91影视

In this problem, consider the free body diagram of the snowboarder when he is on the incline. Use Newton鈥檚 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 ${渭}_{\mathrm{k}}=0.18$.

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

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:

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

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

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

$\begin{array}{c}危F=ma\\ W\sin 胃-F_{\mathrm{f}}=ma\\ mg\sin 胃-{渭}_{\mathrm{k}}{F}_{\mathrm{N}}=ma\end{array}$

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

$N=mg\cos 胃$

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

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

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

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

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

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

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

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

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

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

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

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:

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

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:

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

Thus, 235.98 m is the required distance.