Đáp án:
\(\begin{array}{l}
3.B\\
{W_d}' = \frac{1}{2}m'v{'^2} = \frac{1}{2}.\frac{m}{2}.{(2v)^2} = 2.\frac{1}{2}m{v^2} = 2{W_d}\\
4.C\\
{W_d} = \frac{1}{2}m{v^2} \Rightarrow v = \sqrt {\frac{{2{W_d}}}{m}} = \sqrt {\frac{{2.8}}{1}} = 4m/s\\
5.A\\
{W_d} = \frac{1}{2}m{v^2} = \frac{1}{2}.0,{5.2^2} = 1J\\
{W_t} = mgh = 0,5.10.0,5 = 2,5J\\
W = {W_d} + {W_t} = 1 + 2,5 = 3,5J\\
6.C\\
A = {F_k}s.\cos 60 = 50.8.\cos 60 = 200J
\end{array}\)
