# 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 9 kg and speed of 4 m/s collides with and compresses the spring until it stops moving. How much will the spring compress?

Mar 9, 2016

Compression: $x = \setminus \sqrt{\frac{m}{k}} {v}_{0} = \setminus \sqrt{\frac{9 \setminus \quad k g}{6 \setminus \quad \frac{N}{m}}} \left(4 \setminus \quad \frac{m}{s}\right) = 4.899 \setminus \quad m$

#### Explanation:

Mechanical Energy Conservation: When the forces acting on a system are purely conservative, its mechanical energy remains conserved.

$\setminus \Delta E = \setminus \Delta K + \setminus \Delta U = 0$

$\setminus \Delta E$ - Change in total mechanical energy;
$\setminus \Delta K$ - Change in kinetic energy;
$\setminus \Delta U$ - Change in potential energy;

In this case when the kinetic energy of the block decreases, the potential energy in the spring increases by the same amount to keep the total mechanical energy of the system a constant.

$m$ - mass of the block; $\setminus q \quad {v}_{0}$ - speed of the block.
$k$ - spring constant; $\setminus \quad \setminus q \quad x$ - compression of the spring;

Decrease in KE : \Delta K = -1/2mv_0^2;
Increase in PE : $\setminus \Delta U = + \frac{1}{2} k {x}^{2}$

$\setminus \Delta E = 0 \setminus q \quad \implies \setminus \Delta U = - \setminus \Delta K$
1/2 kx^2 = 1/2 mv_0^2; \qquad x = \sqrt{m/k}v_0 = \sqrt{(9\quad kg)/(6\quad N/m)}(4\quad m/s) = 4.899\quad m