changhein height when its speed changes from 50m/s to 40m/s


Homework Equations


Ek=mv2/s



The Attempt at a Solution


A)EK=mv2/2 B)i didnt get question B
=2.0kg*60.0ms
=3600J
 
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Well, for (a) you have calculated the kinetic energy of the mass as it leaves the ground. This is not the answer to the question, but it will help. Do you know how to calculate potential energy at a height h above the ground? Can you say anything about the sum of the potential and kinetic energies of the mass at anyone point throughout its flight?

edit: of course there are many ways to do the problem, as Kurdt points out below! Pick which one you prefer.
 
Last edited:
This is more to do with the kinematic equations than anything else. What will the speed be at 20m above ground level and thus what will the kinetic energy be. Then part b is obtained directly from one of the kinematic equations.
 

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