A system of two blocks are connected by a string. These blocks are horizontally accelerated by an applied force of magnitude F on the frictionless horizontal surface. If the right block is 5kg, and the left block is 3kg, what is tension in the string between the blocks is:

(A) 3 F
(B) 5 F
(C) 3/8 F
(D) 1/3 F

We need to apply the Newton's second Law [tex]F=m*a[/tex]. First we need to add both block masses as:[tex]m_{T}=m_{r} +m_{l}=3+5=8(Kg)[/tex]. Then we can find the acceleration of the system as:[tex]a_{T}=\frac{F}{m_{T} }=\frac{F}{8}[/tex]. Therefore solving for T from the free body diagram, we find: [tex]F_{r} =m_{r}*a_{T}=\frac{5*F}{8}[/tex], and adding forces on axle x we get: [tex]F_{r}= \frac{5*F}{8} =F-T[/tex]. Now we can calculate the tension in the string as:[tex]T=F-\frac{5*F}{8}=\frac{3F}{8}[/tex].