# A spring with a constant of 6/7 (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 3/4 m/s collides with and compresses the spring until it stops moving. How much will the spring compress?

Oct 18, 2017

$\sqrt{\frac{189}{480}} m$

#### Explanation:

Kinetic energy of object = (mv^2)/2 = (3/5 kg × (3/4 ms^-1)^2) / 2 = 27/160 J

Potential energy stored in spring = (kx^2)/2 = (6/7 kgs^-2 × x^2) / (2) = (3x^2)/7 kg.s^-2

According to law of Conservation of Energy
Total energy of object = Total energy absorbed by spring

$\frac{27}{160} J = \frac{3 {x}^{2}}{7} k g . {s}^{-} 2$

x = sqrt((7 × 27/160) / 3) m = sqrt(189/480) m

Oct 18, 2017

The compression is $= 0.63 m$

#### Explanation:

The spring constant is $k = \frac{6}{7} k g {s}^{-} 2$

The kinetic energy of the object is

$K E = \frac{1}{2} m {u}^{2}$

The mass is $m = \frac{3}{5} k g$

The speed is $u = \frac{3}{4} m {s}^{-} 1$

$K E = \frac{1}{2} \cdot \frac{3}{5} \cdot {\left(\frac{3}{4}\right)}^{2} = \frac{27}{160} J$

This kinetic energy will be stored in the spring as potential energy.

$P E = \frac{27}{160} J$

The spring constant is $= \frac{6}{7} k g {s}^{-} 2$

So,

$\frac{1}{2} k {x}^{2} = \frac{27}{160}$

${x}^{2} = 2 \cdot \frac{\frac{27}{160}}{\frac{6}{7}} = \frac{189}{480} {m}^{2}$

$x = \sqrt{\frac{189}{480}} = 0.63 m$/