Question

A skateboarder, with an initial speed of$\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{m}\mathbf{/}\mathbf{s}$rolls virtually friction free down a straight incline of length 18 m in 3.3 s. At what angle is the incline oriented above the horizontal?

Short answer

The angle of inclination of the inclined plane is $12.30°$.

Step-by-step solution

Step 1. Understanding the forces acting on the skateboarder

A skateboarder is moving down an inclined plane, which is inclined at some angle. First, draw the free body diagram. The weight of the skateboarder is acting in the downward direction.

Resolve the weight components along with the horizontal and vertical directions.

Use the second equation of motion to determine the acceleration of the skateboarder. Then, with the help of Newton’s second law, determine the angle of inclination of the inclined plane.

Step 2. Identification of given data 

The given data can be listed below as:

• The initial speed of the skateboarder is $v_i=2.0\mathrm{m}/\mathrm{s}$.
• The final speed of the skateboarder is $v_f=0\mathrm{m}/\mathrm{s}$.
• The length of the inclined plane is $L=18\mathrm{m}$.
• The time taken by the skateboarder is $t=3.3\mathrm{s}$.
• The acceleration due to gravity is $g=9.81\mathrm{m}/{\mathrm{s}}^2$.

Step 3. Free body diagram representation 

The free body diagram of the inclined plane can be shown as:

Here, a is the acceleration of the skateboarder, $F_N$is the normal reaction force on both the board and skateboarder g is the acceleration due to gravity, mg is the weight of the skateboarder, and $\theta$is the angle of inclination of the inclined plane.

Step 4. Determination of the acceleration of the skateboarder 

From Newton’s second law, the distance moved by the skateboarder can be expressed as:

$L=v_{i}t+\frac{1}{2}a{t}^2$

Here, L is the distance moved by the skateboarder, which is equal to the length of the inclined plane,t is the time taken by the skateboarder to reach the bottom, and $v_i$is the initial velocity of the skateboarder.

Substitute the values in the above expression.

$\begin{array}{c}18\mathrm{m}=2\mathrm{m}/\mathrm{s}\times 3.3\mathrm{s}+\frac{1}{2}a{\left(3.3\mathrm{s}\right)}^2\\ \frac{1}{2}a{\left(3.3\mathrm{s}\right)}^2=18\mathrm{m}-2\mathrm{m}/\mathrm{s}\times 3.3\mathrm{s}\\ a=\frac{11.4\mathrm{m}\times 2}{{\left(3.3\mathrm{s}\right)}^2}\\ a=2.09\mathrm{m}/{\mathrm{s}}^2\end{array}$

Step 5. Determination of the angle of inclination of the inclined plane

At the equilibrium condition, the forces along the horizontal direction can be expressed as:

$\begin{array}{c}\sum F_x=ma\\ mg\sin \theta =ma\\ \sin \theta =\frac{a}{g}\\ \theta ={\sin }^{-1}\left(\frac{a}{g}\right)\end{array}$

Substitute the values in the above expression.

$\begin{array}{c}\theta ={\sin }^{-1}\left(\frac{2.09\mathrm{m}/{\mathrm{s}}^2}{9.81\mathrm{m}/{\mathrm{s}}^2}\right)\\ =12.30°\end{array}$

Thus, the angle of inclination of the inclined plane is $12.30°$.