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

##### 2 Answers
Mar 10, 2018

Spring will compress by $\text{2.19 m}$

#### Explanation:

Kinetic energy of object = Energy stored in spring

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

From above equation

$x = v \sqrt{\frac{m}{k}} = \text{2 m/s" × sqrt("6 kg"/("5 kg/s"^2)) = "2.19 m}$

Mar 10, 2018

The compression is $= 2.19 m$

#### Explanation:

Mass of the object is $m = 6 k g$

Speed of the object is $v = 2 m {s}^{-} 1$

The kinetic energy of the object is

$K E = \frac{1}{2} m {v}^{2} = \frac{1}{2} \cdot 6 \cdot {\left(2\right)}^{2} = 12 J$

This energy will be stored in the spring

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

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

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

Therefore,

$12 = \frac{1}{2} \cdot 5 \cdot {x}^{2}$

${x}^{2} = \frac{24}{5} = 4.8 {m}^{2}$

The compression of the spring is

$x = \sqrt{4.8} = 2.19 m$