Question

A 75-kg snowboarder has an initial velocity of at the top of a 28° incline (Fig. 4–75). After sliding down the 110-m-long incline (on which the coefficient of kinetic friction is${\mu }_k=0.18$), the snowboarder has attained a velocity v. The snowboarder then slides along a flat surface (on which${\mu }_k=0.15$) and comes to rest after a distance x. Use Newton’s second law to find the snowboarder’s acceleration while on the incline and while on the flat surface. Then use these accelerations to determine x.

Short answer

The obtained results of acceleration for the inclined and flat surfaces are $3.04\mathrm{m}/{\mathrm{s}}^2$and $-1.47\mathrm{m}/{\mathrm{s}}^2$, respectively. The value of the distance is 235.98 m.

Step-by-step solution

Step 1. Draw the free body diagram of the snowboarder

In this problem, consider the free body diagram of the snowboarder when he is on the incline. Use Newton’s second law for determining the acceleration of the snowboarder.

Given data:

The mass of the snowboarder is $m=75\mathrm{kg}$.

The initial velocity of the snowboarder is $v=5\mathrm{m}/\mathrm{s}$.

The kinetic coefficient of friction after sliding down is ${\mu }_{\mathrm{k}}=0.18$.

The kinetic coefficient of friction on a flat surface is ${\mu }_{\mathrm{k}}=0.15$.

The distance traveled by the snowboarder is $x$.

The given length is $d=110\mathrm{m}$.

The free body diagram of the snowboarder on the inclined is as follows:

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

$\begin{array}{c}\Sigma F_{\mathrm{y}}=0\\ F_{\mathrm{N}}-W\cos \theta =0\\ F_{\mathrm{N}}=mg\cos \theta \end{array}$

Here, $F_{\mathrm{N}}$ is the normal force, W is the weight, and g is the gravitational acceleration.

Step 2. Determine the acceleration in the case of the inclined plane

The relation of force from Newton’s second law along the plane is given by:

$\begin{array}{c}\Sigma F=ma\\ W\sin \theta -F_{\mathrm{f}}=ma\\ mg\sin \theta -{\mu }_{\mathrm{k}}{F}_{\mathrm{N}}=ma\end{array}$

The relation of force from Newton’s second law, perpendicular to the plane, is given by:

$N=mg\cos \theta$

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

$\begin{array}{c}mg\sin \theta -{\mu }_{\mathrm{k}}\left(mg\cos \theta \right)=ma\\ a=\left(\sin \theta -{\mu }_{\mathrm{k}}\cos \theta \right)g\\ a=\left(\sin 28°-\left(0.18\right)\cos 28°\right)\left(9.81\mathrm{m}/{\mathrm{s}}^2\right)\\ a=3.04\mathrm{m}/{\mathrm{s}}^2\end{array}$

Thus, $a=3.04\mathrm{m}/{\mathrm{s}}^2$is the required acceleration.

Step 3. Determine the forces in the vertical and horizontal directions for the flat surface

The free body diagram for the flat surface is as follows:

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

$\begin{array}{c}\Sigma F_{\mathrm{y}}=0\\ F_{\mathrm{N}}-W=0\\ F_{\mathrm{N}}=mg\end{array}$

If the acceleration of the snowboarder on the flat surface is $a\text{'}$, the relation of force from Newton’s second law in the horizontal direction is given by:

$\begin{array}{c}\Sigma F=ma\text{'}\\ -F_{\mathrm{f}}=ma\text{'}\\ -{\mu }_{\mathrm{k}}{F}_{\mathrm{N}}=ma\text{'}\end{array}$

Step 4. Determine the acceleration in the case of the flat surface

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

$\begin{array}{c}-{\mu }_{\mathrm{k}}mg=ma\text{'}\\ -{\mu }_{\mathrm{k}}g=a\text{'}\\ a\text{'}=-\left(9.81\mathrm{m}/{\mathrm{s}}^2\right)\left(0.15\right)\\ a\text{'}=-1.47\mathrm{m}/{\mathrm{s}}^2\end{array}$

Thus, $a\text{'}=-1.47\mathrm{m}/{\mathrm{s}}^2$is the required acceleration.

Step 5. Determine the speed of the snowboarder at the end of the slope

The relation to determine the final speed of the snowboarder at the end of inclined plane ($v\text{'}$) is given by:

$v{\text{'}}^2=v^2+2ad$

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

$\begin{array}{c}{v\text{'}}^2={\left(5\mathrm{m}/\mathrm{s}\right)}^2+2\left(3.04\mathrm{m}/{\mathrm{s}}^2\right)\left(110\mathrm{m}\right)\\ v\text{'}=26.34\mathrm{m}/\mathrm{s}\end{array}$

Step 6. Determine the distance covered by the snowboarder 

The relation to determine the distance of the snowboarder is given by:

${v_{\mathrm{f}}}^2=v{\text{'}}^2+2a\text{'}x$

Here, $v_{\mathrm{f}}$is the final speed, whose value is zero.

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

$\begin{array}{c}{\left(0\right)}^2={\left(26.34\mathrm{m}/\mathrm{s}\right)}^2+2\left(-1.47\mathrm{m}/{\mathrm{s}}^2\right)x\\ x=235.98\mathrm{m}\\ \mathrm{x}\approx 236\mathrm{m}\end{array}$

Thus, 235.98 m is the required distance.