Hi there!
We can find the compressed distance by using the equation for Spring Potential Energy:
[tex]U = \frac{1}{2}kx^2[/tex]
U = Potential Energy (3.2 J)
k = Spring Constant (529 N/m)
x = Compression distance (? m)
Rearrange and solve for 'x'.
[tex]x^2 = \frac{2U}{k}\\\\x = \sqrt{\frac{2U}{k}}\\\\x = \sqrt{\frac{2(3.2)}{529}} = .11 m[/tex]
Now, we can find the velocity using the Work-Energy theorem. There is work done by friction in this instance, however.
[tex]E_i = E_f\\[/tex]
Initially, there is only Spring Potential Energy. Then, some of that energy is converted to kinetic energy, while some is lost to friction.
Thus:
[tex]U = K + W_f\\\\\frac{1}{2}kx^2 = \frac{1}{2}mv^2 + \mu mgd[/tex]
Rearrange to solve for velocity.
[tex]\frac{1}{2}kx^2 - \mu mgd = \frac{1}{2}mv^2\\\\kx^2 - 2\mu mgd = mv^2\\\\v^2 = \frac{kx^2 - 2\mu mgd}{m}\\\\v = \sqrt{\frac{kx^2 - 2\mu mgd}{m}}[/tex]
Plug in the values and solve. (d = compression distance)
[tex]v = \sqrt{\frac{529(.11^2) - 2(.18)(1.25)(9.8)(0.11)}{1.25}} = \boxed{2.175 \frac{m}{s}}[/tex]
