A block of mass 0.256 kg is placed on top of a light, vertical spring of force constant 4 850 N/m and pushed downward so that the spring is compressed by 0.097 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.)

9.09m

Explanation:

vertical spring of force constant K =4 850 N/m

✓the mass under gravitational potential energy since it's on the top of the light

✓vertical spring of force constant 4 850 N/m is given which is under elastic potential energy

We can equate the elastic potential energy and the gravitational potential energy to have

mgh= ½·k·x²

From here we can make "h" the subject of the formula which is our maximum height above the point of release

h= (½·k·x²)/mg

h= 1/2[ ·k·x²)/mg]

Where k= vertical spring of force constant K =4 850 N/m

m= mass of block= 0.256 kg

g=acceleration due to gravity

x= spring compression= 0.097 m.

Substitute this values we have

h= [0.5× 4850 ×(0.097)^2]/(9.81×0.256)

h=9.0854m

maximum height above the point of release it rise is 9.09m

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