A mass of 2.33 kilograms is placed on a horizontal frictionlesssurface against an uncompressed spring...

A mass of 2.33 kilograms is placed on a horizontal frictionless surface against an uncompressed spring with spring constant 1,022.7 N/m. The mass is pushed against the spring until the spring is compressed a distance 0.35 m and then released. How high (vertically) in m does the mass rise from the original height before it stops (momentarily).

Expert Solution

First use conservation of energy to find the speed just after the block separates from the spring

1/2kx² = 1/2mv²

v = sqrt ( kx² / m )

v = sqrt ( 1022.7 * 0.35² / 2.33)

v = 7.332 m/s

now, again use conservation of energy.

at the position where block stops (momentarily) , its K.E is zero. At the bottom , its P.E is zero,

so,

1/2mv² = mgh

1/2v² = gh

h = v² / 2g

h = 7.332² / 2 * 9.8

h = 2.743 m

genius_generous answered 2 years ago

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