Question

An object is held at rest on top of a smooth plane inclined at 30° to the horizontal. when released, it takes 5s for it to slide down the plane. Calculate the (a) distance covered and (b) height of the plane. (g = 10m/s²)​

Answer 1

To find the distance an object slides down a 30° inclined plane in 5 seconds, we use kinematic equations with the acceleration due to gravity. The height of the plane is calculated using the distance found and the sine of the incline's angle.

Step-by-step explanation:

The question involves calculating the distance covered by an object sliding down a smooth inclined plane and the height of that plane. The incline is at a 30° angle, and the acceleration due to gravity (g) is given as 10 m/s². The object takes 5s to slide down the plane.

Distance covered

To find the distance covered, we use the formula for the distance covered by an object under uniform acceleration:

d = ut + ½at²

Since the initial velocity (u) is 0 (because the object starts from rest), and the acceleration (a) is g sin θ, where θ is the angle of the incline, the distance (d) will be:

d = ½g sin(30°)t²

Height of the plane

The height (h) of the plane is related to the distance (d) via the sine of the angle of the incline:

h = d sin(30°)

Answer 2

(a) To calculate the distance covered, we can use the equation:

distance = 1/2 * acceleration * time²

The acceleration of the object down the inclined plane can be found using trigonometry:

acceleration = g * sin(30°) = 5 m/s²

So the distance covered is:

distance = 1/2 * 5 m/s² * (5 s)² = 62.5 m

(b) To calculate the height of the plane, we can use the equation:

height = distance / sin(30°)

Substituting the value of distance we calculated in part (a), we get:

height = 62.5 m / sin(30°) ≈ 125 m

Therefore, the height of the plane is approximately 125 meters.