m = 2kg; Rơi tự do nên
\(\eqalign{ & h = {{g{t^2}} \over 2} \cr & A = P.h.\cos {0^0} = mg{{g{t^2}} \over 2} = {{m{g^2}{t^2}} \over 2} \cr & {A_P} = {{2.9,{8^2}.1,{2^2}} \over 2} = 138,3(J) \cr} \)
\({P_{tb}} = {{{A_P}} \over t} = {{138,3} \over {1,2}} = 115,25\)(W)
\({P_{tt}} = \overrightarrow P .\overrightarrow v = P.v = mg.gt = m{g^2}t \)
\(= 2.9,{8^2}.1,2 = 230,5\)(W).
