Answer to Question #148415 in Mechanics | Relativity for linacnkruger

Question #148415

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


1
Expert's answer
2020-12-04T07:32:27-0500

Let`s solve this problem.

we have given:

m=2kg,

k=200N/m,

v=5mps.

we need to find l --> elongation.

In this case, if we spent maximum energy, then we reach maximum elongation. It means:

Ekinetic=k×\times l

m×v22\frac{m \times v^2}{2}=k×\timesl, we can find l (elongation) easily:

l=m×v22×k\frac{m \times v^2}{2\times k} =2×522×200\frac{2\times 5^2}{2\times 200} =0.125 m,

maximmum elongation is 0.125 m.


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