FIGURE $P7.27$shows a $6.0N$ force pushing two gliders along an air track. The $200g$ spring between the gliders is compressed. How much force does the spring exert on

(a) glider A and

(b) glider B?

The spring is firmly attached to the gliders, and it does not sag.

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.

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

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

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

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.

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

This is force due to spring on glider B .

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

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

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.