A spring with a constant of $\frac{2}{5} \frac{k g}{s^{2}}$ $2/5 (kg)/s^2$ is lying on the ground with one end attached to a wall. An object with a mass of $1 k g$ $1 kg$ and speed of $2 \frac{m}{s}$ $2 m/s$ collides with and compresses the spring until it stops moving. How much will the spring compress?

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A spring with a constant of $\frac{2}{5} \frac{k g}{s^{2}}$ $2/5 (kg)/s^2$ is lying on the ground with one end attached to a wall. An object with a mass of $1 k g$ $1 kg$ and speed of $2 \frac{m}{s}$ $2 m/s$ collides with and compresses the spring until it stops moving. How much will the spring compress?

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Answer 1 · 2016-05-16T13:32:36

$x = 2 \sqrt{\frac{5}{2}}$ $x = 2sqrt(5/2)$ [m]

Explanation:

The moving object has a kinetic energy given by $E_{c} = \frac{1}{2} m v^{2}$ $E_c =1/2 m v^2$
All this energy is absorbed by the spring as potential energy as $E_{p} = \frac{1}{2} K x^{2}$ $E_p = 1/2K x^2$ . Equating the energies
$\frac{1}{2} m v^{2} = \frac{1}{2} K x^{2}$ $1/2 m v^2 = 1/2K x^2$
and solving regarding $x$ $x$ gives
$x = v \sqrt{\frac{m}{K}} = 2 \sqrt{\frac{5}{2}}$ $x = v sqrt(m/K) = 2sqrt(5/2)$ [m]

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