Question

A 1.2 kg block is held at rest against the spring with a force constant k= 730 N/m. Initially, the spring is compressed a distance d. When the block is released, it slides across a surface that is frictionless except for a rough patch of width 5.0 cm that has a coefficient of kinetic friction = 0.44. Find d, such that the block's speed after crossing the rough patch is 2.3 m/s.

Answer 1

Answer:

Compression distance: $d \approx 0.102\,m$

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 ($K$), in joules, at the expense of elastic potential energy ($U$), in joules, and overcoming work losses due to friction ($W_{l}$), in joules:

$K + W_{l} = U$ (1)

By definitions of translational kinetic and elastic potential energies and work losses due to friction, we expand the equation described above:

$\frac{1}{2}\cdot m \cdot v^{2} +\mu\cdot m\cdot g \cdot s = \frac{1}{2} \cdot k \cdot d^{2}$ (2)

Where:

$m$ - Mass of the block, in kilograms.

$v$ - Final velocity of the block, in meters per second.

$\mu$ - KInetic coefficient of friction, no unit.

$g$ - Gravitational acceleration, in meters per square second.

$s$ - Width of the rough patch, in meters.

$k$ - Spring constant, in newtons per meter.

$d$ - Compression distance, in meters.

If we know that $m = 1.2\,kg$, $v = 2.3\,\frac{m}{s}$, $\mu = 0.44$, $g = 9.807\,\frac{m}{s^{2}}$, $s = 0.05\,m$ and $k = 730\,\frac{N}{m}$, then the compression distance of the spring is:

$\frac{1}{2}\cdot m \cdot v^{2} +\mu\cdot m\cdot g \cdot s = \frac{1}{2} \cdot k \cdot d^{2}$

$m\cdot v^{2} + 2\cdot m\cdot g \cdot s = k\cdot d^{2}$

$d = \sqrt{\frac{m\cdot (v^{2}+2\cdot g\cdot s)}{k} }$

$d \approx 0.102\,m$