[tex]<b><u>Answer</u></b>[/tex]
The magnitude of the acceleration to the nearest hundredth is [tex]3.39ms^{-2}[/tex]
[tex]<b><u>Explanation</u></b>[/tex]
We have to analyze the external forces acting on the two solid blocks.
The tension,[tex]T[/tex] in the string, the weight [tex]w_1=12.4g[/tex] and [tex]w_2=25.5g[/tex] are all  illustrated in the diagram in the attachment.
Recall from Newton's law of motion that,
[tex]F=ma[/tex], where [tex]m[/tex] is mass measured in kilograms and [tex]a[/tex] is the acceleration in [tex]ms^{-2}[/tex].
For the [tex]12.4kg[/tex] mass the resultant upward force is
[tex]T-12.4g[/tex].
Applying [tex]F=ma[/tex] to the [tex]12.4kg[/tex] mass gives
[tex]T-12.4g=12.4a...eqn(1)[/tex].
For the [tex]25.5kg[/tex] mass the resultant downward force is
[tex]25.5g-T[/tex]
Applying [tex]F=ma[/tex] to the [tex]25.5kg[/tex] mass gives,
[tex]25.5g-T=25.5a...eqn(2)[/tex]
Adding the two equations gives us,
[tex]25.5g-12.4g=25.5a+12.4a[/tex]
This simplifies to
[tex]13.1g=37.9a[/tex]
We divide both sides by [tex]37.9[/tex] to obtain,
[tex]\frac{13.1g}{37.9}=a[/tex]
We were given [tex]g=9.8ms^{-2}[/tex]
We substitute this value to obtain,
[tex]a=\frac{13.1}{37.9} \times 9.8 [/tex]
[tex]a=3.3873 \ms^{-2}[/tex]
Hence the magnitude of the acceleration to the nearest hundredth is [tex]3.39ms^{-2}[/tex]
