Đáp án:

\(\begin{array}{l}
a.W = 300J\\
b.{h_{\max }} = 30m\\
c.{F_c} = 5N
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
a.\\
W = mgh + \frac{1}{2}m{v^2} = 1.10.10 + \frac{1}{2}{.1.20^2} = 300J\\
b.\\
W = {W_{t\max }} = mg{h_{\max }}\\
300 = 1.10.{h_{\max }}\\
 \Rightarrow {h_{\max }} = 30m\\
c.\\
W' - W = {A_c}\\
mgh' - W = {F_c}.s.\cos 180\\
1.10.23,33 - 300 = {F_c}(23,33 - 10).cos180\\
 \Rightarrow {F_c} = 5N
\end{array}\)

a.

$W_{A}$ = $Wđ_{A}$ + $Wt_{A}$

<=> $W_{A}$ = $\dfrac{1}{2}$m$v_{A}^2$ + mg$z_{A}$

<=> $W_{A}$ = $\dfrac{1}{2}$.$20^2$ + 10.10

=> $W_{A}$ = 300 (J)

b.

$W_{A}$ = $W_{B}$ = $Wt_{B}$

<=> 300 = mg$z_{B}$ 

<=> 300 = 10.$z_{B}$

=> $z_{B}$ = 30 (m)

c.

$W_{sau}$ - $W_{trước}$ = $A_{Fc}$

<=> mgh’ - 300 = Fc.s.cos$\alpha$

<=> 10.23,33 - 300 = Fc.13,3.cos180

=> Fc = 5N