Bài 4.113

a.

• wA = wB

<=> wđA + wtA = wđB + wtB

<=> mgzA = 12,5

<=> 0,25.10.zA = 12,5

=> zA = 5 (m)

• wđB = $\frac{1}{2}$m$vB^2$

=> 12,5 = $\frac{1}{2}$.0,25.$vB^2$

=> vB = 10 m/s

b.

wB = wC

<=> 12,5 = wđC + wtC = wdD + wdD

<=> 12,5 = 2.$\frac{1}{2}$m$vC^2$

<=> 12,5 = 0,25$vC^2$

=> vC = 5√2 m/s

c.

wB = wD

<=> 12,5 = wdD + wtD = 2wtD + wtD

<=> 12,5 = 3mgzD

<=> 12,5 = 3.0,25.10.zD

=> zD = 1,6m

Bài 4.114

a.

wA = wđA + wtA = $\frac{1}{2}$m$vA^2$

<=> wA = $\frac{1}{2}$.$20^2$

=> wA = 200 J

b.

wA = wB

<=> 200 = wđB + wtB = mgzB

<=> 200 = 10.zB

=> zB = 20m

c. 

 Vmax khi h=0

=> Vmax = 20m/s

Đáp án:

\(\begin{array}{l}
a.\\
{h_{\max }} = 5m\\
{v_{\max }} = 10m/s\\
b.v = 5\sqrt 2 m/s\\
c.h' = \frac{5}{3}m\\
4.114\\
a.W = 200J\\
b.{h_{\max }} = 20m\\
c.{v_{\max }} = 20m/s
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
a.\\
{W_{d\max }} = {W_{t\max }} = mg{h_{max}}\\
12,5 = 0,25.10.{h_{\max }}\\
 \Rightarrow {h_{\max }} = 5m\\
{W_{d\max }} = \frac{1}{2}mv_{\max }^2\\
12,5 = \frac{1}{2}.0,25.v_{max}^2\\
 \Rightarrow {v_{\max }} = 10m/s\\
b.\\
{W_t} = {W_d} = \frac{W}{2} = \frac{{{W_{d\max }}}}{2}\\
\frac{1}{2}m{v^2} = \frac{{12,5}}{2}\\
\frac{1}{2}.0,25.{v^2} = \frac{{12,5}}{2}\\
 \Rightarrow v = 5\sqrt 2 m/s\\
c.\\
2{W_t}' = {W_d}' \Rightarrow {W_t}' = \frac{W}{3}\\
mgh' = \frac{{12,5}}{3}\\
0,25.10.h' = \frac{{12,5}}{3}\\
 \Rightarrow h' = \frac{5}{3}m\\
4.114\\
a.\\
W = {W_{d\max }} = \frac{1}{2}mv_{\max }^2 = \frac{1}{2}{.1.20^2} = 200J\\
b.\\
W = {W_{t\max }}\\
200 = mg{h_{\max }}\\
200 = 1.10.{h_{\max }}\\
 \Rightarrow {h_{\max }} = 20m\\
c.\\
{W_{d\max }} = \frac{1}{2}mv_{\max }^2\\
200 = \frac{1}{2}.1.v_{\max }^2\\
 \Rightarrow {v_{\max }} = 20m/s
\end{array}\)