Answer
(a) $a = 1.7 ~m/s^2$
(b) F = 430 N
(c) $a = 1.7 ~m/s^2$
F = 230 N
Work Step by Step
(a) To find the acceleration, let's consider the system of both boxes. The total mass is 190 kg.
$\sum ~F = ma$
$F - F_f = ma$
$a = \frac{F-F_f}{m} = \frac{650 ~N - (190 ~kg)(9.80 ~m/s^2)(0.18)}{190 ~kg}$
$a = 1.7 ~m/s^2$
(b) Let's consider the system of the 125-kg box. Let $F$ be the force that the 65-kg box exerts on the 125-kg box.
$F - F_g = ma$
$F = F_g + ma$
$F = (125 ~kg)(9.80 ~m/s^2)(0.18) + (125 ~kg)(1.7 ~m/s^2)$
$F = 430 ~N$
The boxes exert a force of 430 N on each other.
(c) We can use the same equation as part (a) to find that the acceleration is $a = 1.7 ~m/s^2.$ Now let's consider the system of the 65-kg box. Let $F$ be the force that the 125-kg box exerts on the 65-kg box.
$F - F_g = ma$
$F = F_g + ma$
$F = (65 ~kg)(9.80 ~m/s^2)(0.18) + (65 ~kg)(1.7 ~m/s^2)$
$F = 230 ~N$
The boxes exert a force of 230 N on each other.
