6. (a) A loaded test-tube of total mass m floats in water and is in equilibrium when a length d is submerged, as shown - Leaving Cert Applied Maths - Question 6 - 2015

When the test-tube is pushed down by a small distance $x$, the new length submerged becomes $d + x$. The upward force can be expressed as:

$F_i = mg - k(d + x) = mg - kd - kx$

Since $F = mg$, it follows that:

$F_i = -kx$

This leads to:

$a = -\frac{k}{m} x$

This relationship shows that the motion will be harmonic since it fits the form of simple harmonic motion. The period of this motion is given by:

$T = 2\pi \sqrt{\frac{m}{k}}$

Substituting for $k$:

$T = 2\pi \sqrt{\frac{m}{\frac{mg}{d}}} = 2\pi \sqrt{\frac{d}{g}}$

Starting with energy conservation, the skier's energy at point A is a combination of kinetic and potential energy. At point B, the skier loses contact, meaning the forces must balance out:

$\frac{1}{2} mv^2 + mg h = \frac{1}{2} mv^2 (at B)$

The radius of the arc $R$ is:

$mg \cos(\alpha) - R = \frac{mv^2}{R}$

To find $\alpha$, we set $R = 0$, and applying the trigonometric identities with given conditions:

$\frac{1}{2} (mg \cos \alpha) + mg = \frac{\frac{1}{2} (8)^2}{7}$

Through algebraic manipulation, we find:

$\cos(\alpha) = 0.9291$

Thus,

$\alpha = \cos^{-1}(0.9291) \approx 21.7°$