physics

have a question involving the spring costant:

A 2-kg block is attached to a horizontal ideal spring with a spring constant of 200N/m. When the spring has its equilibrium length the block is given a speed of 5 m/s. What is the maximum elongation of the spring? The answer is supposed to be in m.
I used 1/2mv^2=1/2KX^2
When I solved I got .25m, this time. This is not even close to what the book gives.

x^2= mv^2/K

I made an error before, forgot m.

x= v sqrt (m/k)

I get .5m

How does the book get an answer of .05m? I'm confused

x=v sqrt (m/K) = 5* sqrt (2/200)
=5 sqrt (1/100)= .5m

Could it be the book is wrong? Heaven forbid. Trust the force, as Obiwan Kenobi stated.

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asked by Diane

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