A rectangular block of wood of mass 20 kg and height 2 m floats in a liquid - Leaving Cert Applied Maths - Question 6 - 2013

When the block is pushed down, it experiences a restoring force. The net force can be expressed as:

$F = 20g - 400x$

Where x is the displacement from equilibrium. The acceleration can be determined by:

$a = \frac{F}{m} = -20x$

Thus, the angular frequency (ω) of simple harmonic motion is given by:

$ω = \sqrt{\frac{20}{20}} = \sqrt{1} = 1\text{ rad/s}$

The period T of the motion is:

$T = \frac{2\pi}{ω} = 2\pi\text{ s}$

To find the specific time when pushed down 0.3 m, we calculate:

1. The angular frequency:

$ω = \sqrt{\frac{20\text{ N}}{20\text{ kg}}} = \sqrt{1} = 1\text{ rad/s}$

2. The period:

$T = 2\pi/1 = 2\pi ≈ 6.28 ext{ s}$

Using the geometry of the forces in equilibrium:

$\alpha = 60^{\circ}$

From the vertical components of the tensions:

1. $T_1 \cos 60^{\circ} - T_2 \cos 60^{\circ} = mg$
2. $T_1 - T_2 = 2mg$

For the horizontal component, when the mass is in circular motion:

$T_1 \sin 60^{\circ} + T_2 \sin 60^{\circ} = m r ω^2$

Using these equations, we combine them algebraically to express T1 and T2 in terms of m and ω.

Using kinematic equations, we denote:

• Initial velocity, u = 0
• Acceleration, a = g
• Distance fallen, s = 2l.

The equation is:

$s = ut + \frac{1}{2} a t^2$

Substituting the known values gives:

$2l = 0 + \frac{1}{2} g t^2$

Rearranging, we find:

$t = \sqrt{\frac{4l}{g}} = 2\sqrt{\frac{l}{g}}$

Thus, the time before the mass strikes the horizontal surface is approximately 2\sqrt{\frac{l}{g}}.