A crate of oranges slides down an inclined plane without friction. If it is released from rest and reaches a speed of \(5.832 \mathrm{~m} / \mathrm{s}\) after sliding a distance of \(2.29 \mathrm{~m},\) what is the angle of inclination of the plane with respect to the horizontal?

Short Answer

Expert verified
Answer: The angle of inclination of the plane is approximately \(48.7^{\circ}\).

Step by step solution

01

List the given information

Initial velocity \((v_i) = 0 \mathrm{~m/s}\) Final velocity \((v_f) = 5.832 \mathrm{~m/s}\) Distance \((d) = 2.29 \mathrm{~m}\)
02

Find the acceleration

To find the acceleration, we can use the kinematic equation: $$ v_f^2 = v_i^2 + 2ad $$ Since initial velocity \((v_i)\) is 0, the equation is reduced to: $$ v_f^2 = 2ad $$ Rearrange the equation to solve for acceleration: $$ a = \frac{v_f^2}{2d} $$ Plug in the values: $$ a = \frac{(5.832)^2}{2(2.29)} $$ and calculate the acceleration: $$ a \approx 7.432 \mathrm{~m/s^2} $$
03

Find the angle of inclination

Now, we will use the acceleration value we found in the previous step to find the angle of inclination. We know that on an inclined plane, the component of gravitational acceleration \(g\) along the plane is given by: $$ a = g \sin{\theta} $$ where \(\theta\) is the angle of inclination of the plane. We can rearrange the equation to solve for the angle of inclination: $$ \theta = \sin^{-1}(\frac{a}{g}) $$ Plug in the values: $$ \theta = \sin^{-1}(\frac{7.432}{9.81}) $$ and calculate the angle of inclination: $$ \theta \approx 48.7^{\circ} $$ The angle of inclination of the plane with respect to the horizontal is approximately \(48.7^{\circ}\).

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