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

**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://ift.tt/1rzL6jp

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:

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

##\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.

http://ift.tt/1mV5NPd

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