Maximum Spring Elongation for a Block on a Horizontal Ideal Spring

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SUMMARY

The discussion centers on calculating the maximum elongation of a horizontal ideal spring attached to a 2-kg block with a spring constant of 200 N/m and an initial speed of 5 m/s. Using the conservation of mechanical energy, the correct formula is applied: 0.5kx² = 0.5mv². The initial calculation yields an elongation of 0.5 meters, but the expected answer is 5 meters, suggesting a potential error in the problem statement regarding the initial velocity, possibly indicating it should be 50 m/s instead.

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Homework Statement



A 2-kg block is attached to a horizonal ideal spring with a spring constant of 200N/m. When the spring has its equilibrium length the block is given a speed of 5m/s. What is the maximum elongation of the spring.

Homework Equations



Conservation of mechanical energy, PEspring=0.5kx^2 KE=0.5mv^2

The Attempt at a Solution



It seems like a fairly simple question to me, applying conservation of mechanical energy at equilibrium and at maximum elongation we get 0 + 0.5kx^2 = 0 + 0.5mv^2, so x =√(mv^2/k) = 0.5 meters, however the solution says the answer should be 5 meters, can anyone tell me what I'm missing here?
 
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Your result looks good. Maybe a typo in the question? Perhaps they meant for the initial velocity to be 50 m/s.
 

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