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Homework Statement
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?
Homework 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)##
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##
##\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.