Answer
(a) $a = 3.96~m/s^2$
(b) $F_N = 21.7~N$
Work Step by Step
(a) $\sum F = ma$
$F_{thrust} - mg = ma$
$a = \frac{F_{thrust} - mg}{m} = \frac{(1720~N)-(125~kg)(9.80~m/s^2)}{125~kg}$
$a = 3.96~m/s^2$
(b) We can find the mass of the power supply.
$mg = 15.5~N$
$m = \frac{15.5~N}{9.80~m/s^2} = 1.58~kg$
We can find the normal force that the floor exerts on the power supply.
$\sum F = ma$
$F_N - mg = ma$
$F_N = ma + mg = (1.58~kg)(3.96~m/s^2)+(1.58~kg)(9.80~m/s^2)$
$F_N = 21.7~N$
