Two points A and B are 6 m apart on a smooth horizontal surface - Leaving Cert Applied Maths - Question 6 - 2018

Starting with the conservation of energy, we know:

$\frac{1}{2} mu^2 + 0 = \frac{1}{2} mv^2 + mg(2d).$

This simplifies to:

$mu^2 = mv^2 - 4mgd.$
By considering the tension $T$ and using:

$T + mg = \frac{mu^2}{d} - 4mg,$
we can express $T$:

$T = \frac{mu^2}{d} - 5mg.$
For the condition of motion:

$T ≥ 0$ leads to:

$\frac{mu^2}{d} ≥ 5mg.$
Thus:

$u^2 ≥ 5gd.$

Considering the tension while P moves in a vertical circle:

$T_{min} = \frac{mu^2}{d} - 5mg$ and substituting:

$T_{min} = 0$ gives:

$\frac{m(6gd)}{d} - 5mg = 0 \Rightarrow 6mg - 5mg = mg$ which simplifies to:

$T_{max} = 7mg.$
Since $T_{max} = kT_1$, we set:

$T_1 = 6mg,$ therefore:

$k = 7.$