Question

A small block of mass \(m\) slides down from the top of a frictionless inclined surface, while the inclined plane is moving towards left with constant acceleration \(a_0\). The angle between the inclined plane and ground is \(\theta\) and its base length is \(L\). Assuming that initially the small block is at the top of the inclined plane, the time it takes to reach the lowest point of the inclined plane is _______. 

• A. \(\displaystyle \sqrt{\frac{4L}{g\sin 2\theta-a_0(1+\cos 2\theta)}}\)
• B. \(\displaystyle \sqrt{\frac{2L}{g\sin\theta-a_0\cos\theta}}\)
• C. \(\displaystyle \sqrt{\frac{4L}{g\cos^2\theta-a_0\sin\theta\cos\theta}}\)
• D. \(\displaystyle \sqrt{\frac{2L}{g\sin 2\theta-a_0(1+\cos 2\theta)}}\)

Hint:

For problems involving accelerating frames, switch to the non-inertial frame and include the pseudo force opposite to the frame’s acceleration.

Answer 1

The Correct Option is B

To find the time \( t \) it takes for a block to slide down a frictionless inclined plane, while the inclined plane itself is moving with constant acceleration \( a_0 \) to the left, we need to analyze the forces and accelerations involved. Given the angle of inclination \( \theta \) and the base length \( L \), the problem can be solved as follows:

Step 1: Analyze the Forces

• The block experiences gravitational force \( mg \) downwards.
• The component of gravitational force along the incline is \( mg \sin \theta \).
• The normal force \( N \) acts perpendicular to the inclined plane.
• The inclined plane has an acceleration \( a_0 \) to the left.

Step 2: Calculate the Net Acceleration

In the non-inertial frame of the inclined plane, the effective acceleration of the block along the plane is affected by both gravity and the pseudo force due to the inclined plane's acceleration.

• The effective acceleration along the incline is: \(a = g \sin \theta - a_0 \cos \theta\)

Step 3: Apply Kinematic Equation

The kinematic equation for the motion along the incline is given by:

\(s = ut + \frac{1}{2} a t^2\)

Since the block starts from rest \( u = 0 \), the equation simplifies to:

\(L = \frac{1}{2} (g \sin \theta - a_0 \cos \theta) t^2\)

Step 4: Solve for Time \( t \)

Rearranging the above equation to solve for time \( t \), we get:

\(t^2 = \frac{2L}{g \sin \theta - a_0 \cos \theta}\)

Thus, the time taken is:

\(t = \sqrt{\frac{2L}{g \sin \theta - a_0 \cos \theta}}\)

The correct answer is therefore:

\(t = \sqrt{\frac{2L}{g \sin \theta - a_0 \cos \theta}}\)