Question

A spring is used to stop a 50-kg package which is moving down a 20º incline. The spring has a constant k = 32 kN/m and is held by cables so that it is initially compressed 50 mm. Knowing that the velocity of the package is 2 m/s when it is 8 m from the spring and neglecting friction, determine the maximum additional deformation of the spring in bringing the package to rest.

Answer 1

Answer:

$\Delta x_{1} \approx 0.259\,m$

Explanation:

The system formed by the spring and the package is described by the means of the Principle of Energy Conservation:

$(50\,kg)\cdot (9.807\,\frac{m}{s} )\cdot(8\,m+\Delta x)\cdot \sin 20^{\textdegree}+\frac{1}{2}\cdot (50\,kg)\cdot (2\,\frac{m}{s})^{2} +\frac{1}{2}\cdot (32000\,\frac{N}{m}) \cdot (0.05\,m)^{2} = \frac{1}{2}\cdot (32000\,\frac{N}{m}) \cdot (0.05\,m+\Delta x)^{2}$

After expanding and symplifying the expression, a second-order polynomial is found:

$1481.677 + 167.710\cdot \Delta x = 16000\cdot [2.5\times 10^{-3}+0.1\cdot \Delta x + (\Delta x)^{2}]$

$1481.677 + 167.710\cdot \Delta x = 40+1600\cdot \Delta x + 16000\cdot(\Delta x)^{2}$

$16000\cdot (\Delta x)^{2}+1432.29\cdot \Delta x - 1441.677 = 0$

The roots of the polynomial are:

$\Delta x_{1} \approx 0.259\,m$

$\Delta x_{2}\approx - 0.348\,m$

Physically speaking, the first root is the only real solution.