A block of mass 3kg when released on an inclined plane describes a circle of radius 15m in the vertical plane in reaching the bottom. What is the maximum height of the incline?
Well, that seems like a rather enthusiastic block! I guess it really wanted to show off its circular moves on the inclined plane. But don't worry, I won't let it hog the spotlight. Let's calculate that maximum height for you.
To do that, we need to consider the forces acting on the block. We have the force of gravity pulling it downwards and the normal force acting perpendicular to the inclined plane. The component of gravity that acts along the incline is responsible for the circular motion of the block.
The block needs a net inward force to keep it moving in a circle, and that force is the difference between the component of gravity along the incline and the static friction force. At the maximum height, the static friction is just enough to provide the inward force needed for circular motion, so the block doesn't slip.
Using some trigonometry and a sprinkle of physics equations, we can determine that the maximum height (h) of the incline is given by the equation:
h = r - r cos(theta)
Where r is the radius of the circle (15m) and theta is the angle of the incline with respect to the horizontal.
So, if we know the value of theta, we can plug it into this equation and calculate the maximum height of the incline.