Question

Find time taken by the block to reach the bottom.

• A. \(\left[\dfrac{4L}{\cos\theta\,(g\sin\theta - a_0\cos\theta)}\right]^{1/2}\)
• B. \(\left[\dfrac{2L}{\cos\theta\,(g\sin\theta - a_0\cos\theta)}\right]^{1/2}\)
• C. \(\left[\dfrac{8L}{\cos\theta\,(g\sin\theta - a_0\cos\theta)}\right]^{1/2}\)
• D. \(\left[\dfrac{L}{\cos\theta\,(g\sin\theta - a_0\cos\theta)}\right]^{1/2}\)

Hint:

In accelerating frames, always add a {pseudo force} and then resolve all forces {along the direction of motion}.

Answer 1

The Correct Option is B

Concept:
The block slides down an inclined plane inside a frame accelerating horizontally with acceleration \(a_0\).

We analyze motion in the non-inertial frame of the wedge.
A pseudo force \(ma_0\) acts on the block opposite to the direction of acceleration.
Effective acceleration along the incline determines the time of descent.

Step 1: Resolve Forces Along the Incline
Forces acting along the incline:

Component of gravity along plane: \(mg\sin\theta\)
Component of pseudo force along plane: \(ma_0\cos\theta\) (opposing motion)
Hence, net acceleration of block along incline: \[ a = g\sin\theta - a_0\cos\theta \]
Step 2: Determine Distance Along the Incline
From geometry: \[ \text{Length of incline} = \frac{L}{\cos\theta} \]
Step 3: Apply Equation of Motion
Starting from rest: \[ s = \frac{1}{2}at^2 \] Substitute values: \[ \frac{L}{\cos\theta} = \frac{1}{2}(g\sin\theta - a_0\cos\theta)t^2 \]
Step 4: Solve for Time
\[ t^2 = \frac{2L}{\cos\theta\,(g\sin\theta - a_0\cos\theta)} \] \[ \boxed{t = \left[\dfrac{2L}{\cos\theta\,(g\sin\theta - a_0\cos\theta)}\right]^{1/2}} \]