# 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 1/3 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 12, 2017

The spring is compressed 0.433 m.

#### Explanation:

The object originally possesses kinetic energy. When it has come to a stop, with the spring compressed a distance x, 100% of the object's kinetic energy will have been 100% transferred to the spring. The spring's potential energy will equal the number of Joules that the object had originally in the form of kinetic energy. This is the principle of conservation of energy.

The object's initial kinetic energy, KE_i:

$K {E}_{i} = \frac{1}{2} \cdot m \cdot {v}^{2} = \frac{1}{2} \cdot \frac{1}{3} k g \cdot {\left(\frac{3}{4} \frac{m}{s}\right)}^{2}$

$K {E}_{i} = \frac{1}{6} k g \cdot \frac{9}{16} {\left(\frac{m}{s}\right)}^{2} = 0.0937 k g \cdot {m}^{2} / {s}^{2}$
Note: that combination of units is eqivalent to the Joule.

The spring's final spring potential energy, SPE_f:

$S P {E}_{f} = \frac{1}{2} \cdot k \cdot {x}^{2} = \frac{1}{2} \cdot 1 \frac{k g}{s} ^ 2 \cdot {x}^{2}$

$S P {E}_{f} = \frac{1}{2} \frac{k g}{s} ^ 2 \cdot {x}^{2}$

Remember that the principle of conservation of energy says that $S P {E}_{f} \text{ will be equal to } K {E}_{i}$. Therefore

$\frac{1}{2} \frac{k g}{s} ^ 2 \cdot {x}^{2} = 0.0937 k g \cdot {m}^{2} / {s}^{2}$

Cancelling where possible and solving for ${x}^{2}$,

$\frac{1}{2} \frac{\cancel{k g}}{\cancel{{s}^{2}}} \cdot {x}^{2} = 0.0937 \cancel{k g} \cdot {m}^{2} / \cancel{{s}^{2}}$

${x}^{2} = 2 \cdot 0.0937 {m}^{2}$

$x = \sqrt{2 \cdot 0.0937} \text{ m} = 0.433 m$

I hope this helps,
Steve