Answer
The work done on the cab by the force from the cable is $~~25,900~J$
Work Step by Step
We can find the acceleration of the cheese:
$ma = F_N-mg$
$a = \frac{F_N}{m}-g$
$a = \frac{3.00~N}{0.25~kg}-9.8~m/s^2$
$a = 2.2~m/s^2$
Thus the acceleration of the cab must also be $a = 2.2~m/s^2$
Let $M$ be the mass of the cab.
We can find the force $F_T$ that the cable exerts on the cab:
$F_T-Mg = Ma$
$F_T = M(a+g)$
$F_T = (900~kg)(2.2~m/s^2+9.8~m/s^2)$
$F_T = 10,800~N$
We can find the work done on the cab by the force from the cable:
$W = F_T~d$
$W = (10,800~N)(2.40~m)$
$W = 25,900~J$
The work done on the cab by the force from the cable is $~~25,900~J$
