Answer
(a) $F_T = 1700 ~N$
(b) The mechanical advantage is 5.7
(c) When the rope makes an angle of $30^{\circ}$, the effort is counterproductive.
Work Step by Step
(a) Let $F_T$ be the force of tension in the rope. Note that the student's push $F_p$ is opposed by the vertical (in the image) component from both sides of the rope:
$2F_T ~sin(\theta) = F_p$
$F_T = \frac{F_p}{2 ~sin(\theta)} = \frac{300 ~N}{2 ~sin(5^{\circ})} = 1700 ~N$
The rope is pulling on the car with a force of 1700 N.
(b) The mechanical advantage is $\frac{1700 ~N}{300 ~N}$ which is about 5.7.
(c) If it is counterproductive, then the force applied by the student is less than the force of tension pulling on the car. Let's equate the two forces $F_p$ and $F_T$ to find the angle where the effort starts to become counterproductive:
$2F_T ~sin(\theta) = F_p$
$sin(\theta) = \frac{1}{2}$
$\theta = sin^{-1}(\frac{1}{2}) = 30^{\circ}$
When the rope makes an angle of $30^{\circ}$, the effort is counterproductive.
