91Ó°ÊÓ

The gliders are on the air track, friction is negligible.

Mass of right glider A, $m=0.4kg$

Mass of right spring, $ms=0.2kg$

External force $F_{ext}=6N$

The gliders are on the track, friction is negligible. Hence all move with constant acceleration and let it be a.

$$\begin{gathered} \left(M+m+m_s\right)a=F_{ext} \\ a=\frac{F_{ext}}{\left(M+m+m_s\right)}……\left(1\right) \\ {F}_{SA}=F_{ext}-ma……\left(2\right) \end{gathered}$$

Substituting the given data in the equation $\left(1\right)$

role="math" localid="1649077951078" $$\begin{gathered} a=\frac{6}{\left(0.6+0.4+0.2\right)} \\ a=5m/s^2……\left(3\right) \end{gathered}$$

Substituting the given data in the equation$\left(3\right)$value in the equation$\left(2\right)$

$$\begin{gathered} -F_{SA}=6-\left(0.4\right)\left(5\right) \\ {F}_{SA}=-4N \end{gathered}$$

The force exerted by the spring on glider A is$4N$with the opposite direction of the external force.

The gliders are on the air track, friction is negligible.

Mass of right glider B, $m=0.6kg$

Mass of right spring,$m_s=0.2kg$

External force $F_{ext}=6N$

The gliders are on the track, friction is negligible. Hence all move with constant acceleration and let it be a.

$$\begin{gathered} \left(M+m+m_s\right)a=F_{ext} \\ a=\frac{F_{ext}}{\left(M+m+m_s\right)}……\left(1\right) \\ {F}_{SB}=ma……\left(3\right) \end{gathered}$$

This is force due to spring on glider B .

Substituting the given data in the equation $\left(1\right)$

$$\begin{gathered} a=\frac{6}{\left(0.6+0.4+0.2\right)} \\ a=5m/s^2……\left(3\right) \end{gathered}$$

Substituting the given data in the equation $\left(3\right)$value in the equation $\left(2\right)$

localid="1649679014912" role="math" $$\begin{gathered} F_{SB}=\left(0.6kg\right)\left(5m/s^2\right) \\ {F}_{SB}=3N \end{gathered}$$

Itis $3N$in the direction of external force.