A box slides down a frictionless inclined plane.Starting from rest,the box slides a distance of 0.73 m and reaches a speed of 2.1 m/s.What is the angle of inclination of the plane with respect to the horizontal?
A)3.0 degrees
B)18 degrees
C)20 degrees
D)70 degrees
E)72 degrees

Question

Quizplus · Accepted Answer

The angle of inclination can be found using the conservation of energy principle. The potential energy lost by the box is converted into kinetic energy. The change in potential energy ($\Delta PE$) is $mgh$, and the kinetic energy ($KE$) gained is $\frac{1}{2}mv^2$, where $m$ is the mass of the box, $g$ is the acceleration due to gravity (9.8 m/s$^2$), $h$ is the height the box descends, and $v$ is the final velocity of the box. Since the box starts from rest, its initial kinetic energy is 0.Given that the box slides a distance of 0.73 m down the incline and reaches a speed of 2.1 m/s, we can equate the change in potential energy to the gain in kinetic energy:$$mgh = \frac{1}{2}mv^2$$The mass ($m$) cancels out, and we don't need to know it. We need to express $h$ in terms of the distance slid down the incline ($d$) and the angle of inclination ($\theta$). Since $h = d \sin(\theta)$, we can rewrite the equation as:$$gd\sin(\theta) = \frac{1}{2}v^2$$Solving for $\theta$, we get:$$\sin(\theta) = \frac{v^2}{2gd}$$Plugging in the given values ($v = 2.1$ m/s, $d = 0.73$ m, and $g = 9.8$ m/s$^2$):$$\sin(\theta) = \frac{(2.1)^2}{2 \times 9.8 \times 0.73}$$$$\sin(\theta) \approx 0.309$$Taking the inverse sine (arcsin) of both sides:$$\theta \approx \arcsin(0.309)$$$$\theta \approx 18^\circ$$Therefore, the angle of inclination of the plane with respect to the horizontal is approximately 18 degrees.

A Box Slides Down a Frictionless Inclined Plane