A spring with a constant of $1 (kg)/s^2$ is lying on the ground with one end attached to a wall. An object with a mass of $8 kg$ and speed of $9 m/s$ collides with and compresses the spring until it stops moving. How much will the spring compress?

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Answer 1 · 2016-05-03T09:44:10

$25.5m$

Conservation of energy tells us that there must be exactly the same amount of energy before a reaction as after.

Before the reaction energy is kinetic, since the object is moving, and energy is given by

$E=1/2mv^2$ ,

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

After the reaction, the energy is potential, stored in the spring, and given by

$E=1/2kx^2$ ,

where $x$ is the distance the spring compresses and $k$ is the spring constant.

Since these two equations must be equal (conservation of energy above),

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

$therefore mv^2=kx^2$

The question is asking you about how much the spring compresses, so you rearrange to make $x$ the subject.

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

Put in the values you know and solve

$sqrt((8*9^2)/1)=x$

$sqrt(648)=x=25.5m$

A spring with a constant of 1 (kg)/s^21 (kg)/s^2 is lying on the ground with one end attached to a wall. An object with a mass of 8 kg8 kg and speed of 9 m/s9 m/s collides with and compresses the spring until it stops moving. How much will the spring compress?