Đáp án:

 a,$\begin{array}{l}
k = 50\left( {N/m} \right)\\
{F_{dh}} = 2\left( N \right)
\end{array}$

b.$26cm$

c.$l = 0,28m$;$\omega  = 8,45\left( {rad/s} \right)$

Giải thích các bước giải:

$\begin{array}{l}
a.\Delta {l_1} = 4cm = 0,04m\\
k = \frac{{{m_1}g}}{{\Delta {l_1}}} = \frac{{0,2.10}}{{0,04}} = 50\left( {N/m} \right)\\
{F_{dh}} = {P_1} = {m_1}g = 0,2.10 = 2\left( N \right)\\
b.\Delta {l_2} = \frac{{\left( {{m_1} + {m_2}} \right)g}}{k} = \frac{{\left( {0,2 + 0,1} \right).10}}{{50}} = 0,06m = 6cm\\
{l_2} = {l_0} + \Delta {l_2} = 20 + 6 = 26cm\\
c.\cos {60^0} = \frac{{{m_1}g}}{{{F_{_{dh}}}}} \Rightarrow k.\Delta l = \frac{{{m_1}g}}{{\cos {{60}^0}}} \Rightarrow 50.\Delta l = \frac{{0,2.10}}{{0,5}} \Rightarrow \Delta l = 0,08 = 8cm\\
l = {l_0} + \Delta l = 20 + 8 = 28cm = 0,28m\\
{F_{ht}} = m{\omega ^2}R\\
R = l.\sin {60^0} = 0,14\sqrt 3 m\\
\tan {60^0} = \frac{{{F_{ht}}}}{{mg}} \Rightarrow m{\omega ^2}R = mg\tan {60^0}\\
 \Rightarrow {\omega ^2}.0,14\sqrt 3  = 10.\sqrt 3 \\
 \Rightarrow \omega  = 8,45\left( {rad/s} \right)\\
\end{array}$