A spring with a constant of 12 (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 3 m/s collides with and compresses the spring until it stops moving. How much will the spring compress?

Mar 26, 2018

The spring will compress $2.12$ metres.

Explanation:

We can either do this using conservation of energy or kinematics. I'll use conservation of energy.

The object's potential energy will be at $0$ (because it's on the floor). It's kinetic energy will be given by $\frac{1}{2} m {v}^{2}$. Once it hits the spring, some of it's kinetic energy will be transferred to the spring. Therefore our equation will be

$\frac{1}{2} m {v}_{\text{initial"^2 = 1/2mv_"final}}^{2} + \frac{1}{2} k {x}^{2}$

We know that ${v}_{\text{initial}} = 3 \frac{m}{s}$ and v_"final = 0 m/s. Therefore we get:

$\frac{1}{2} \left(6\right) {\left(3\right)}^{2} = \frac{1}{2} \left(6\right) {\left(0\right)}^{2} + \frac{1}{2} \left(12\right) {x}^{2}$

$27 = 6 {x}^{2}$

$x = \sqrt{\frac{27}{6}} \approx 2.12 m$

Hopefully this helps!

Mar 26, 2018

$\implies x = \sqrt{4.5} \text{ m" approx 2.1 " m}$

Explanation:

Quantities given:

• $k = 12 \text{ kg" * "m/s}$
• $m = 6 \text{ kg}$
• $v = 3 \text{ m/s}$

We can calculate the kinetic energy of the object at the moment before it contacts the spring:

$\implies {E}_{k} = \frac{1}{2} m {v}^{2}$

The potential energy of a spring is given as:

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

When the object comes to rest, it will be because the object has lost all of its kinetic energy and the energy has been transferred into the spring.

$\implies {E}_{k} = {E}_{p}$

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

$\implies m {v}^{2} = k {x}^{2}$

We can solve for $x$, which is the displacement of the spring (assuming its initial displacement was $0$):

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

Substituting known values:

$\implies x = \sqrt{\frac{\left(6\right) {\left(3\right)}^{2}}{12}} = \sqrt{4.5} \approx 2.1 \text{ m}$