Hi there!
a)
We can begin by using the work-energy theorem. We initially have ALL Spring Potential Energy, which is entirely converted to kinetic energy. Thus:
[tex]E_I = E_F\\\\U = K \\\\\frac{1}{2}kx^2 = \frac{1}{2}mv^2[/tex]
Rearrange to solve for speed:
[tex]kx^2 = mv^2\\\\v^2 = \frac{kx^2}{m}\\\\v = \sqrt{\frac{kx^2}{m}}[/tex]
Plug in the values and solve.
[tex]v = \sqrt{\frac{770 * (0.42^2)}{3.25}} = \boxed{6.465 \frac{m}{s}}[/tex]
b)
We can find the final kinetic energy by subtracting the work done by friction from the original kinetic energy.
[tex]\frac{1}{2}mv_i^2 - W_f= \frac{1}{2}mv_f^2\\\\\frac{1}{2}mv_i^2 - \mu mgd = K_f[/tex]
Solve by plugging in values:
[tex]K_f = \frac{1}{2}(3.25)(6.465^2) - .23(9.8 * 3.25 * 1.05) = \boxed{60.227 J}[/tex]
c)
When the mass stops sliding, it will have NO kinetic energy. (All potential). Thus:
[tex]K = U \\\\\frac{1}{2}mv_f^2 = mgh[/tex]
Rearrange and solve for 'h', the VERTICAL distance the block moves up the incline.
[tex]\frac{1}{2}v_f^2 = gh \\\\h = \frac{v_f^2}{2g} = 1.89 m[/tex]
To find 'D', we must use trigonometry to solve.
Recall:
[tex]sin\theta = \frac{O}{H}\\\\sin\theta = \frac{h}{D}\\\\D = \frac{h}{sin\theta} = \frac{1.89}{sin(36.5)} = \boxed{3.179 m}[/tex]
d)
The velocity when the block slides down to its original position will be the SAME as the velocity on its way up, which is 6.09 m/s.
e)
When the block slides all the way down, the block has the same velocity as above.
Therefore, its original kinetic energy is equal to:
[tex]K_i = \frac{1}{2}mv^2 = \frac{1}{2}(3.25)(6.09^2) = 60.22 J[/tex]
Now, we must calculate its final kinetic energy after some is lost to the friction area.
[tex]K_i - W_f = K_f\\\\60.22 - .230(3.25 * 9.8 * 1.05) = \frac{1}{2}(3.25)v_f^2\\\\v_f = 5.686 \frac{m}{s}[/tex]
Now, this kinetic energy is converted to elastic potential energy when the block is stopped, so:
[tex]K = U \\\\\frac{1}{2}mv^2 = \frac{1}{2}kx^2\\\\mv^2 = kx^2\\\\x = \sqrt{\frac{mv^2}{k}} = \boxed{.369 m = 36.9 cm}[/tex]
Therefore, x₂ is LESS than x₁, or the resulting compression distance is LESS than the original.
