A uniform 3.00 kg rope, 24.0 m long lies on the ground at the top of a vertical cliff. A mountain climber at the top lets down half of it to his partner to help him climb the cliff. What is the change in Potential energy of the rope during this maneuver ?

Let the cliff be $h = 12.0 \, m$ high to make the calculation simple.
$U_0 = mgh$
half the rope is at $h = 12 m$ and the center of mass of the other half is at $\dfrac{h}{2} = 6 \, m$
$U_f = \dfrac{m}{2}gh + \dfrac{m}{2}g \cdot \dfrac{h}{2} = \dfrac{3mgh}{4}$
$\Delta U = U_f  U_0 = \dfrac{3mgh}{4}  mgh = \dfrac{mgh}{4} = 88.2 \, J$