a) Khi lúc thả rơi:

`W_c = W_đ+W_t = 0 + mgh = 20*10*m = 200m`

b) `v^2-u^2=2as`

`=> v^2 = 2*10*20`

`=> v^2 = 400`

`=> v =20m//s`

c)

`W_đ = 1/2W_c = 100m`

`=> 1/2mv^2 = 100m`

`=> v^2 = 200`
`=> v = 10sqrt(2)m//s`

`W_t = W_c-W_đ = 100m`

`=> mgh = 100m`

`=> 10*h=100`

`=> h=10m`

d)

`W_đ = 2W_t`

`=> W_đ =400/3m, W_t = 200/3m`

`=> 1/2mv^2 = 400/3m`

`=> v^2 = 800/3`
`=> v = 20sqrt(2/3)m//s`

`=> mgh = 200/3m`

`=> 10*h=200/3`

`=> h=20/3m`

e)

`W_t = 1/3W_c = 200/3m`

`=>` Giong phần d

f) 

`W_đ = 0.5W_t`

`=> W_đ = 200/3m, W_t = 400/3m`

`=> 1/2mv^2 = 200/3m`

`=> v^2 = 400/3`
`=> v = 20/sqrt(3)`

`=> mgh = 400/3m`

`=> 10*h=400/3`

`=> h=40/3m`

Đáp án:

a) \(200m\)

b) \(20\left( {m/s} \right)\)

c) \(\left\{ \begin{array}{l}
{v_1} = 10\sqrt 2 \left( {m/s} \right)\\
{h_1} = 10\left( m \right)
\end{array} \right.\)

d) \(\left\{ \begin{array}{l}
{h_2} = \dfrac{{20}}{3}\left( m \right)\\
{v_2} = \dfrac{{20\sqrt 6 }}{3}\left( {m/s} \right)
\end{array} \right.\)

e) \(\dfrac{{20}}{3}\left( m \right)\)

f) \(\left\{ \begin{array}{l}
{h_2} = \dfrac{{20}}{3}\left( m \right)\\
{v_2} = \dfrac{{20\sqrt 6 }}{3}\left( {m/s} \right)
\end{array} \right.\)

Giải thích các bước giải:

a) Cơ năng của vật là:

\(W = mgH = m.10.20 = 200m\)

b) Vận tốc khi chạm đất là:

\(V = \sqrt {\dfrac{{2W}}{m}}  = \sqrt {\dfrac{{2.200m}}{m}}  = 20\left( {m/s} \right)\)

c) Bảo toàn cơ năng: 

\(\begin{array}{l}
\left\{ \begin{array}{l}
{W_d} = \dfrac{W}{2} = 100m\\
{W_t} = \dfrac{W}{2} = 100m
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
\dfrac{1}{2}mv_1^2 = 100m\\
mg{h_1} = 100m
\end{array} \right.\\
 \Rightarrow \left\{ \begin{array}{l}
{v_1} = 10\sqrt 2 \left( {m/s} \right)\\
{h_1} = 10\left( m \right)
\end{array} \right.
\end{array}\)

d) Ta có: 

\(\begin{array}{l}
\left\{ \begin{array}{l}
{W_d} + {W_t} = 200m\\
{W_d} = 2{W_t}
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
{W_t} = \dfrac{W}{3} = \dfrac{{200}}{3}m\\
{W_d} = \dfrac{{2W}}{3} = \dfrac{{400}}{3}m
\end{array} \right.\\
 \Rightarrow \left\{ \begin{array}{l}
mg{h_2} = \dfrac{{200}}{3}m\\
\dfrac{1}{2}mv_2^2 = \dfrac{{400}}{3}m
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
{h_2} = \dfrac{{20}}{3}\left( m \right)\\
{v_2} = \dfrac{{20\sqrt 6 }}{3}\left( {m/s} \right)
\end{array} \right.
\end{array}\)

e) Ta có: 

\({W_t} = \dfrac{W}{3} \Rightarrow {h_3} = \dfrac{H}{3} = \dfrac{{20}}{3}\left( m \right)\)

f) Bảo toàn cơ năng: 

\(\begin{array}{l}
\left\{ \begin{array}{l}
{W_d} + {W_t} = 200m\\
{W_d} = 2{W_t}
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
{W_t} = \dfrac{W}{3} = \dfrac{{200}}{3}m\\
{W_d} = \dfrac{{2W}}{3} = \dfrac{{400}}{3}m
\end{array} \right.\\
 \Rightarrow \left\{ \begin{array}{l}
mg{h_2} = \dfrac{{200}}{3}m\\
\dfrac{1}{2}mv_2^2 = \dfrac{{400}}{3}m
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
{h_2} = \dfrac{{20}}{3}\left( m \right)\\
{v_2} = \dfrac{{20\sqrt 6 }}{3}\left( {m/s} \right)
\end{array} \right.
\end{array}\)