Asked by Paige
You are trying to package a 50g egg so that it won't break after you drop it from a height of 10 meters. You decide to use wood as a nice soft packaging to protect the egg. When the egg hits the ground, the wood "softens" the landing so that it takes 5 milliseconds for the egg to come to rest. What is the average force on the egg when it lands?
Answers
Answered by
Damon
F delta t = change in momentum
F = .050 kg * v /5*10^-3 seconds
we need v
use conservation of energy
(1/2) m v^2 = m g h
v = sqrt (2gh) =sqrt (196) = 14 m/s
so
F = .05*14/5*10^-3 = 50*14/5 = 140 Newtons
F = .050 kg * v /5*10^-3 seconds
we need v
use conservation of energy
(1/2) m v^2 = m g h
v = sqrt (2gh) =sqrt (196) = 14 m/s
so
F = .05*14/5*10^-3 = 50*14/5 = 140 Newtons
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