Question

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?

Answer 1

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:

E_kinetic=k$\times$ l

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

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

maximmum elongation is 0.125 m.