Kinetic energy when a object is dropped from certain distance

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SUMMARY

The discussion focuses on calculating the kinetic energy and velocity of a 0.06 kg egg dropped from a height of 0.5 meters just before it impacts a frying pan. The gravitational potential energy (GPE) is equated to kinetic energy (KE) using the formula GPE = KE, leading to the equation mgh = 1/2 mv². By substituting the values, the kinetic energy can be determined as 0.294 Joules, and the velocity just before impact is calculated to be approximately 3.13 m/s.

PREREQUISITES
  • Understanding of gravitational potential energy (GPE) and kinetic energy (KE)
  • Basic knowledge of Newton's laws of motion
  • Familiarity with the formula for acceleration due to gravity (9.8 m/s²)
  • Ability to manipulate algebraic equations
NEXT STEPS
  • Study the principles of energy conservation in physics
  • Learn how to apply kinematic equations for freely falling objects
  • Explore the concept of work done by gravitational forces
  • Investigate real-world applications of kinetic energy calculations
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Students studying physics, educators teaching mechanics, and anyone interested in understanding the principles of energy and motion in real-world scenarios.

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Drop a 0.06kg egg from 0.5m above the frying pan.

what is its kinetic energy when it hits the frying pan?

what is its velocity the instant before it hits the pan?
 
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Hints:
Considering energy: How much work does gravity do on the object as it falls?
Considering kinematics: What's the acceleration of any freely falling body?
 
I get the hint, 9.8 m/s2. But, how do I solve? Do i go:
0.06x0.5 = .5(0.06) x Vsquared because GPE = KE, and then just cancel?
 
Almost. GPE = KE --> mgh = 1/2mv^2. (If you included units in your work, you'd realize that something was missing.)
 

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