A spring with a constant of $6 (kg)/(s^2)$ is lying on the ground with one end attached to a wall. An object with a mass of $8 kg$ and speed of $5 m/s$ collides with and compresses the spring until it stops moving. How much will the spring compress?

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Answer 1 · 2017-03-05T11:12:46

$5.77m$

Assuming no energy is lost, the energy stored by the spring will equal the kinetic energy beforehand.

Potential energy in a spring is

$E = 1/2kx^2$

where $k$ is the spring constant and $x$ is the extension.

Kinetic energy is given by

$E = 1/2mv^2$

where $m$ is mass and $v$ is velocity.

If all the energy is conserved between the two,

$1/2kx^2 = 1/2mv^2$

The question asks for the extension, $x$ , so we rearrange to find $x$ :

$1/2kx^2 = 1/2mv^2$

$kx^2 = mv^2$

$x^2 = (mv^2)/k$

$x = sqrt((mv^2)/k)$

We know from the question that $k = 6, m = 8$ and $v = 5$ , so

$x = sqrt((8xx5^2)/6) = sqrt(33.3) = 5.77m$

A spring with a constant of 6 (kg)/(s^2)6 (kg)/(s^2) is lying on the ground with one end attached to a wall. An object with a mass of 8 kg 8 kg and speed of 5 m/s 5 m/s collides with and compresses the spring until it stops moving. How much will the spring compress?