# a block of mass .250 kg is placed on top of a light, vertical spring of force constant 5000 N/m and pushed dow?

a block of mass .250 kg is placed on top of a light, vertical spring of force constant 5000 N/m and pushed downward so that the spring is compressed by .100 m. at the block is released from rest, it travels upward and then leaves the spring. to what height above the point of release does it rise

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• Anonymous
8 years ago

Use this form to determine the angular velocity...

ω = √(K/m)

So...

x = Acos(√(K/m)t)

Perform the differentiation twice to obtain the acceleration...

v = -A√(K/m)sin(√(K/m)t)

a = -AK/m * cos(√(K/m)t)

If the block leaves the spring, then a = -g, so..

-g = -AK/m * cos(√(K/m)t)

The special trick here to find the height instead of the tedious substitution is to let x = Acos(√(K/m)t), the original equation we have found. That gives us...

g = K/m * Acos(√(K/m)t)

g = Kx/m

x = gm/K

Therefore...

x = 9.81 * .25/5000

≈ 4.905 * 10^(-4) m

Good luck!

Source(s): τ