Newton's Second Lawcontact force

Contact Force

Force FF is applied upon m1m_1 and m2m_2.
Graph 2
a). Find aa.
b). Find contact force FcF_c

Solution

a. F=(m1+m2)a⇒a=Fm1+m2=101+2≈3.3m/s2b. Fc=m1a≈3.3N\begin{aligned} \text{a. } F &= (m_1 + m_2) a\\ \To a &= \frac{F}{m_1 + m_2} = \frac{10}{1 + 2} \approx 3.3 m/s^2\\ \text{b. } F_c &= m_1 a \approx 3.3 N \end{aligned}