Answer
(a) $6.46N$
(b) $3.91N$
Work Step by Step
(a) According to Newton's 2nd law of motion
$F=ma$
We plug in the known values to obtain:
$7.5N=(3.20Kg+1.3Kg+4.9Kg)a$
$7.5N=(9.4Kg)a$
$\implies a=\frac{7.5N}{9.4Kg}=0.797m/s^2$
Now the contact force between box 1 and 2 is given as
$F_{12}=(3.2Kg+4.9)(0.797m/s^2)$
$F_{12}=6.46N$
(b) We can find the contact force between boxes 2 and 3 as follows:
$F=ma$
We plug in the known values to obtain:
$F=(0.797m/s^2)(4.9Kg)$
$F=3.91N$
