### Child's Toy

(a) Look at all three cars as one system.
$\sum F_{x} = m_{total}a_{x}$
$6.5N = (3.1kg+2.4kg+1.2kg)a_{x}$
$0.97\frac{m}{s^{2}} = a_{x}.$
(b) Now look at only the third car.
$\sum F_{x}=m_{3}a_{x}=(1.2kg)(0.97\frac{m}{s^{2}})$
and you can solve to get F=1.2N pointing to the right.

(c) Look at only the second and third cars.
$F=(m_{2}+m_{3})a_{x}=(2.4kg+1.2kg)(0.97\frac{m}{s^{2}})=3.5N.$