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

To find the time taken by the block to reach the bottom, we analyze the motion of the block on an inclined plane when a horizontal acceleration \(a_0\) is applied. The block moves with a constant acceleration along the incline.

Step 1: Resolve accelerations along the incline

• The component of gravitational acceleration along the incline is \(g\sin\theta\).
• The component of horizontal acceleration along the incline is \(a_0\cos\theta\), acting opposite to motion.
• Hence, the net acceleration of the block down the incline is:

\(a = g\sin\theta - a_0\cos\theta\)

Step 2: Apply the equation of motion

• The block starts from rest and travels a distance \(L\) along the incline.
• Using the equation of motion for constant acceleration:

\(s = \dfrac{1}{2}at^2\)

Substituting \(s = L\) and the value of acceleration:

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

Step 3: Solve for time

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

\(t = \left[\dfrac{2L}{g\sin\theta - a_0\cos\theta}\right]^{1/2}\)

Therefore, the time taken by the block to reach the bottom of the incline is:

\(\boxed{t = \left[\dfrac{2L}{g\sin\theta - a_0\cos\theta}\right]^{1/2}}\)