A 72-kg water skier is being accelerated by a ski boat on a flat (“glassy”) lake. The coefficient of kinetic friction between the skier’s skis and the water surface is ${\mu }_k=0.25$(Fig. 4–74). (a) What is the skier’s acceleration if the rope pulling the skier behind the boat applies a horizontal tension force of magnitude$F_T=240\mathrm{N}$to the skier$\left(\theta =0°\right)$? (b) What is the skier’s horizontal acceleration if the rope pulling the skier exerts a force of$F_T=240\mathrm{N}$on the skier at an upward angle$\theta =12°$? (c) Explain why the skier’s acceleration in part (b) is greater than that in part (a).

Draw the free body diagram for the skier in part (a) at zero upward angle and in part (b) at 12-degree upward angle. In determining the acceleration, use Newton’s second law for horizontal and vertical directions.

Given data:

The mass of the water skier is $m=72\mathrm{kg}$.

The kinetic coefficient of friction between the skier and the water is ${\mu }_k=0.25$.

The magnitude of tension force is $F_T=240\mathrm{N}$.

The upward angle made by the skier is $\theta =12°$.

The free body diagram of the skier is as follows:

The relation to calculate the forces in the vertical direction is given by:

$\begin{array}{c}\Sigma F_y=0\\ N-W=0\\ N=mg\end{array}$

Here, $N$is the normal force, $W$is the weight, and g is the gravitational acceleration.

The relation to calculate the forces in the horizontal direction is given by:

$\begin{array}{c}\Sigma F_x=ma\\ F_T-F_{fr}=ma\\ F_T-{\mu }_{k}N=ma\end{array}$

Here, a is the acceleration of the skier.

On plugging the values in the above relation, you get:

$\begin{array}{c}{F}_T-{\mu }_{k}mg=ma\\ \left(240\mathrm{N}\right)-\left(0.25\times 72\mathrm{kg}\times 9.81\mathrm{m}/{\mathrm{s}}^2\right)=\left(72\mathrm{kg}\right)a\\ a=0.88\mathrm{m}/{\mathrm{s}}^2\end{array}$

Thus, $a=0.88\mathrm{m}/{\mathrm{s}}^2$is the acceleration of the skier.

The free body diagram of the skier is as follows:

The relation to calculate the forces in the vertical direction is given by:

$\begin{array}{c}\Sigma F_y=0\\ N+F_T\sin \theta -W=0\\ N=mg-F_T\sin \theta \end{array}$

The relation to calculate the forces in the x-direction of the block is given by:

$\begin{array}{c}\Sigma F_x=ma\\ F_T\cos \theta -F_{fr}=ma\\ F_T\cos \theta -{\mu }_{k}N=ma\end{array}$

On plugging the values in the above relation, you get:

$\begin{array}{c}{F}_T\cos \theta -{\mu }_k\left(mg-F_T\sin \theta \right)=ma\\ a=\left[\frac{F_T\left(\cos \theta +{\mu }_k\sin \theta \right)-{\mu }_{k}mg}{m}\right]\\ a=\left[\frac{\left(240\mathrm{N}\right)\left(\cos 12°+\left(0.25\right)\sin 12°\right)-\left(0.25\right)\left(72\mathrm{kg}\right)\left(9.81\mathrm{m}/{\mathrm{s}}^2\right)}{\left(72\mathrm{kg}\right)}\right]\\ a=0.9\mathrm{m}/{\mathrm{s}}^2\end{array}$

Thus, $a=0.9\mathrm{m}/{\mathrm{s}}^2$is the acceleration of the system.

In pat (b), the value of the acceleration is higher than part (a) because the upward tilt of the tow rope decreases the amount of frictional force by reducing the normal force.

Thus, the acceleration in part (b) is greater than acceleration in part (a).