# Isolated Systems, with $\Delta E_{th}$

1. Jul 5, 2014

### Zondrina

1. The problem statement, all variables and given/known data

A $m = 2.0 kg$ box slides along a floor with speed $v_1 = 4.0 \frac{m}{s}$. It then runs into and compresses a spring until the box stops. Its path to the initially relaxed spring is frictionless $(F_f = 0)$, but as it compresses the spring, kinetic friction with a magnitude of $F_k = 15 N$ acts on the box.

If $k = 10 000 \frac{N}{m}$, what is the compression of the spring when the box stops? What is the coefficient of kinetic friction?

2. Relevant equations

Here's a FBD of the box i drew: http://gyazo.com/886409de36e80192b30e873d0160970a

Up and right are positive.

$ΔE_T = ΔE_{mech} + ΔE_{th} + ΔE_{int} = W, \quad (1)$

3. The attempt at a solution

The scenario indicates that $F_N$ and $F_g$ do no work at any time as there is no vertical displacement. $F_s$ and $F_k$ do work on the box when it comes into contact with the spring wall.

The earth, box and spring-wall form an isolated system. We know that in an isolated system, the total energy $E_T$ cannot change. That is, $ΔE_T = 0$. Hence $(1)$ can be re-written as:

$ΔE_{mech} + ΔE_{th} + ΔE_{int} = 0$​

The internal energy in this case is zero, and we can reduce the equation further to:

$ΔK + ΔU + F_k d = 0$
$\frac{1}{2} m (v_f^2 - v_i^2) + \frac{1}{2}kd^2 + F_k d = 0$
$\frac{1}{2} (2.0) (0^2 - (4.0)^2) + \frac{1}{2}(10 000) d^2 + (15) d = 0$
$5000d^2 + 15d - 16 = 0$​

which is quadratic in $d$. Solving I get $d = 0.055 m$.

Now we know that $\mu_k = \frac{F_k}{F_N}$. Hence:

$\sum F_y = 0 \Rightarrow F_N = F_g = (2.0)(9.81) N$

Therefore, $\mu_k = \frac{F_k}{F_N} = \frac{15}{(2.0)(9.81)} = 0.764 = 0.76$.

If someone could help me by looking over this and telling me if it sounds okay, it would be appreciated.

2. Jul 5, 2014

### Nathanael

It all looks good to me, well done.

3. Jul 5, 2014

### Zondrina

Thank you.

I'm trying to get into the physics groove again for the summer.