A 4.20kg block hangs from a spring with spring constant 2300N/m . The block is pulled down 6.80cm from the equilibrium position and given an initial velocity of 1.60m/s back toward equilibrium.
The amplitude is not 6.80 though I thought it was. I am not sure how to calculate it though.
The amplitude is not 6.80 though I thought it was. I am not sure how to calculate it though.
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The amplitude would be 6.8 if the object was simply released.
But it was given extra ENERGY.
the key is to find the total energy at the beginning ( 1/2 k x^2 + 1/2 m v^2)
then work out how far the spring will stretch when ALL this energy is in the spring.
i.e 1/2 k z^2 = 1/2 k x^2 + 1/2 m v^2
where z is the maximum amplitude.
But it was given extra ENERGY.
the key is to find the total energy at the beginning ( 1/2 k x^2 + 1/2 m v^2)
then work out how far the spring will stretch when ALL this energy is in the spring.
i.e 1/2 k z^2 = 1/2 k x^2 + 1/2 m v^2
where z is the maximum amplitude.
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kx^2 +mv^2= 2300*0.068^2+4.2*1.60^2 = 21.3872
kz^2= 21.372 so z = sqrt( 21.372/2300) = 0.964 m Just as I said it would.
The arithmetic you used is faulty.
kz^2= 21.372 so z = sqrt( 21.372/2300) = 0.964 m Just as I said it would.
The arithmetic you used is faulty.
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Note you are not using SI which is part of the problem. k = 2300N/m , x= 0.068m, m = 4.2 and v = 1.6
With these values the formula I showed you works perfectly.
With these values the formula I showed you works perfectly.
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