a block of mass 0.260 kg is placed on top of a light, vertical spring of force constant 5 025 n/m and pushed downward so that the spring is compressed by 0.106 m. after the block is released from rest, it travels upward and then leaves the spring. to what maximum height above the point of release does it rise? (round your answer to two decimal places.)

A block of mass 0.260 kg is placed on top of a light, vertical spring of force constant 5 025 n/m and pushed downward. After the block is released, the spring will travel upward to the maximum height of 10.86 meters.

The relation between the potential energy on a string and the length it stretches is given by:

E = 1/2 . kx²

Where:

k = spring constant

x = spring stretch or compression

In the given problem:

k = 5,025 N/m

x = 0.106 m

Hence,

E = 1/2 . 5025 . (0.106)²

m . g . h = 1/2 . 5025 . (0.106)²

Where;

m = mass

h = height from the release point.

Hence,

0.260 x 10 x h = 28.23

h = 10.86 m

Therefore, the spring will travel upwards up to 10.86 m

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