Chapter 5.3, Problem 41SSC

To determine

Magnitude and direction of skier’s acceleration.

Answer to Problem 41SSC

  $a=3.2{\text{m/s}}^2$ and the direction is up the incline.

Explanation of Solution

Given:

Mass of Skier, $m=63\text{kg}$

Angle of inclination, $\theta =14°$

Coefficient of kinetic friction, $\mu =0.27$

Tension of rope, $F_T=512\text{N}$

Formula Used:

From Newton’s second law of motion,

  $F=ma$

Where, $F$ is force, $m$ is mass and $a$ is acceleration of object.

Calculation:

The free body diagram of water skier is:

  

As Skier’s motion is upward,therefore, force of kinetic friction $F_f$ will be downward.

Here, $F_N=F_g\cos \theta =mg\cos \theta$ ( $F_N$ is normal force)

Now, net force in the direction of motion is, $F_T-F_{g\parallel }-F_f$

Here, $F_{g\parallel }$ is Skier’s weight parallel to inclination.

So, from Newton’s second law

  $\begin{array}{l}{F}_T-F_{g\parallel }-F_f=ma\\ a=\frac{F_T-F_{g\parallel }-F_f}{m}\\ \left(\because F_{g\parallel }=mg\sin \theta \text{and }{F}_f={\mu }_k{F}_N\right)\\ a=\frac{F_T-mg\sin \theta -{\mu }_k{F}_N}{m}\\ \left(\because F_N=mg\cos \theta \right)\\ a=\frac{F_T-mg\sin \theta -{\mu }_{k}mg\cos \theta }{m}\\ \Rightarrow a=\frac{F_T-mg\sin \theta -{\mu }_{k}mg\cos \theta }{m}\end{array}$

On putting the values,

  $\begin{array}{l}a=\frac{512-\left(63\times 9.8\times \sin 14°\right)-\left(0.27\times 63\times 9.8\times \cos 14°\right)}{63}\\ \Rightarrow a=3.2{\text{m/s}}^2\end{array}$

The direction is up the incline.

Conclusion:

Therefore, the acceleration of Stacie is $a=3.2{\text{m/s}}^2$ and the direction is up the incline.