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.
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?