# A spring with a constant of 6/5 (kg)/s^2 is lying on the ground with one end attached to a wall. An object with a mass of 3/5 kg and speed of 5/4 m/s collides with and compresses the spring until it stops moving. How much will the spring compress?

Mar 2, 2018

$\frac{5 \sqrt{2}}{8} \text{m}$

#### Explanation:

Kinetic energy of object = Energy stored in spring

$\frac{1}{2} m {v}^{2} = \frac{1}{2} k {x}^{2}$

$x = v \sqrt{\frac{m}{k}} = \frac{5}{4} \text{m/s"sqrt((3/5"kg")/(6/5 "kg/s"^2)) = 5/(4sqrt(2)) "m" = (5sqrt(2))/8 "m}$

Mar 2, 2018

The compression is $= 0.88 m$

#### Explanation:

Mass of the object is $m = \frac{3}{5} k g$

Speed of the object is $v = \frac{5}{4} m {s}^{-} 1$

The kinetic energy of the object is

$K E = \frac{1}{2} m {v}^{2} = \frac{1}{2} \cdot \frac{3}{5} \cdot {\left(\frac{5}{4}\right)}^{2} = 0.46875 J$

This energy will be stored in the spring

$E = \frac{1}{2} k {x}^{2}$

The spring constant is $k = \frac{6}{5} k g {s}^{-} 2$

Let the compression of the spring be $= x m$

Therefore,

$0.46875 = \frac{1}{2} \cdot \frac{6}{5} \cdot {x}^{2}$

${x}^{2} = \frac{5 \cdot 0.46875}{3} = 0.78 {m}^{2}$

The compression of the spring is

$x = \sqrt{0.78} = 0.88 m$