Answer:
Compression distance: [tex]d \approx 0.102\,m[/tex]
Explanation:
According to this statement, we know that system is non-conservative due to the rough patch. By Principle of Energy Conservation and Work-Energy Theorem, we have the following expression that represents the system having a translational kinetic energy ([tex]K[/tex]), in joules, at the expense of elastic potential energy ([tex]U[/tex]), in joules, and overcoming work losses due to friction ([tex]W_{l}[/tex]), in joules:
[tex]K + W_{l} = U[/tex] (1)
By definitions of translational kinetic and elastic potential energies and work losses due to friction, we expand the equation described above:
[tex]\frac{1}{2}\cdot m \cdot v^{2} +\mu\cdot m\cdot g \cdot s = \frac{1}{2} \cdot k \cdot d^{2}[/tex] (2)
Where:
[tex]m[/tex] - Mass of the block, in kilograms.
[tex]v[/tex] - Final velocity of the block, in meters per second.
[tex]\mu[/tex] - KInetic coefficient of friction, no unit.
[tex]g[/tex] - Gravitational acceleration, in meters per square second.
[tex]s[/tex] - Width of the rough patch, in meters.
[tex]k[/tex] - Spring constant, in newtons per meter.
[tex]d[/tex] - Compression distance, in meters.
If we know that [tex]m = 1.2\,kg[/tex], [tex]v = 2.3\,\frac{m}{s}[/tex], [tex]\mu = 0.44[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]s = 0.05\,m[/tex] and [tex]k = 730\,\frac{N}{m}[/tex], then the compression distance of the spring is:
[tex]\frac{1}{2}\cdot m \cdot v^{2} +\mu\cdot m\cdot g \cdot s = \frac{1}{2} \cdot k \cdot d^{2}[/tex]
[tex]m\cdot v^{2} + 2\cdot m\cdot g \cdot s = k\cdot d^{2}[/tex]
[tex]d = \sqrt{\frac{m\cdot (v^{2}+2\cdot g\cdot s)}{k} }[/tex]
[tex]d \approx 0.102\,m[/tex]
