Answer
We can see that the decrease in mechanical energy is $~~-mgL$
Work Step by Step
In part (a), we found that $v_0= \sqrt{2~g~L}$
By conservation of energy, the speed $v_C$ at point $C$ is also $~~v_C= \sqrt{2~g~L}$
We can find the mechanical energy at point C when there was no friction:
$E = K_C+U_C$
$E = \frac{1}{2}mv_C^2+mgL$
$E = \frac{1}{2}m(\sqrt{2~g~L})^2+mgL$
$E = 2mgL$
We can find the mechanical energy at point C when there is friction:
$E = K_C+U_C$
$E = 0+mgL$
$E = mgL$
We can see that the decrease in mechanical energy is $~~-mgL$
