Kinetic energy when a object is dropped from certain distance

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When a 0.06 kg egg is dropped from a height of 0.5 m, its gravitational potential energy (GPE) converts to kinetic energy (KE) as it falls. The GPE can be calculated using the formula GPE = mgh, which results in 0.294 J. At the moment before impact, the egg's velocity can be determined using the equation KE = 1/2 mv^2, leading to a velocity of approximately 3.42 m/s. The discussion emphasizes the relationship between GPE and KE, highlighting the importance of including units in calculations. Understanding these principles is crucial for solving problems related to kinetic energy and motion.
Nikola Tesla
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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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