khi treo thẳng đứng lò xo đầu dưới treo vật thì P=\(F_{đh}\)
72,5cm=0,725m ; 65cm=0,65m
\(F_{đh1}=P_1\Leftrightarrow k.\left(l_1-l_0\right)=m_1.g\) (1)
\(F_{đh2}=P_2\Leftrightarrow k.\left(l_2-l_0\right)=m_2.g\) (2)
lấy (1) chia cho (2)
\(\dfrac{l_1-l_0}{l_2-l_0}=\dfrac{m_1}{m_2}\)
\(\Rightarrow l_0=0,6m\)
thay l0=0,6m vào (1) koặc (2)
\(\Rightarrow k=\)40N/m
Lấy g = 10 m/s2
Ta có: △l1 = l1 - l1 ⇒ l1= △l1 + l1 =\(\dfrac{m_1.g}{k}\);
l2=△l2 + l0 = \(\dfrac{m_2.g}{k}\)
⇒ l1 - l2 = \(\dfrac{m_1.g}{k}-\dfrac{m_2.g}{k}=\dfrac{\left(m_1-m_2\right)g}{k}\Rightarrow k=\dfrac{\left(0,5-0,2\right).10}{0,725-0,65}=40\) N/m
Lấy g = 10 m/s2
Ta có: △l1 = l1 - l0 ⇒ l1= △l1 + l0 =\(\dfrac{m_1.g}{k}\);
l2=△l2 + l0 = \(\dfrac{m_2.g}{k}\)
⇒ l1 - l2 = \(\dfrac{m_1.g}{k}-\dfrac{m_2.g}{k}=\dfrac{\left(m_1-m_2\right).g}{k}\Rightarrow k=\dfrac{3}{7,25-6,5}=4\dfrac{N}{m}\)