Question

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).

Answer 1

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