Answer
See work.
Work Step by Step
a. Use equation 6.2. The angle between the force and the displacement is $0^{\circ}$.
$$(1350N)(5.00 \times 10^3 m)cos0^{\circ}=6.75 \times 10^6 J$$
Use equation 6.2. The angle between the force and the displacement is $35^{\circ}$.
$$(1350N)(5.00 \times 10^3 m)cos35^{\circ}=5.53 \times 10^6 J$$
b. Use equation 6.2. The angle between the force and the displacement is $180^{\circ}$.
$$(1350N)(5.00 \times 10^3 m)cos180^{\circ}=-6.75 \times 10^6 J$$
Use equation 6.2. The angle between the force and the displacement is $145^{\circ}$.
$$(1350N)(5.00 \times 10^3 m)cos145^{\circ}=-5.53 \times 10^6 J$$
Note that when the car and truck are taken together as a system, the tension does no net work.
c. Use equation 6.2. The angle between the force and the displacement is $90^{\circ}$. The cosine is zero so there is no work done. Whenever a force acts perpendicular to the direction of motion, its contribution to the net work is zero.
