Answer
(a) $1.04N$
(b) $3.59N$
Work Step by Step
We know that
$a=\frac{F}{m}$
We plug in the known values to obtain:
$a=\frac{7.5N}{1.3Kg+3.2Kg+4.9Kg}=0.7978m/s^2$
Now the contact force between boxes 1 and 2 can be determined as
$F_{12}=ma$
We pug in the known values to obtain:
$F_{12}=(1.30Kg)(0.7978m/s^2)$
$F_{12}=1.04N$
(b) The contact force between boxes 2 and 3 can be determined as
$F_{23}=ma$
We plug in the known values to obtain:
$F_{23}=(3.2Kg+1.3Kg)(0.79m/s^2)$
$F_{23}=3.59N$
