# 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 1 m/s collides with and compresses the spring until it stops moving. How much will the spring compress?

Mar 15, 2016

Compression of the spring $= 0.82 m$ rounded to second place of decimal.

#### Explanation:

Suppose the spring be compressed by $X$ meter.

Force which compressed the string is the force applied by the moving body which came to rest after acting on the spring.
The speed of the object changed from $1 m {s}^{-} 1$ to zero.
Using the expression from linear motion
${v}^{2} - {u}^{2} = 2 a s$,
where $v$ is final velocity, $u$ initial velocity, $a$ acceleration and $s$ distance moved.
Inserting the given values we obtain

${0}^{2} - {1}^{2} = 2 a X$,

recall that the compressing objects moves distance $X$ before it came to stop. This is the amount of compression of the spring produced by it. Solving for acceleration we obtain
$a = - \frac{1}{2 X}$, $- v e$ sign indicates that the acceleration is in a direction opposite to direction of movement of object or it is retardation.

We know that Force $F = m a$
$\therefore F = 8 \times \left(- \frac{1}{2 X}\right) = - \frac{4}{X}$

From Hooke's law for springs we know that
Reaction Force $F = - k X$, where $k$ is the spring constant.
Now inserting the various values in the equation we obtain
$- \frac{4}{X} = - 6 X$, solving for $X$
${X}^{2} = \frac{4}{6}$
or $X = \sqrt{\frac{4}{6}} = 0.82 m$ rounded to second place of decimal.