Answer
(a) $a = 0.88 ~m/s^2$
(b) $a = 0.98 ~m/s^2$
(c) The smaller force of friction results in a faster rate of acceleration.
Work Step by Step
(a) $ma = \sum F$
$ma = F_T - F_f$
$a = \frac{F_T-F_f}{m}$
$a = \frac{240 ~N - (72 ~kg)(9.80 ~m/s^2)\cdot 0.25}{72 ~kg} = 0.88 ~m/s^2$
(b) We can find the normal force $F_N$ which the water exerts on the skier.
$F_N + F_T ~sin(\theta) = mg$
$F_N = (72 ~kg)(9.80 ~m/s^2) - 240 ~sin(12^{\circ})$
$F_N = 656 ~N$
We can use a force equation to find the acceleration.
$ma = \sum F$
$ma = F_T ~cos(\theta) - F_f$
$a = \frac{F_T ~cos(\theta) - F_N \cdot \mu_k}{m}$
$a = \frac{240 ~N \cdot cos(12^{\circ}) - (656 ~N)\cdot 0.25}{72 ~kg} = 0.98 ~m/s^2$
(c) The vertical component of the tension reduces the normal force between the skier and the water. This results in a smaller force of friction. Although the horizontal component of the tension is also slightly reduced, the smaller friction force results in a faster rate of acceleration.
