mgh-μmgcosα? L=Ek-0
So ek = mgh-μ GX; Therefore, the kinetic energy of sliding along the AD slope to the bottom is larger, so A is correct;
B. Power of gravity when sliding to the bottom: p = mgvcos (90-α) = mgvsin α;
When sliding to the bottom along the AD slope, the terminal speed is large, the slope angle is large, and the instantaneous power of gravity is large; Therefore, B is correct;
C, according to the functional relationship, the reduction of the internal energy of the system is equal to the increase of the internal energy of the system, which is equal to the work done by the system to overcome friction, as follows:
△E=Q=μmgcosα? l =μmgx; Same; So, C is wrong and D is right;
Therefore, ABD ..