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

Concept: When a block slides on an accelerating inclined plane, it is convenient to analyze the motion in the non-inertial frame of the inclined plane. In this frame, a pseudo force \(ma_0\) acts on the block opposite to the direction of acceleration of the plane.
Step 1: Forces in the non-inertial frame In the frame of the inclined plane:
Gravitational force \(mg\) acts vertically downward.
Pseudo force \(ma_0\) acts horizontally to the right.
Normal reaction balances perpendicular components.
Resolve forces along the plane. Component of gravity along plane: \[ mg\sin\theta \] Component of pseudo force along plane (opposing downward motion): \[ ma_0\cos\theta \]
Step 2: Effective acceleration of the block Net acceleration of the block along the incline: \[ a_{\text{eff}}=g\sin\theta-a_0\cos\theta \]
Step 3: Distance travelled along the incline The length of the incline is: \[ s=\frac{L}{\cos\theta} \]
Step 4: Use kinematics The block starts from rest, so: \[ s=\frac12 a_{\text{eff}} t^2 \] \[ \frac{L}{\cos\theta}=\frac12\left(g\sin\theta-a_0\cos\theta\right)t^2 \]
Step 5: Solve for time \[ t^2=\frac{2L}{\cos\theta\left(g\sin\theta-a_0\cos\theta\right)} \] \[ t=\sqrt{\frac{2L}{g\sin\theta-a_0\cos\theta}} \] Final Answer: \[ \boxed{t=\sqrt{\frac{2L}{g\sin\theta-a_0\cos\theta}}} \]