State the principle of conservation of energy - Leaving Cert Physics - Question a - 2019

First, we need to calculate the height (h) of the mass at position B using:
$h = 0.8 - 0.8 imes ext{cos}(35^ ext{o}) = 0.145 ext{ m}$
Now, applying the conservation of energy, we have:
mgh + rac{1}{2}mv^2 = mgh + rac{1}{2}mv^2
Substituting values gives us:
$v^2 = 2gh + v_0^2$
Where:

• $g = 9.8 ext{ m s}^2$,
• $v_0 = 4 ext{ m s}^{-1}$, therefore:
  $v^2 = 2(9.8)(0.145) + 4^2$
  Calculating gives:
  $v = 3.63 ext{ m s}^{-1}$

The centripetal acceleration ($a$) is given by the formula:
a = rac{v^2}{r}
Where:

• $v = 3.63 ext{ m s}^{-1}$,
• $r = 0.8 ext{ m}$, hence:
  a = rac{3.63^2}{0.8} = 16.5 ext{ m s}^{-2}
  This acceleration is directed towards the center.

A labelled diagram should include:

• Weight (W), acting downwards, which equals $mg$,
• Tension (T), acting upwards along the string,
  Label each force with the correct notation. Incorrect forces can be penalized with a deduction of marks.